Eric The Blue

Socratic method

Just bought a cup of coffee for $2.97. Paid with a five. The young employee entered $10 on the cash register. Realizing her mistake, she summoned a colleague for help. I thought maybe for some inscrutable reason the transaction had to be voided but, no, it was just that she was now confused about how much change I deserved: garbage in, garbage out.

They discussed the problem in my presence. It was fun to mark the Eureka expressions when I suggested they subtract five dollars from whatever the register indicated. Seven-oh-three? Ok, then, two-oh-three, right? (Didn't want them to think I was pulling a fast one.) "You're so smart!" one of them commented. I put a dollar in the tip jar and swaggered off.

You know you're old when you think the young are not going to be able to keep the world going. Somehow, they will. Maybe one of those kids puzzling at the register will be president! The little episode reminds me of the following clip from The Howard Stern Show.

 

 

 

 

 

Math and media

With concern high about the ability of the Omicron variant to dodge existing vaccines, it's too bad the Guardian newspaper had to run an article beneath the headline, "More than half Omicron cases in England are in the double jabbed." The graf in the story containing the details reads:

Twelve of the 22 cases occurred more than 14 days after the individual had received at least two doses of vaccine. Two cases were more than 28 days after a first dose of vaccine. Six were unvaccinated, while two had no available data. 

Let's set aside from consideration the two cases for which vaccination status isn't known. The first and most obvious thing to be said is that a sample size of 20 is so small as to be practically irrelevant. The networks don't call New Hampshire for one candidate or the other after the ballots have been tabulated from that one village that votes at midnight. They report the vote totals but don't pretend they are significant. But the Guardian's headline, the linguistic equivalent of wild hand waving, pretends that the very first returns are significant. 

The second thing to be said relates to the so-called base rate fallacy, which I've written about here and here. If everyone were vaccinated, so that every Omicron case was a breakthrough infection, this would not prove the vaccines were worthless against Omicron, right? What if universal vaccination reduced the total case load by 90 per cent? You have to take account of vaccination rates. When you do that, the scary-sounding results from the absurdly small sample size aren't that scary. Here is how I think that works out:

A chart in the Guardian story indicates that, in England, 69.3 percent of the population is fully vaccinated. But the fine print acknowledges that people under 12 aren't yet eligible. According to this BBC story, 90 percent of the eligible population have received at least one shot, and 80 percent have received two. Let's run some numbers using the 80 percent figure. Out of every thousand people, 800 will have been fully vaccinated, and 200 will be either unvaccinated or partly vaccinated—almost all unvaccinated. Of 20 positive Omicron tests in the population, 12 are for fully vaccinated people, and the other 8 are from the unvaccinated group. For purposes of gauging vaccine efficacy, then, the numbers that matter are 12-out-of-800 among the vaccinated, and 8-out-of-200 among the unvaccinated, or 1.5 percent versus 4 percent. The disease is more than 2.5 times more prevalent in the unvaccinated group. Yet the headline, while technically true, encourages people to conclude that, based on what is known so far, existing vaccines are essentially worthless against Omicron. "More than half the cases are among the fully vaccinated!"

 

Mathing

At supper, I asked the kids if they had any jokes and was rewarded with:

Why are math books depressed?

They have so many problems.

Volunteered by fifth grader. Fortunately, eighth grader later in the evening reported being over challenged by one such problem, which spared me the agony of watching either election returns or a dull World Series game, the outcome of which (I'm talking about the ballgame only) did not interest me in the least. She was supposed to calculate the area of a regular hexagon that's 20 centimeters on a side.

I hemmed and hawed embarrassingly for a couple of minutes. In my defense, she had the advantage of stewing about it for 15 minutes, or more likely 15 seconds, before submitting the problem to dad and expecting instant insight. My first idea was to think of the hexagon as being inscribed within a rectangle, then subtracting the areas of four identical right triangles at the corners of that rectangle. But what is known about the triangles? Only that their hypotenuse is 20 cm and their angles 30º--60º--90º. That's enough, but it's a trigonometry problem, and her class is geometry. I asked whether the phrase "cosine of thirty degrees" meant anything to her and got the gaze I associate with a prizefighter regarding his opponent while the referee gives instructions before the fight. 

I could see from the other problems that the lesson seemed to be "applications of the Pythagorean Theorem," and what ended up working was to carve up the hexagon into triangles. She knew that the interior angles were all 120º—goes up by 180º for every side added, so 540º for a pentagon (108 per angle), 720º for a hexagon (120 per angle), &c. The "spokes" bisect these angles, so the resulting triangles have to be equilateral:

We don't have right triangles yet, but each equilateral triangle can be divided into two identical right triangles that look like this: 

The area of these right triangles, each one-twelfth of the hexagon, is given by the product of half the base and the height. Half the base is 5. She knew how to use the Pythagorean Theorem to find h:

10² + h² = 20²------->h = √300 = 10√3.

So the area of each right triangle is 50√3 and the area of the whole hexagon is (12)(50)(√3), or 600√3 sq cm. She endured me pointing out that the same strategy can be used to derive a general formula for the area of a regular hexagon with the length of a side equal to s, for the triangles then look like:

And, by the Pythagorean Theorem:

(s/2)² + h² = s²------>h² = s² - (s/2)² = s² - s²/4 = 3s²/4----->h = s√3/2.

The area of each right triangle is then half the base (s/4) times the height (s√3/2), which equals s²√3/8. Since the regular hexagon is made of 12 of these right triangles, the general formula for its area is:

12 s² √3/8 = (3/2) (s²) (√3). 

As to the so-called election "results," I now sleep in camo and have loaded my weapons into the minivan. I assume we'll be marching on Richmond but am awaiting specific instructions from Soros and Hillary. 

COVID and the base rate fallacy

I didn't know, when I wrote this back on August 9, that the bit of Covid innumeracy I was trying to describe has a name: the base rate fallacy. A week earlier, on August 1, Damon Jones, using numbers that make the point perhaps more clearly than I did, had illustrated the fallacy in this brief Twitter thread. I'll summarize for those who don't care to follow the link.

Suppose that 9 out of 10 people who haven't been vaccinated end up in the hospital with Covid. Suppose, further, that there is a vaccine, and that only 1 in 10 of the vaccinated get a case of Covid bad enough to require hospitalization. That's a helluva vaccine, right? Your chance of going to hospital drops from 90 percent all the way down to 10 percent. It's such a good vaccine that 90 percent of the population gets vaccinated. Only 10 percent don't, for whatever reason they find persuasive—perhaps because they've heard the vaccinated can get sick, too.

Now let's consider the question of who ends up in the hospital with Covid:

Out of every 100 people, 90 will be vaccinated, and, of those, 10 percent, or 9, will end up in the hospital. Ten people will not be vaccinated, and 90 percent of them, or 9, will end up in the hospital. So 18 people in the hospital, half vaccinated, half unvaccinated. It can be terribly misleading to denigrate the efficacy of vaccines by citing the percentage of people in the hospital who've been vaccinated! Don't fall for it. Get vaccinated.

As Jones notes, it's also useful to consider what happens, in the above scenario, if only half the population is vaccinated. Then out of 50 vaccinated people, 5 end up in hospital, and, of the 50 unvaccinated ones, 45 are hospitalized. So now 90 percent of the hospitalized are unvaccinated. There are also 50 people in the hospital, as opposed to just 18. 

Exponential

My older girl’s math teacher is trying to impress upon her seventh graders the power of exponential growth.

A traveling farm worker offers to work in a farmer’s field every day for a month. He wants to be paid 1 cent for the first day, 2 cents for the second day, 4 cents for the third, and so on—each day, twice as much as the day before. The farmer accepts after calculating that he would not have to pay the man as much as a dollar a day until the start of the second week. Was this wise? How much will the worker earn on day 10? Day 20? About how much, total, for the month?

If you’re familiar with this or similar exercises, you know that the upside for the farmer, assuming he honors the agreement, is that he can probably leave off worrying about the estate tax. If the work day is long, the daily wage attains the current federal minimum at about the month’s midpoint. By day 20, however, the worker gets more than $5000. On the last day of the month (30), he gets more than $5 million. For the month, his pay is about $10.74  million.

I plan to reveal my view of Israel’s treatment of its neighbors on the day this blog logs its 2^20 page view. That should give it more than enough time to settle and annex all the disputed territory, and my anti-Semitism will remain a secret to all reading these words.

Bonus problem

It would be too cold to go anywhere even if there were anywhere to go, the bathrooms are clean, and I'm feeling the force of the maxim concerning how all the ills of the world may be traced to the inability of people to be alone with their thoughts. I'm on a medication, a blood thinner, that is not supposed to be used with alcohol, so probably it's lucky that my 7th-grader has a math problem she can't get:

Oliver, Ros, and Alice want to start a petting zoo. They decide the best animals to have are hippos, monkeys, and zebras. Oliver decides to buy 7 hippos, 10 monkeys, and 3 zebras for a total of $151. Ros decides to buy 5 hippos, 2 monkeys, and 9 zebras for a total of $185. Alice buys 4 hippos, 6 monkeys, and 5 zebras for a total of $137. What is the cost for each animal? Set up and solve this system of equations using any method of your choice. Show your work!

Actually, this was a question on a test, and Lydia showed it to me as an example of how outrageously unreasonable her teacher is. As she explains it, you get graded on a 4-point scale, and for a 4 you have to get the "bonus problem" right. The above was the bonus problem. How can anyone except "Mickey, the class genius," possibly get a 4 when "Ms. Novak's never showed us what to do if there are three variables!" Etc., etc., Ms. Novak is a tyrant, etc. I say maybe Ms. Novak wants to see who can apply what they know to a new kind of problem and who will just throw their hands in the air and moan. Now I'm in the same moral category as Ms. Novak, but hell if I'm going to agree that the teacher's expectations are too high. Put on your thinking cap, Whiney McWhinerson!

She'd actually made a good start—the "setting up the system of equations" part:

7h + 10m + 3z = 151
5h + 2m + 9z = 185
4h + 6m + 5z = 137

It was at this point that she'd thrown her arms in the air, or maybe just hoped for partial credit on the dreaded bonus problem. She can solve the problems with two equations and two variables, and the method is pretty much the same: you "manipulate" the equations to eliminate variables and make apt "substitutions." The permitted "manipulations" are multiplying any of the equations by the same constant, and adding (or subtracting) one equation from another (which amounts to adding the same quantity to both sides of an equation). The more variables, the more "manipulations," but it can be fun if it's cold out and you're a big nerd. 

Like, with these equations, it might be useful to subtract the third one from the second, because then you will get an expression for h that can be substituted into other equations:

5h + 2m + 9z = 185
4h + 6m + 5z = 137
  h – 4m + 4z = 48-------->h = 48 + 4m – 4z

Now, in the first original equation, substitute "new h" for "plain old h":

7(48 + 4m – 4z) + 10m + 3z = 151
336 + 28m – 28z + 10m + 3z = 151
38m – 25z = –185
25z - 38m = 185

We can then substitute "new h" for "plain old h" in, say, the third original equation:

4(48 + 4m - 4z) + 6m + 5z = 137
192 + 16m - 16z + 6m + 5z = 137
22m - 11z = -55
2m - z = -5
z = 2m + 5

That worked out nice!—an expression for z that can be substituted into the equation highlighted above:

25(2m + 5) - 38m = 185
50m + 125 - 38m = 185
12m = 60
m = 5

If m is 5, z is 15 (since z = 2m + 5—we're "back substituting" now). And if m is 5 and z is 15, we can substitute those values into one of the original equations to find that h is 8. To be OCD (obsessive compulsive disorder) about it, substitute all 3 values into the original 3 equations to make sure it checks out, and it does. Hippos are $8, monkeys are $5, and $15 for a zebra.

Only two hours now till the Gopher basketball game, and some of that time will be taken up by making supper, and eating it, and cleaning up afterwards. Pump out those vaccines, man—though I won't need one if we lose to Nebraska. 

Homework

Whew! This morning's effort at "snow" removal was like using a shovel to empty a swimming pool. I feel like my aged cat, who used to entertain himself by jumping from the floor to the top of the refrigerator but who I now lift onto the bathroom vanity so that he can drink from the running faucet while I brush my teeth.

With steady application, however, I can still help my 7th-grader work through to the solution of a vexing math problem. Word problems in particular cause her to throw her arms in the air and yell, with her expressive face, because I would frown if she used her larynx, "WTF!?"

A man receives income from two investments at simple interest rates of 6½% and 8%, respectively. He has twice as much invested at 6½% as at 8%. If his annual interest from the two investments is $698.25, find how much is invested at each rate.

First, since this is the Minneapolis Public Schools, I think the problem should commence, "A person of unspecified gender mired in end-stage capitalism . . ."—but let's accept it on its own unwoke terms. "Calm down," I say to her, since in my experience that works with women of all ages, "and give thanks that it's simple interest. You just need to change the words into equations."

Look of expressive despair converted to blank disdain, but, with coaxing, she writes down, after a couple false starts:

0.065x + 0.08y = 698.25

And what else do you know? "There's twice as much at 6½%,"  she says. A little light in her eyes now. That look must be why some people stay in education despite the low pay. She writes down

x = 2y

and now knows just what to do:

0.065(2y) + 0.08y = 698.25
0.13y + 0.08y = 698.25
0.21y = 698.25
y = 698.25 ÷ 0.21 = 3325

She's really happy when she punches the last button on her calculator and sees the nice round figure, "3325." That's when you know you're right. If it was fractions of a cent, or even of a dollar, back to the drawing board. But this is math class, not real life, and I remember that moment of tension followed immediately by either despair [3325.4148629713] or satisfaction [3325]. Homework help almost makes me feel young again. I remind her that she needs to say how much is invested at each rate.

y  = 3325 and x = 2y so x = 3325 x 2 = 6650⇒the dude had $6650 at 6½% and $3325 at 8%. 

I make a joke about how "the dude" needs a different financial advisor, but she saved the hardest problem for last and is already on her phone, calling her bff, whom she never sees anymore except on Facetime—for the same reason that school is in her bedroom. I had a little fun with the details of the problem but am impressed that school kids in Minneapolis are learning this stuff in 7th grade. I mean, I'd be surprised if the people Hollywood stars and hedge fund managers hire to take college entrance exams for their kids couldn't do problems like this, but presumably they're not  7th graders. At this rate, she's on her own within a year or two. 

Funniest thing I read today: Women on a dating app switch their political filter to "conservative" in order to find men who offer to send them pics of themselves inside the Capitol on January 6. The women then forward the pics along with their correspondent's dating profile to the FBI. I wonder if the FBI is embarrassed not to have thought to do this itself. More women should work in law enforcement. For some reason, they seem to have a better handle on how stupid these losers are apt to be. 

Distance learning in snow

Yesterday, I ran the lawn mower over my yard just to pick up leaves. Today, I shoveled maybe two tons of wet snow off my driveway and walk. To every neighbor who approached for small talk, I'd say, "Apparently raking season is over." They'd laugh, which gave me courage to repeat the line to the next one. A dog walker observed that if you don't like the weather in Minnesota, wait an hour. He was in Garrison Keillor's demographic group, age and race, but it wasn't him. 

Just for shits and giggles, here is a math problem I helped the seventh grader with today. I don't think I had to work on problems like this till I was in at least eighth grade, maybe ninth or even tenth—too long ago to remember for sure. 

Easy mistake: forgetting to change the direction of the inequality when multiplying both sides by a negative number. Gave her the tip of checking her work with an easy point, like (0, 0), to make sure she had identified the correct side of her graph. She said the teacher had already suggested that. I asked why in that case she hadn't done it. She said, "There's twelve more problems!"

Distance learning

First day of school, a hundred percent "distance learning" into the foreseeable future, and I have a pass as the kids are with their alternate custodian.  But I did receive a FaceTime call that displayed the following 4th-grade puzzler:

I'll admit that I dicked around with this for awhile, to no avail, until it occurred to me that since the "average" value of the numbers is 5, and 4 x 5 = 20 > 17, it's probably the case that the boxes at the vertices, which are overused compared to the others, should be filled with low numbers. It then took but a minute to come up with:

After a decent interval, I took a picture of my solution and texted it to said 4th-grader. Here's what that ended up looking like:

At least she acknowledged the difficulty. Maybe someone helped her. Didn't dare ask. 

 

Last 6th-grade math problem of the year

My 6th-grader, Lydia, did not exhibit much interest in the pictured "exceeding standard question"  about the regular octagon. She's cool with "meets requirements," especially when her friends are facetiming her. You can see the blank space for work and the perfunctory line drawn between points H and F, which was supposed to persuade me that she'd made an effort. Here are the steps of the approach I described to her while she glanced back and forth between the paper and her dinging phone.

1. You know the interior angles of a triangle sum to 180 degrees. It goes up by 180 degrees for every side added: for a quadrilateral (such as a rectangle) the interior angles sum to 360 degrees, for a pentagon 540 degrees, etc. A formula is: Sum of Interior Angles = 180(# of sides - 2).

2. So the interior angles of an octagon add up to 180 x 6 = 1080 degrees. Since it's a regular octagon (all sides equal length), the eight angles all have the same measure of 1080/8 = 135 degrees.

3. You're supposed to find the measure of ∠HAF, which can be thought of as the sum of ∠HAG and ∠GAF. ∠HAG is 22.5 degrees (because ∠GHA is 135 degrees and ∠HAG is half the remaining 45 degrees of triangle AHG). For the same reason, ∠BAC is also 22.5 degrees. 

4. Is this enough to persuade you that the 135 degrees of ∠HAB is made of six equal "little angles" all with vertex A? [Bored look.] Note that 22.5 x 6 = 135. [Completely impassive.]

5. If not, the hypothesis can be checked. Since the octagon is "regular," opposite sides are parallel and ∠AFE is 90 degrees. This means that ∠GFA is 45 degrees (since we know that ∠GFE is 135 degrees). Also, since ∠HGF is 135 degrees, and ∠HGA is 22.5 degrees, ∠AGF must be 135 - 22.5 = 112.5 degrees.

6. Therefore ∠GAF = 180 - 45 - 112.5 = 22.5 degrees. And ∠HAF = ∠HAG + ∠GAF = 22.5 + 22.5 = 45 degrees.

Lydia misses baseball, because in 5th grade, when the Twins were playing on TV, it was her homework, not mine.

Flipping coins: an academic paper

I found this academic paper, on a topic at the intersection of philosophy and math, a hundred times more interesting than one expects an "academic paper" to be. If a tossed coin lands heads 92 times in a row, would you be surprised? The author of the paper, Martin Smith, argues that you should not be surprised, and makes a stronger case than I for one would have thought possible. 

The article, very clearly written, is only about 7 pages long, but I understand if you don't want to click and read so will try to give the main points myself in a lot less than 7 pages. We should make a distinction between what you'd predict and what would surprise you. Suppose you determined to flip a coin just five times. Would you predict that it would land heads every time? No, because most likely it wouldn't. To be precise, the chance of this is only 1/2⁵, or 1/32, a titch above 3%. Pretty unlikely. But suppose you then flipped the coin five times and got five heads. Should you be surprised? No, because there had to be some specific result, and HHHHH is just as likely as, say, THHTH. If you're not surprised by THHTH, then you shouldn't be surprised by HHHHH, either. You wouldn't have predicted five heads in a row. Most likely, something else happens. But "something else" encompasses all the other possibilities. If you actually flip the coin five times, there has to be some specific result, and HHHHH is as likely as any other. Five heads in a row doesn't require an explanation.

Neither does 92 heads in a row, for the same reason. That is, the argument doesn't change on account of the number of trials. (Smith chose 92 because, in a Tom Stoppard play, one of the characters flips a coin in a gambling game—heads he loses—and as the curtain rises is trying to make sense out of having lost 92 games in a row.)

My reasoning faculties find the argument to be sound. All the rest of me says to keep looking for the flaw. This isn't an unusual sensation in human affairs and reminds me of a story about the paradox of Zeno, who postulated that you could never traverse the distance between two points, since you'd first have to cover half the distance, and then half the balance, and on and on, with the result that, at the end of time, you'd still be some distance away from the destination. To demonstrate, a teacher had a boy and a girl face each other from opposite ends of a classroom. The boy then advanced toward the girl, the teacher measuring off half the distance each time. After a few advances, the boy was getting pretty close, and grinned at the girl. Trusting Zeno, she said, "You'll never get here." He answered, "Right, but I'll get close enough for all practical purposes." Maybe he was being a jokester, maybe he sensed that there was something unfair about chopping up distance but not time. 

One's strong impression of reality, or whatever you want to call it, might make you doubt Zeno, but it isn't always a good guide to the truth. Apparently it's true that all matter is made of just over a hundred different kinds of atoms—these tiny particles we can't see that jiggle about constantly, because they attract one another until, having been drawn together, they instead repel. I anyway think the atomic hypothesis is true, though it's not on account of my "strong impression of reality" that I've reached this conclusion. Crediting only my own impressions, I'd say there are many hundreds, maybe thousands, of different substances just in my house alone. 

I'm inclined to give my assent to Smith, partly because it seems people's probabilistic impressions are unusually susceptible to error, so that what seems true might very well not be. There is, for example, an entire literature about "the Monty Hall problem"—so called, I think, because it's hard to make people believe that the best strategy in a certain game is indeed the best strategy. If you're of a certain age, like mine, you'll know that there used to be a game show hosted by the TV personality Monty Hall. In the big game at the end of the show, there would be three prizes concealed behind three doors. Behind two of the doors would be something like a live, impassive goat. But behind the third door there'd be something like a sports car. The contestant gets the prize behind the door she chooses. The wrinkle is that, after she's made her choice, Monty reveals the prize behind one of the unselected doors. This is always a goat, so it's evident that Monty knows where the big prize is and wants to heighten the suspense. He then offers the contestant a deal (the name of the show was "Let's Make a Deal"): if he wants, he can take the prize behind the remaining closed door instead of the one behind the door he originally chose. 

Should the contestant switch? Does it make a difference? In experiments, more than 80 percent of people stick with their original choice (according to the Wikipedia article on the Monty Hall Problem). It's not that hard to show, however, that by switching a contestant doubles the chance of winning the valuable prize. After making your original choice you have a one-third chance of winning—that's not controversial. It seems people must reason that the chance they've chosen correctly rises after Monty reveals a goat. This reasoning would be sound if Monty revealed at random what was behind one of the unselected doors. Remember, though, it never happens that he reveals a sports car and the contestant immediately loses. It's always a goat, followed by the offer to switch. The likelihood that the contestant has chosen correctly therefore remains one-third. Monty is in effect divulging that the two-thirds chance the contestant was wrong is not evenly distributed between two unselected doors. Rather, all two-thirds attaches to the unselected door Monty hasn't yet opened. By switching, the chance of winning the car goes from one-third to two-thirds.

I can't think of an experimental test for Smith's argument, but, if you doubt the one I've sketched out above for the Monty Hall Problem, you might accurately simulate the game by taking a couple of deuces (goats) and an ace (sports car) from a deck of cards.  After "shuffling," deal face down one card to yourself and two to someone you are co-quarantining with. Now, to make it more interesting, you each ante a dollar and the person with the ace wins the pot. This is obviously a bad game for you, right? Suppose you make a rule that, before seeing who has the ace, your co-quarantiner has to look at his two cards and turn a deuce face up. It's an illusion to suppose this changes anything, right? It's still a bad game for you. But if there were yet another rule, one requiring you and the co-quarantiner to now swap the remaining downturned cards before looking to see who has the ace, the game would become a money-maker for you. If you're still not persuaded, just play a bunch of games and see who wins how often under what rules. 

What I find intriguing about Smith's argument concerns what, if true, it suggests about our minds—namely, that they seem to be repulsed by randomness, prefer order, and therefore are constantly exercising themselves to discover patterns. It doesn't mean there aren't patterns. It means that detecting a pattern when there is none is a common mistake compared to failing to detect a pattern. I think this would explain a lot, like the hold of elaborate conspiracy theories on the human mind. Oswald was not enough of a marksman to have shot Kennedy! There must have been additional gunmen in Dealey Plaza and a conspiracy involving Lyndon Johnson, J. Edgar Hoover, and the Soviet Union! But what if Oswald just got lucky? (It happens, I once hit two doubles in the same baseball game.) A basketball player misses a foul shot that would have won the game. She choked! But wait, she doesn't make them all when practicing in her driveway. What if the fateful, errant shot was just one of those that she happens to miss? We're told that everything happens for a reason. I'll subscribe, if "random chance that might not have been realized but was" is allowed as a reason.

Day 1 of distance learning

Around a half day into "distance learning," the dropout rate is high—fifty percent, if you must know the exact statistic. In the admittedly unlikely event that you are looking to while away some quarantine time by contemplating a few 6th-grade math problems, the student who has not dropped out is working on the above four. Notice—what I do not necessarily like, but it's how it is—that there's no effort even to pretend that we are not here just coaching kids to perform satisfactorily on a standardized test. The problems aren't dumb problems, but they are exactly the kind that get asked on the standardized test to determine whether a 6th-grader "meets the standard," "exceeds the standard," etc. "You have to know this stuff because our school will look bad if you don't!" If it's math, but not on the "big test" all 6th graders take . . . some other day. It seems there's something to be said for having an agile mind—bringing to bear what you know on a new or unfamiliar kind of problem. But the idea here is to make sure there won't be any unfamiliar problems on Test Day.

Whatevs, I'm a "school nerd," as the little wenches around here keep reminding me.

Actually, the first question, demonstrating mere "emerging understanding," is an example of what I'm talking about. I guess it's considered the easiest and most basic because you are supplied with the formula for the area of a trapezoid. All you have to do is put the right numbers from the figure into the formula and then do the arithmetic correctly. But supposing, I asked my marginally interested daughter, you were not given the formula and had forgotten what it was? No need to panic. Think of the figure as a 10.2 x 12 rectangle with a 5 x 12 x 13 right triangle sitting atop it. So it's (area of rectangle) + (area of right triangle) = (10.2)(12) + (6)(5) = 122.4 + 30 = 152.4 sq cm. I thought it might make an impression on her when you arrive at the same answer my way as she had when plugging numbers into the formula, but it didn't. Whatevs.

Whatevs is the word for day 1 of distance learning.

To me, the question to prove you're "proficient" doesn't seem any harder than the ones for "emerging proficiency" and "partial proficiency." But the last one, to exceed the standard, is definitely the hardest, no? The idea is that a triangular prism has been cut and laid out flat, which makes it arguably easier to work out the surface area: the sum of the areas of three rectangles and two triangles. We're also supposed to figure out the volume, however. My 6th-grader tries to remember all the formulas she thinks she needs to know. Seems better just to remember that, in all these solids, the volume is given by the product of the area of a face and the length of the lines connecting it to it to its "partner" on the other side. So, for example, the volume of a Campbell's soup can (aka a "right circular cylinder") is the product of its height and the area of either circular end. In this case, the identical triangles, with areas of 6 sq cm, would be connected in the reconstituted prism by the 3 cm edges of the center square in the flat net. The volume of the prism must therefore be 18 cubic centimeters? Unless Homeschool Dad cannot himself exceed the 6th-grade standard.

 

 

 

Politics! Math!

I was out and about yesterday and so am not in position to award the gold medal for most absurd defense of President Trump during the full House's impeachment debate, but while driving I did hear on the car radio one guy quote Jesus on the cross, "Forgive them Father, for they know not what they do," before assuring Democrats that, following Christ's example, he was praying for them in just the same way. I thought this was, you know, over the top by an inch or two, and this morning, trying to track down the identity of the biblical scholar, I kept getting referred by Google to some other representative who had compared the Democrats unfavorably to Pontius Pilate. If Minority Leader McCarthy hadn't been working so hard on his own floor speech, it might have occurred to him to import a Gospel choir to sing, during the interval when representatives were voting electronically to impeach Trump, "Where Were You When They Crucified My Lord?"

I was sitting at the dining room table being perplexed by a math test that Lydia, my 6th grader, had brought home from school. The problem she got wrong is pictured above. Let's see, if Jim had $4  Peter would have $7, then Peter gives Jim three-fourteenths of $7, which is a buck fifty, and they both have $5.50, so that must be the answer, except that if Jim had $8 then Peter had $14, and Peter would then give Jim $3, in which case they'd both have $11, so I guess there are at least two "right" answers, and if there are two maybe there are a billion more! 

Poor Lydia. As she pointed out, "I tried," but she's only in 6th grade and could never have figured out that the correct answer is: "My math teacher might be a little weak at math, maybe he should read up on the New Testament and run for Congress, he's more than smart enough and the salary is $174,000 per year."  

UPDATE: A fellow math geek writes to defend Lydia's teacher, on the theory that the above problem is a kind of trick question, and that the right answer is that Jim could have started with any amount of money—which is indeed the right answer to the problem as it was set out on Lydia's test.

I meant to insult Bible-thumping congressmen, not middle school math teachers, who, being overworked and underpaid, might pull the deficient (or trick) problem from some source available to math teachers. If so, and the source provides answers to its suggested problems, I'd be curious about the solution it gives for the one about Jim and Peter. I told my geeky friend that I think the direction "Show all work" indicates the problem is deficient, not a "trick," but I could be wrong. If instead of a test question this problem had been homework, and Lydia had asked for help, we could have gotten to the bottom of everything by turning in a paper that said:

Jim could have started with any amount of money. Let x stand for however much he had to start, then Peter started with 7x/4 dollars. Subtract from Peter's 7x/4 dollars (3/14)(7x/4) and add to Jim's x dollars the same amount. Peter then has 7x/4 — (3/14)(7x/4) = 7x/4 — 21x/56 = 14x/8 — 3x/8 = 11x/8. And Jim has the same amount: x + (3/14)(7x/4) = x + 21x/56 = x + 3x/8 = 11x/8. 

I think that's a good answer. It's "correct" if it's a "trick" question and, if it's not meant as a trick question, it shows that something's gone wrong. That the fractions are a little unwieldy helps to disguise what I take to be an error. Suppose the problem had been:

Peter has twice as much money as Jim. After he gives a quarter of his money to Jim, they have the same amount. How much did Jim have to start?

Same principle, but it's easier to see that Jim could have started with any amount.

Another "family math" problem

"Family math": that's what my 6th-grader's math teacher calls these somewhat elaborate word problems that she hands out once per month. At first, I was alarmed, because: is my 6th-grader really supposed to be able to figure this out? But now that I know it's called "family math," I've formed the theory that the teacher knows these problems are over challenging for kids this age, and the idea, perhaps, is just to get them talking with a parent about a math problem. If so, here is the way it works at this house: "Dad, the family math problem is due Friday. Have you figured it out yet?" 

Here's the problem due on Friday:

You have five bales of hay. For some reason, instead of being weighed individually, they were weighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1 and 5, bales 2 and 3, bales 2 and 4, and so on. The weights of each of these combinations were written down and arranged in numerical order, without keeping track of which weight matched which pair of bales. The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.

Your initial task is to find out how much each bale weighs. In particular, you should determine if there is more than one possible set of weights, and explain how you know. Once you are done looking for solutions, look back over the problem to see if you can find some easier or more efficient way to find the weights.

First observation would be that it's good there are ten data points, because there are indeed ten ways of "choosing two from five." If we complete the job of setting them out, using letters instead of numbers (so that the bale identifiers don't get mixed up with their weights), they are a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, and d+e. 

Second observation is that the heaviest bale has to weigh at least 46 kilograms (else there is no way for the heaviest result of 91 kilograms) and the lightest bale can be no more than 40 kilograms (else there is no way for the lightest result of 80 kilograms). 

Third observation is that the ten data points range from a low of 80 to a high of 91—not much variation, indicating that there isn't very much variation in the weights of the individual bales, either. 

Fourth observation is that no result occurs twice but two possible results within the range do not occur at all—namely, 81 and 89. I think this must mean that the weights of the individual bales, though tightly bunched, are nonconsecutive. (Notice, for example, that if you are making all possible simple sums of 1, 2, 3, and 4, the results are 3, 4, 5, 5, 6, and 7—two 5s and no "gaps." We have gaps but no double results.) Also, if two individual bales had the same weight, then there would have to be a "double result"—for example, if a and b are both 40, then a+c=b+c, and you'd have the same result twice.

Last observation is that one of our gaps (81) occurs near the low end of the range of data points and the other (89) occurs toward the high end of the range. Since the gaps in the outputs occur near the low end and near the high end of the range, the gaps in the inputs are likely also toward the bookends.

If we put this all together, it seems a reasonable guess might be that the two lightest bales weigh 39 and 41 kilograms (thus yielding the 80 result) and that the two heaviest weigh 44 and 47 kilograms (thus yielding the 91 result). Let's see if we can make that work.

If our individual bales a through e are in ascending order of weight, then we are guessing that a is 39, b is 41, d is 44, and e is 47. So,

a+b=39+41=80

a+c=39+c

a+d=39+44=83

a+e=39+47=86

b+c=41+c

b+d=41+44=85

b+e=41+47=88

c+d=c+44

c+e=c+47

d+e=91 

The six completed sums—80, 83, 85, 86, 88, and 91—match up with six of the results that we're given. The four incomplete sums are 39+c, 41+c, 44+c, and 47+c. We're hoping to match them to the four results that haven't been accounted for, which are 82, 84, 87, and 90. And they can be matched if c is 43. One solution, therefore, is that the five bales weigh 39, 41, 43, 44, and 47 kilograms.

When looking for more solutions, it seems there's a pretty small number of possibilities that don't contradict any of the above observations. For example, if a is 36 and b is 44 (thus yielding the lightest result of 80), the range becomes too wide and there is no way to hold the heaviest result to just 91. If a is 37 and b is 43, the others could be 44, 45, and 46—but you can't have both 44 and 45, since 89 is not among the results. You have to run through the possibilities related to a being 38 and b being 42, but none of them can be made to work. I think the solution we've found must be the only one.

Not going to be easy making Lydia take an interest in this.

UPDATE: She was less resistant than I guessed and, in talking through it with her, I thought of a more "elegant" method. It seems almost trivial to note that a+b has to be 80, and d+e has to be 91. An "observation" that I should have made, as it's at least as helpful as any I mentioned, is that a+c has to be the second lightest result (82) and c+e has to be the second heaviest result (90). The possible weights for bales (a, b), then, are (40, 40), (39, 41), (38, 42), (37, 43), etc. But (40, 40) doesn't work, since if a=b then, for example, a+c=b+c, and it doesn't, since there are no "double" results—each result is unique. If a is 39, then b is 41, and c is 43, so that a+c=82. Further, if c is 43, then e is 47, so that c+e=90; and in that case, d has to be 44 in order to make d+e=91. That's the solution we arrived at above.

Next check (38, 42) for (a, b). Then c has to be 44, d has to 45, and e has to be 46. But in that case, c+d=89, and 89 is not one of the results.

Next check (37, 43) for (a, b). Then, for a+c to be 82, c has to be 45. But now there is no way to hold the heaviest result to only 91. The same problem arises for (36, 44), (35, 45), (34, 46), etc. So the only solution is a=39, b=41, c=43, d=44, and e=47.

Football! Math! Football math!

An upside of the Gophers' disappointing loss in Iowa City yesterday is that, assuming Wisconsin wins at home next weekend against Purdue, the regular season will end with the Badgers playing here in Minneapolis on Saturday, November 30, not only for Paul Bunyan's axe but also for the championship of the Big Ten West. That'll be the second consecutive Gopher home football game with more championship significance than any in program history since the 1960s.

Of the various "what ifs" that people love to discuss after a close loss, the one that sticks in my mind relates to the plays at the end of the first half. The Gophers, trailing 20-3, had the ball inside Iowa's 10-yard line with 10 seconds left and no time outs. We threw into the end zone, where Iowa's d-back interfered with our guy—actually, he interfered with him for the entire pass route and kept it up while the ball was in the air. So the referee called the obvious interference and, by rule, we get the ball, first down, on the 2-yard line. There were now 4 seconds left in the half. We decided to kick a field goal, which our kicker bounced off the inside of one of the uprights, ricocheting it through for 3 points: 20-6 at the half. We ended up losing 23-19.

First thing this makes me think of is that the penalty for defensive pass interference in the end zone at the end of one of the halves seems insufficient. That the offense gets a first down is completely irrelevant if there is only time left for one play anyway. So the defense, to prevent a touchdown, might as well interfere, hold, rough, grab, spit, trip, whatever. If it's not called, great; and, if it is called, you're a few more seconds toward zeros on the clock without a touchdown having been scored. In yesterday's game, the Gophers decided to kick a field goal with the ball on the 2 and 4 seconds left. We were already inside the 10, so holding us to a field goal attempt was the best Iowa could hope for. There was no penalty at all for the touchdown-saving interference.

Second thing I stew about is the decision to kick. It's a hard choice and I'm not saying Fleck was necessarily wrong. But, had we gone for it and scored, then kicked the point, we would have had 4 more points, which in the end is what we lost by—4 points. I tend to think of these hard choices in a mathematical way. A touchdown is twice as many points as a field goal, plus you get the try for a point after. This means that the number of expected points scored is higher if you try for a touchdown, so long as the probability of scoring a touchdown is close to half what the chance of making a field goal is. Just to demonstrate, let's say that the chance of scoring a touchdown on one play from the 2-yard line is 48%, and the chance of kicking for the extra point, or making a field goal of about the same length, is 95%. Then the expected points scored from trying for a touchdown is 

(.48)(6) + (.48)(.95) = 2.88 + .456 = 3.336

whereas the expected points scored from trying a field goal is

(.95)(3) = 2.85 

and 3.336 > 2.85. It's a significant difference. The chance of scoring a touchdown on one play from the 2 would have to be somewhat less than 48% for the expected points scored from a field goal attempt to be greater. (For what it's worth, the success rate for a 2-point conversion after touchdown in the NFL is 47.9%; the try comes from the 2-yard line, just where the Gophers were after the interference penalty.)

In the event, we almost missed the field goal, which brings up a further little bit of mathematical football that I've sometimes wondered about. In professional football, the hash marks on the field line up with the uprights on the goal posts, which are 18.5 feet apart. Width-wise, then, the ball is always pretty close to the middle of the field. In college football, however, the hash marks are 40 feet apart. This means that if the ball is on the left or right hash mark and near the goal line, the angle to the goal posts is rather severe, and I've wondered whether, mathematically, that makes a field goal try more difficult. It happened yesterday that the Gophers had to kick that short field goal before halftime from the right hash mark, and, from the shot behind the goal posts you usually get on TV, it looks like it might have an effect. It's not a crazy thought, right? If you were kicking a "0 yard field goal" from the right hash mark, there'd be no opening to kick the ball through. If you were kicking a "1 yard field goal" from the right hash mark, almost your only chance would be to kick the ball just in front of the right upright and then bounce it off the inside of the left upright and through. How far back must you go before the effect becomes negligible? I was curious enough to start dusting off my high school trigonometry before remembering it's 2019, so instead I googled "trigonometry of field goal kicking," which led me to this article. My takeaways are:

(1) I was wise not to try and figure it out myself; and  

(2) The situation gets better so fast that, considering that the goal posts are at the back of the end zone, and that the try always comes from about 7 yards behind the line of scrimmage, there are no field goal attempts that are short enough so that being on the hash mark means there is less margin for error on the kick. If, for example, you are near the goal line on the hash mark, it would never be wise, from the standpoint of the mathematics, to take a penalty in order to widen the angle: being closer with a slightly smaller angle allows for more error in direction than being a little farther with a wider angle.

 

 

Gophers! More 6th-grade math!

Sunday of what is now in the modern era the 5-day MEA weekend. Hell, why not just give them the whole week off? I am a half inch away from being that parent who regales the kids with tales of having walked miles to school in winter weather unmitigated by global warming—and only two days off for MEA!

Since the end of the long weekend was coming into view, Lydia finally glanced at the math problem she'd been assigned over the break. Take a guess who was looking at it next. Right when Illinois was driving for the winning field goal against the Badgers, too. If next Saturday the Gophers were to beat Maryland, while Ohio State wins at home against Wisconsin—by no means certainties but both events more likely than not to occur—then the Gophers would be in first place of the Big Ten West by two games with four to play. Our schedule gets tough after that, I know. But on the last Saturday of the regular season, we have a home game against Wisconsin, and it's beginning to look like it's not a pipe dream to think that the championship of the Big Ten West could be on the line. How much fun would that be? When I was 9, the Gophers beat Wisconsin on the last weekend of the season to claim a share of the Big Ten title. That was the last time. Fifty-two years.

I am clearly setting myself up to be crushed. Here is Lydia's math problem:

King Arthur was the ruler in Camelot who had all those knights and a round table. He loved inviting the knights over for parties around his round table. If there was something pleasant that the king could give to only one knight—an extra dessert, a dragon to chase, etc.—he had them play a game to determine who would get it.

The game went like this. First, King Arthur put numbers on the chairs beginning with 1 and continuing around the table, one chair for each knight. He had the knights sit down so that every chair was occupied. Then he stood behind the knight in chair 1 and said, "You're in." Next, he moved to the knight in chair 2 and said, "You're out," and the knight left his seat and went off to stand at the side of the room to watch the rest of the game. Next, he moved to the knight in chair 3 and said, "You're in." Then he said "You're out" to the knight in chair 4, and that knight left his seat to stand at the side of the room.

The king continued around the table in this manner. When he came back around to the knight in chair 1, he said either "You're in" or "You're out," depending on what he had said to the previous knight. (If the previous knight was "in," then the knight in chair 1 was now "out," and vice versa.)

The king kept moving around the table, alternately saying, "You're in" or "You're out" to the knights who remained at the table. If a chair was now empty, he simply skipped it. He continued until only one knight was left sitting at the table. That knight was the winner. 

YOUR TASK: Of course, the number of knights varied from day to day, depending on who was sick, who was away chasing dragons, and so on. Sometimes there were only a handful of knights present, but other times there could be a hundred or more. The question for you is: If you knew how many knights were going to be at the table, how could you quickly determine which chair to sit in to win the game?

Develop a general rule, formula, or procedure that will predict the winning seat in terms of the number of knights present. Explain why your rule works.

By the time I'd read all that, the Badgers had lost and the Gopher game was on. It's not like I had an immediate idea about what the answer was, so to put Lydia off—it's her homework, dammit, not mine—I told her to figure out who wins if there's 2 knights, 3, 4, 5, 6 and up to 10. That should keep her busy, I thought, and maybe there will be a pattern. She was gone about a half hour before returning with what I guess she's been taught to call a "t-chart." It looked sort of like this, but with a lot of erasures:

# knights	winning chair
2	1
3	3
4	1
5	3
6	5
7	7
8	1
9	3
10	5

While she was working up her t-chart, I stewed over the problem enough to conclude:

(1) A knight sitting in an even-numbered chair at the start of the game will always be eliminated on the first trip around the table; 

(2) If there are an even number of knights at table, then an even number will be eliminated, and the next round will commence with the knight in chair 1 being "in"; and

(3) Consequently, if the number of knights at table is given by 2ⁿ where n is a positive integer (so the number of knights at the start of the game is 2, 4, 8, 16, etc.), then each successive round always has an even number of knights remaining, and the knight in chair 1 wins.

So that covers a few of the possibilities. What about all the others? Lydia's table suggests that, after the regular "recursions" to chair 1 being the winner, each additional knight at the start of the game just moves the eventual winning chair up to the next odd number. If there are 16 knights to start, sit in chair 1. If there are 17, sit in chair 3. If there are 18, sit in chair 5. Etc.

But that doesn't really satisfy the requirement in the problem concerning a "general rule, formula, or procedure" that allows you to predict the winning chair quickly. Suppose you are told that the game begins with 107 knights sitting around the table. Which chair wins? Quick!

For that, another "t-chart":

Distance from last 2n	Winning chair
0	1
1	3
2	5
3	7
4	9
x	2x + 1

So if there are 107 knights at the start of the game, the winning chair is #87, because 107 is 43 more than 2⁶ = 64, and 1 more than twice 43 is 87.  Lydia will be writing this up at around 9:45 tonight, unless it's at 8:45 tomorrow morning. She leaves for school at 9:10.

More homework

I know it's counterintuitive, but millions of devoted readers have expressed interest in my 6th-grader's math homework, and comparatively few in, say, Israeli electoral politics. Whoever pitched "Are You Smarter than a Fifth Grader" as a game show idea may have been onto something. As it happens, I've been consulted again regarding (b) in the picture. The context—"learning objective," I think they say in ed-speak—concerns the geometry of circles. This mysterious number pi (π) is the ratio of the circle's circumference to its diameter. Half the diameter is the radius. The area of the circle is the product of π and the square of the radius. From the definition of π it follows that the circumference is equal to π times the diameter. I don't think you can see in the picture that they are given the information that the squares are 1 unit long on a side. The problem is to find the area and perimeter of the figures.

Lydia could do (a) and (c). Maybe you can see enough of them to agree, with me, that (b) is indeed a little harder. To find the area, you have to satisfy yourself that there are two partial circles of the same size, plus the square, and no overlap between the three—a little tricky, perhaps, since the circles, if whole, would overlap. The radii of the circles are the same as the side of the square, length 1 unit, and, since it's a square, the missing portion from each circle is a 90-degree sector. Then you just have to put all that together. The individual circles, if whole, would have an area of π square units. Since 90 of the 360 degrees (one-fourth) are missing, they both have an area of 0.75 π square units, and together, an area of 1.5 π square units. Plus 1 square unit for the square containing the two missing sectors: 1 + 1.5 π, or about 5.71 square units. 

I beg Lydia to consider, once she has an answer, whether it makes any sense, a habit that tends to guard against errors in number grinding or keying on a calculator. I might as well beg Kyle Gibson, through the tv screen, to "work quickly and throw strikes." In this case though, it seems that a useful observation might be that, since the circles if completed would slightly overlap, there is a portion of the square within both circles. Therefore the final answer should be a little less than the sum of the areas of two full circles, which would be 2π, or about 6.28, and 5.71 satisfies that check.

Regarding the perimeter: it's three-fourths of the way around a circle with circumference of 2π, and then the same trip again, so (1.5)(π)(2), or 3π, about 9.42.

Homework

In case you're wondering what kinds of math problems sixth graders in the Minneapolis public schools are working on these days, I recently heard Lydia FaceTiming a friend about what turned out to be:

A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocked over all the containers of eggs. Though she is unhurt, every egg is broken. So she goes to her insurance agent, who asks her how many eggs she had. She says she doesn't know, but she remembers some things from various ways she tried packing the eggs.  

When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over.

How many eggs were in her cart?

First, let me put my Old Fogeyism on hold and admit I think it's cute when kids use their smart phones to put their heads together about math homework. It's when they spend hours playing Roblox that I get pissed off, except for when it allows me to watch Twins games on tv without interruption. Lydia and her friend didn't have a better idea than to "horse it": that is, work through all the multiples of 7 until they came to one that satisfied all the other described properties. This was proving tedious, and leading to "Eurekas" that were then discovered to be mirages, and ultimately there was frustration and despair. Their worksheet more or less predicted this process, for another instruction stated: "Describe what you did in attempting to solve the problem. . . .  Include things that didn't work out or gave you unexpected results."

The "trick" is to use the supplied information to proceed with efficiency, instead of "horsing it" and just hoping you will soon hit on the answer. If there's one egg left when they're placed in groups of five, then the answer must end with 1 or 6 (since multiples of 5 are 5, 10, 15, 20, &c). But it can't end in 6, because when placed in groups of two there is also one left over. So the answer ends with 1 and is divisible by 7. The first possibility is 21, which doesn't work. The next possibility is 21 + x, where x is a multiple of 7 (so that the number is still divisible by 7) and a multiple of 10 (so that the number still ends with 1)—in other words, x is 70 and the next possibility is 91, which fails the test since 91/4 yields a remainder of 3. What about 161? No, divide it by 3 and you have 2 left over, not 1. We're still "horsing it," sort of, but considering only one out of every ten multiples of 7 is way more efficient. The next possibility, 231, doesn't work on account of being divisible by 3—no remainder at all. But 301: ding! ding! ding! ding! Sixth grade gonna be fun for dad!

I think Lydia is okay with the help and simultaneously a little exasperated about   the old man being such a dweeb. I tried to explain that I come by it honestly. When as a kid I asked her grandpa for math help, he would peruse my book, to get the lay of the land, and then begin grousing about how he had half a mind to call school since it was obvious the author of the textbook "knows nothing of mathematics." Good grief, I just want help with this one problem, don't make it an issue in the next school board election, for God's sake. He would regale my sister and me at supper with math anecdotes, like the one about the child prodigy who, during an elementary school arithmetic lesson, was disrupting the class by tapping his pencil loudly on his desk. The teacher told him to stop several times before finally admonishing him, "I don't want to hear another peep from you until you've summed the first hundred integers." He stared at the teacher defiantly, still tapping his pencil. "I mean it!" the teacher said. "Five thousand fifty!" he said.

And my dad explained that rather than thinking "one plus two is three and three is six and four is ten"—"horsing  it," and soon having to sharpen your pencil while risking confusion and error—the "prodigy" thought of it as:

(1+100) + (2+99) + (3+98) + … + (50+51). 

That is, 50 paired sums of 101, so that the solution is 50 x 101 = 5050. And my sister and I, forebears of Lydia: "Great, what's for dessert?"

James Holzhauer, Jeopardy prodigy

Like retired dweebs everywhere I watch Jeopardy on TV and am enjoying the run of the current champion, James Holzhauer, a professional sports bettor from Las Vegas who has won a total of $1,225,987 in his sixteeen consecutive victories.  James's dominance makes me wonder whether the bell-curve distribution applies to Jeopardy contestants.  Before him, the record take from a single game was $77,000.  James has gone over $100,000 several times now.  His average amount won per game is within a few hundred dollars of the previous single-game record.  

There are it seems to me only a few elements to Jeopardy stardom.  The most obvious is having a head full of knowledge on diverse topics.  Another is physical reflexes.  From what I understand, after the question is asked there is a short period of time in which you cannot "ring in."  Moreover, if you do ring in early, you are locked out for another flash of time--long enough so that one of your opponents will likely beat you to the draw.  Of course, if you're too slow you'll lose out, too.  So there is an art and rhythm to working the signaling device, and if you're not good at that you'll never win: the other contestants have heads full of knowledge, too, else they wouldn't be on Jeopardy.  

But these elements are on the bell-curve continuum.  James is way out to the right on both, and this can plausibly account for 16 wins in a row, but not for the way in which he laps the field in race after race.  The credit for that, I think, must go to superior strategy.  James's Jeopardy run has elicited a lot of commentary, much of it relating to his aggressive betting, which is the hallmark of his in-game strategy, but I think there is a psychological aspect to it that I haven't seen anyone discuss.  Since I don't want to get into the details of Jeopardy rules, how you win and all that, consider the following simpler game that mimics a common Jeopardy dilemma--it's no doubt the source for the name, Jeopardy:

Suppose you're given a chance to win money by answering a lot of questions.  If you choose to answer, and you're correct, you win $500, but you have to pay $500 if you're wrong.  So you answer the question if you can, and pass if you don't know.  There are a lot of questions.  You're doing really well and have given enough correct answers to earn $20,000.  Now suddenly for one question the game changes. Before hearing the question, you have to bet any amount, up to the whole $20,000 you've won, that you can answer it correctly.  After you make your wager, the question is read, and if you answer correctly you add whatever you bet to your 20k.  Naturally you lose the same amount if you answer incorrectly.  No passing allowed on this question.  If you don't know, and you bet a lot, you're going to lose a lot.  Of course there is a high potential upside, too.  How much should you bet?

It obviously depends on how likely you are to answer correctly.  But let's say that so far you've known about two-thirds of the answers, and you trust that the game is honest--the question will be about as hard as all the others that you could pass on if you wanted.  In that case, shouldn't you bet the whole $20,000, assuming your goal is to win as much money as possible?  It's a matter of "expected return."  If you have a two-thirds chance of answering correctly, and you play three times, betting everything each time, you'd win $40,000 twice and nothing once for a total of $80,000.  Divide by three games to get an expected return per game of $26,667.  If you bet anything less than the whole $20,000, your expected return is less.  For example, if you are really conservative, and bet nothing, your expected (and actual) return is just $20,000.  So long as you have a better than even chance of answering correctly, bet everything to maximize your expected return.

If you were playing this game in "real life," personal considerations might require a more conservative approach.  If, for example, you start the game with only the shirt on your back and an $18,000 debt to a loan shark, due and payable in a week, you should probably bet $2,000 at the most, even if you're pretty sure you'll know the answer.  But the details of how the Jeopardy game work cut in another direction.  It's more like three people are playing the above game simultaneously, and only the one who accrues the most money gets to keep the money--the other two get comparatively worthless consolation prizes.  In such a game, betting very aggressively is the best strategy, assuming you know most of the answers--and I assume the vetting process for potential contestants guarantees that they all will know most of the answers.

And here's the thing:  James is a very aggressive bettor, but a lot of contestants aren't.  I think that for most of us regulars there is something psychologically or emotionally repellent about the prospect of losing in one trial everything we'd accumulated over time by getting a lot of things right.  What a blow that would be!  We just can't bring ourselves to risk it, even if, in our head, we know it's the right thing to do.  It's possible to dive deeper into the psychology of Jeopardy.  Speaking for myself, I am the kind of person who'd really hate to lose everything I'd painstakingly acquired.  I also am the kind of person who'd figure out that, to win, I really should take big risks.  So I would force myself to do what I am uncomfortable with doing, and the result of that is I'd probably be less apt to answer the question correctly.  My brain might freeze on a question about the Minnesota Twins!  I think it's interesting that James's day job is  Las Vegas sports bettor, a line of work that I assume involves pretty routinely a mix of big wins and big losses.  Pursuing the optimal strategy doesn't phase him in the least.  

Trumps do mental math

 

Yesterday, I came upon this video of Trump & Son, on the air with Howard Stern, trying to do 17 x 6 in their heads.  Junior first guesses 96, which is close, sort of, considering that 16 x 6 = 96.  His next guess is 94.  Heading the wrong direction, son!  Ivanka, who according to her dad is "good with numbers," good enough to be a candidate for the top job at the World Bank, then butts in to protect her brother, offering the defense that the problem is not "practical."  The current president weighs in, announcing authoritatively that the answer is 1,112.  He pronounces it "eleven-twelve." 

Though it seems impossible, things go downhill from there.

Yes, yes, when young I made jokes about the kids who took the short bus, but I now realize that was wrong, and I wouldn't mention the troubles of the Trumps if it were not for the fact that, on the same day I saw this video, Sarah Huckabee Sanders, chatting on Fox with Chris Wallace about congressional efforts to obtain Trump's tax returns, said:

This is a dangerous, dangerous road and, frankly Chris, I don't think Congress, particularly not this group of Congressmen and women, are smart enough to look through the thousands of pages that I would assume that President Trump's taxes would be.

Having been a Sarah Huckabee Sanders watcher for a couple years now, I know to perk up my ears when she indicates her intention to be "frank," because it's how she signals she's about to add something impolitic to her normal blend of word salad falsehoods.  Wallace should have asked if Trump's returns are also too tricky for the top dogs at the IRS, since they've been "under audit" now since before the president earned his negative-three-million mandate in November, 2016.  

How do you do 17 x 6 in your head?  As the nosy, dweeby parent of some inner city public school students, I know that, in the elementary years, they have units called "mental math" wherein they receive instruction in diverse strategies.  For example, can you count by 17s?--17, 34, 51, that's three of them, so six would be twice 51, or 102.  Ding! Ding! Ding! Ding!  Howard Stern would have been impressed! There is also something called the Distributive Property, under the terms of which you are permitted to conceive of 17 x 6 as

(10 x 6) + (7 x 6),

which is clearly 60 + 42, which is clearly 102.  So all signs point toward 102!  In addition to learning techniques of "mental math," Minneapolis public school students are encouraged to develop what their workbooks call "number sense."  For example, in the fourth grade, while learning to work problems like 993 x 55, students are instructed to take note of the fact that the precise answer they derive from exercising the normal method had better be a little less than 55,000 (since 1000 x 55 = 55,000 and 993 < 1000).  This guards against common mistakes, such as when normal methods tempt you to claim that 17 x 6 = "eleven-twelve."

Riddling with FiveThirtyEight

A cool drizzly October day, the kids in school, Brett Kavanaugh ensconced on the Supreme Court: what is a retired guy to do?  Those unwilling to let go of a possibly bad habit can read up on Kavanaugh biographical tidbits ancillary to his teen-aged acts of worship at the altar of the American masculinity cult--the mystery, for example, of the source of the dough for his distinctly upscale lifestyle, the seven-figure home in Chevy Chase and private school for the kids and the country club membership and season tickets for the Washington Nationals.  His financial disclosure forms do not account for assets or income that might permit all the spending. I, however, am off the case, so that instead I can indulge, with an assist from the FiveThirtyEight Blog, another habit, older and manly, though in a nerdy sort of way.  For in addition to the political and athletic number grinding that the site is famous for, there is The Riddler, a series of mathematical puzzles they lob into cyberspace for the entertainment of their dweeby followers, like me.  Here's one I worked on this morning between trips to the laundry room:

You go to buy a specific car, whose fair price we'll call N.  You have absolutely no idea what N is and the dealer, sadistic capitalist that he is, won't tell you.  The dealer enjoys a good chase, though, so without directly revealing the value of N, he takes five index cards and writes down a number on each of them: N, N + 100, N + 200, N + 300 and N + 400.  Important: the guy's sadistic but not a math major.  The numbers on the cards are numbers, not algebra equations.

He presents the cards to you, one at a time, in random order.  (For example, if the price on the second card is $100 more than the first, you can't be sure if those are the two smallest prices, the two largest, or somewhere in between.)  Each time he shows you a card, you must either pay the price on that card, or ask to see the next card.  You cannot go back to previous cards.  If you make it to the fifth and final card, then that is what you must pay.  If you play the dealer's game optimally, how much should you expect to pay on average above the fair price N?

The real problem is to discover the optimal strategy, right?  If you can figure out what that is, the amount of money above N you'd expect to pay is just arithmetic.  I think you can begin to get a handle on the problem by considering a simpler version: suppose the "game" is the same, but there are only three possible prices, $1, $2, and $3.  If you played randomly, employing no strategy at all, you'd expect to pay on average $2, since you'd pay each of the three possible prices the same number of times.  How to improve on that?  Well, if you noodle around just a little bit, it's not that hard to see that you can do better by never paying the price indicated on the first index card.  There are only six different orders for the three cards:

$1, $2, $3
$1, $3, $2
$2, $1, $3
$2, $3, $1
$3, $1, $2
$3, $2, $1

I've bolded the price you'd pay if you played according to the following (optimal) strategy: (a) never pay the price indicated on the first drawn card; (b) if the price shown on the second card is lower, pay it; and (3) if the price shown on the second card is higher, pay the price shown on the third card.  You can do the arithmetic and satisfy yourself that by this strategy you will pay on average $1.67 instead of $2.00.

Now, bumping up to five possible prices makes the game a lot more complicated.  For one thing, while there are only six ways to order three items, there are 120 ways to order five items.  It's too cumbersome to list all the possible arrangements of five prices.  But, to improve on a random strategy, it's necessary to try to learn something as the game progresses, which means: no matter the number of possible prices, never take the one shown on the first card drawn.  If we try to extrapolate the rule for the simple 3-card game, we'd say that in the 5-card game you should take the second card if it's lower than the first.  If it's higher than the first, proceed to the third card.  But now we have a problem that couldn't arise in the 3-card game.  What if the price indicated on the third card is between the ones shown on the first and second?  In other words, there are two competing strategies.  One would be never to pay the price shown on the first card; instead, as the game progresses, pay the price shown on the first card you reveal that is lower than all the other prices so far revealed.  Another would be never to pay the price shown on the first card; instead, as the game progresses, pay the price shown on the first revealed card that is lower than the price shown on the immediately preceding card. 

At first, I thought the whole challenge might be to figure out which of these possible strategies was superior.  As I was rather tediously trying to work that out, however, it occurred to me that it's somewhat more complicated than that.  The more possible prices there are to pay, the more chance you have to put what you learn to good use.  For example, suppose that the prices shown on the first two cards are $400 apart.  You immediately know that you have revealed the "fair price" and the most marked-up price--the bookends of the range.  If the first card showed the fair price, it's--alas!--too late to select it.  On the bright side, you know that the second lowest price is out there, and you know what it is.  If you don't get it on the third card, proceed until you do get it.  Of course, if the second card is $400 less than the first card, you know it's the fair price and you should take it--the three cards you haven't seen are for $100, $200, and $300 over the fair price.

To avoid working with algebraic expressions, let's designate the five possible prices 0 (for "fair price") and 1, 2, 3, and 4 for $100 over, $200 over, $300 over, and $400 over "fair price," respectively.  Then, of the 120 possible orders, six begin 

0, 4 (in which case we pay 1)

and another six begin

4, 0 (in which case we pay 0).

So we've got the optimal strategy for 12 of the 120 possibilities.  What about the rest?

The strategy will be to consider systematically what the possibilities are after the first two cards are revealed.  Suppose the second card indicates a price $300 more than the first.  That would mean we are dealing either with a sequence beginning 0, 3 or 1, 4.  In the first case, 1, 2, and 4 remain.  In the second case, 0, 2, and 3 remain.  We can't know which case we have, but obviously we should see the third card.  To avoid being stuck with either 4 (first scenario) or 3 (second scenario), we should be looking for a card that is 1 higher than the lowest card we've seen, or better.  If we are in the first scenario, we will then end up with the 1-card.  If we are in the second scenario, we will end up with the 0-card if we are lucky enough to come to it first; otherwise, we'll end up with the 2-card.  At this point, we will have considered twelve more of the 120 possible arrangements of cards, and we will have ended up with the 1-card six times, the 0-card three times, and the 2-card three times.  

Next up would be the case where the second card is $300 less than the first, and then we'd have to take up in turn the cases where the difference between the first and second cards is only $200 or $100.  But this is already getting long, it doesn't get less complicated, and, if you've hung with me this far, you probably want to work it out for yourself anyway.  If you do everything the best you can, I think you can expect to pay this salesman, on average, $90 more than the "fair price."  Gypped, but quite a bit better than the $200 over fair price you'd expect to pay if you just threw up your arms and chose cards at random.

Jottings, cont.

 

Lest anyone doubt my credentials for geekdom, I have my kids working math problems to warm up for the start of school. 

"But, dad, nine months of learning is more than enough!"  

"Shut up and sharpen your pencil." 

The above is Lydia's most recent homemade worksheet.  By chance, the day after she did it, Omarosa released the tape of Lara Trump trying to buy her silence for 15 grand per month.  In the part of the transcript that got my attention, Lara is speaking:

So, I know you, you were making 179 at the White House.  And I think we can work something out where we keep you right along those lines.  Specifically, let me see, I haven't even added up the numbers.  But we were talking about, like, 15k a month.  Let me see what that adds up to.  Times 12.  Yeah.

She's pretty obviously punching keys on her calculator, right?  For 15 x 12?  It's 30 more than 15 x 10, girlfriend.  Did you know that 15 x 10 = 150?  So 180.  I give my 10-year-old harder problems than that.  Maybe when you're a Trump, you're used to working with bigger numbers, and the calculator is just a reflex?  

I guess another question might be how do the small donors feel about having all they can afford being offered to Omarosa to stop her from talking smack about the Trumps for, like, a half hour?  A possible word problem for my next worksheet!  Omarosa is paid $15,000 per month hush money out of funds donated by MAGA types.  If Trevor from Tennessee  mails in a check for $60, for how long does his contribution muzzle her?  Consider a month to be 30 days and that she puts in 16-hour days, 7 days a week, keeping her mouth shut about what corrupt, creepy dopes they are.  

I think the saddest lyric in any Replacements song is from "Here Comes a Regular":  I used to live at home/Now I stay at the house.

The Twins have won 18 of their last 23 home games--and 7 of their last 26 on the road.  

Dweebish

Today our 6-year-old kept asking for my phone.  I was watching a basketball game, or trying to, so around the fourth time I gave it to her.  Phone for peace.  When there was a timeout, I checked in on her, and discovered that she had the calculator app open.  She'd punch in, like, 6 + 4, and then there'd be a pause before she touched the equal key and I could hear her say softly to herself either "Yes!" or "Ach!"  What a little dweeb.

I suppose she comes by it honestly.  About once a week I ask the teenager what the current topic is in math class.  That's always good for the eye roll, and, on rare occasions, an answer, such as, recently, "Circles."  Oh, boy, I am really interested in circles!  Give me an example of a problem!  For the first time in recorded history, she obliged, and soon we were figuring out the area inside a sixty degree sector of a circle of radius 1 but outside the triangle made by the radii and the chord.  That makes it sound harder than it is--it's just the shaded area in the figure, only with the radius of 6 changed to 1.  She's in 8th grade.  Maybe I like this stuff partly because I wish I was in 8th grade myself.  Beats hell out of being 58.  Anyway, for the dweebs, the area of the whole circle would be ­π.  So the area of the sixty degree sector would be a sixth of that, π/6.  The triangle made by the radii and the chord would in this case be equilateral, one unit on all sides, and the area of that would be . . . . well, suppose you don't want to google the formula for area of an equilateral triangle.  The area of any triangle is the product of half the base and the height.  Half the base would be 0.5.  The height may be thought of as one leg of a right triangle where the other leg is 0.5 and the hypotenuse is 1.  By the Pythagorean Theorem, that works out to be √3/2.  So the area of the whole triangle would be half of that, √3/4.  And the area we want is the difference between that and π/6, that is, (π/6 - √3/4). 

I lecture Rianna in the manner of my dad.  "If this answer is correct, it had better be quite a small positive number, right?"  A calculator puts it at just a touch under 0.09, which seems plausible.

Maybe the ACC is not that great?  If Duke wins, it will have two entrants in the Sweet Sixteen, and the Big Ten has three.  Plus, it took a couple of #1 seeds to eliminate Michigan State and Northwestern (sic).

Leibniz!

The German philosopher Gottfried Leibniz (1646-1716), one of the half dozen or so eminent philosophers treated in a new book reviewed here, has I think gotten a bum rap.  For he is today known, if he is known, as the philosopher whose contribution to religious thought was famously, savagely, and effectively satirized by Voltaire in Candide.  Dr Pangloss, one of the characters in that book, held that "this is the best of all possible worlds," and he persisted in this view as Voltaire subjected him to ever more convincing proofs, in the form of awful tragedies begetting unendurable human suffering, that it's not.  Thus "Panglossian: characterized by or given to extreme optimism, especially in the face of unrelieved hardship or adversity."

The view that "this is the best of all possible worlds" was Leibniz's solution to the problem of evil.  You know the puzzle.  God, being merciful, must want to stamp out evil and suffering.  And, being omnipotent, he must be able to do it.  Yet the world is filled with both.  Therefore, this merciful and all-powerful deity doesn't exist.

Leibniz escapes this trap by allowing that God is constrained by the same kind of logical necessities to which we all are subject.  If p is true, then the opposite of p is false.  If you choose p, then the benefits of the opposite of p are lost.  This is as true for God as for everyone else.  So you end up with this picture of the Creator very palely weighing the pros and cons of the various choices before him when designing the universe.  "Human beings should be free.  That means, however, that they might make poor choices that disrupt my beneficent plan.  Okay, maybe I shouldn't let them do that.  No, the consequences of that choice are even worse.  I'll make them free."  And so he proceeds through the cost-benefit analysis of all the decisions that he is obliged to make.  At the end of the line, you are left with "the best of all possible worlds"--the key word being possible.  It is easy to see that this amounts to a denial of God's omnipotence.  It's also easy to see that it invites satire.  And boy, did Voltaire ever walk through that open door!

One of the pleasures of satire is the enjoyable sensation that someone who deserves to get smacked is getting smacked, hard.  As the effectiveness of the satire goes up, the stature of its object goes down. Candide, being highly effective, has left poor Leibniz languishing.  But he is really at the other end of the spectrum from the bumpkin that readers of Candide may suppose him to be.  I'm pretty sure that my account of his solution to the problem of evil isn't completely fair to him.  Voltaire's critique is such a rhetorical triumph that everyone, including me, unconsciously understands the debate in his terms.  When you know a little bit more about Leibniz, you are inclined to believe that anything that would occur to Voltaire would occur to him, too.  Maybe there is more to the story. 

For me, the proof of Leibniz's stature in the pantheon of the world's great geniuses is his work in mathematics.  Years ago, I took "the calculus sequence" at the University of Minnesota.  I still have the textbook.  Ten chapters.  Fall term, Calculus I, covered the first four.  In the winter, Calculus II covered chapters 5 through 7.  In the spring, Calculus III finished the book.  My outcomes, as measured by the grades I received, were (as a calculus student might phrase it) inversely proportional to the effort I put into it.  I aced Calculus I without working very hard.  Calculus II began to be hard for me, and I worked diligently at it, eventually earning a B.  Calculus III consumed my life for a couple of months, and I learned it, sort of, but it would be more true to say that things were moving too fast and I never firmly grasped the material: C.  That was the last math course I ever took.  I wasn't curious about whether the trend would continue.

My experience isn't unique.  It's hard for most people to learn the calculus.  If you know it a little bit, and were impressed by the effort it cost you, you're in position to esteem whoever first invented it--or discovered it, or developed it, or whatever is the proper term for bringing into use new mathematical techniques of considerable scope and utility.  This achievement belongs to Newton and Leibniz, who, each working without knowledge of the other, invented ("discovered"; "developed") calculus.  Newton's achievement may be credited to genius applied to certain problems that arose from his effort to comprehend physical reality--a new tool for a working scientist.  Leibniz, however, was more or less a mathematical hobbyist.   It's as if the calculus just fell out of his mind, the work product of pure thought undertaken for recreational purposes.  That he should be known as the guy Voltaire beat up is an injustice that it seems a merciful and powerful overseer might have avoided.

Of magic pentagons and second graders

One of the really pleasurable sensations of having young children is the humiliation associated with homework help.  Our second grader's current math homework has an "extra credit"--small consolation!--problem concerning "magic pentagons."  The object is to use the numbers 1 through 10, once each, so that the sum of the numbers on the fives sides of the adjoining figure are all the same.  Someone is off to a bad start on this one: since the "average" number is 5.5, and 3 x 5.5 = 16.5, it seems reasonable to assume that the best candidates for the one common sum are 16 and 17--in which case there is already no solution for the "northwesterly" and "northeasterly" sides, 10 being the largest available number.

But my initial efforts, conducted while the second grader was sound asleep, didn't yield any good results, either.  Having after an embarrassing amount of time come up with a solution, I now see that I was handicapped by the assumption that the corner slots, which have to be used for two sums, must probably be in the middling range.  Once I out of frustration dropped that notion, it didn't take much longer to come up with the following solution (proceeding clockwise from the "pinnacle," and as it happens preserving the 1 already filled in): 1, 8, 7, 6, 3, 9, 4, 2, 10, 5.  That is, the sums on the five sides are:

1 + 8 + 7 = 16
7 + 6 + 3 = 16
3 + 9 + 4 = 16
4 + 2 + 10 = 16
10 + 5 + 1 = 16.

Next morning, when I showed the second grader my handiwork, she was visibly pleased, and  painstakingly copied my work into the space provided on her homework sheet.  I asked whether she thought she should add a note indicating that actually her dad had done it.  She thought for only a second before answering, "No."  I thought that was okay because, honestly, I suspect Ms. Kuhlmann is smart enough to know that every kid getting extra credit this week has at least one nerd for a parent.

Right?  If it took me a whole news show plus Jimmy Fallon's monologue to get it, the problem is not really fit for second graders.  Right?  Hoping.

Math in "the real world"

For those of you who may be interested in the cancer-statistics-"Price is Right" nexus, here is an interesting and entertaining article that I for sure would have missed but for the fact that the author, Elisa Long, is Amanda's high-school classmate and FB friend.  She (Elisa, not Amanda) is possibly the only person from Fridley, Minnesota, who has ever taught math at Yale.  But I wonder if she could solve this problem:

A baseball season is 162 games long.  If, after 90 games, a team has 50 wins, what is the probability that for the season this team ends up losing more games than it wins?

Games geeks play

Being a geek, I sometimes, like during halftime of televised football games, find myself inventing little games of chance and then wondering about the best strategy.  For example, supposing someone said that they would give me a hundred thousand dollars or, alternatively, a million dollars if I correctly call a coin flip but nothing if I call it wrong.  What to do?

I find that I have an immediate emotional response that's followed by a more considered reflection nudging me toward the opposite choice, which in turn is followed by a still more considered reflection that pushes me back toward my initial inclination.

My first thought is to take the hundred grand.  The prospect of losing a sum of money that large on a coin flip is too painful to contemplate.  But then the lizard mind is overruled by the one that can do math.  The expected return on the option involving a coin flip ($500,000) is five times greater than the option I'm choosing.  Since $100,000 < $500,000, I should put aside my emotional impulses and make  the obvious right choice.

But then I think:  what exactly is meant by "right choice"?  If my goal is to maximize my expected return, then, yes, I obviously have to try for the million.  It seems, however, that I'm missing another obvious consideration relating to the concept of utility.  For someone who is supposed to pay back, this coming Thursday, $80,000 that he doesn't have to a loan shark with mob connections, taking the sure hundred grand is the easy, right choice.  I think it might be the right choice for me, too.  But I'm not sure it is.  A hundred thousand would be very useful but not really life-changing.  I'd still have to go to work, still have to care what my boss thinks of me. Is that so bad though?  It's kind of a perplexing problem.  If you magnified by ten--I can take a million dollars or a fifty-fifty chance at $10 million--it would be easy.  Put down the coin, hand over the certified check.

In other games, the perplexity is all in the math.  Supposing someone proposed, for a fee, to flip a coin until it turns up heads, then pay me $2 if the first flip is a head, $4 if the first head occurs on the second flip, $8 if on the third, and so on.  (If the first head occurs on the nth trial, I get 2^n dollars.)  I should agree to play this game if the fee is below what price?

 

Bracketology

I'm not too surprised that Warren Buffet could stop feeling nervous about having insured Quicken Loans' billion-dollar payout for a perfect NCAA basketball bracket before the first round of the tournament was completely over.  The chance of a perfect bracket has been widely reported to be 1-in-9.2 quintillion.  The source of that incomprehensible number is 2⁶³, the exponent representing the 63 games in the tournament and the base representing the two possible winners of each of these games.  Since there is obviously not an equal chance of either team winning any particular game, I think this number is too high.  In the first round of the tournament, for example, a #16 seed has never defeated a #1 seed, so if you pick the #1 seed to win their first game in each of the four regions, and an unprecedented upset doesn't occur,  the number of all possible remaining outcomes shrinks to 2⁵⁹. 

But Buffet really wasn't taking any chance at all. The chance of someone winning was more theoretical than actual.  So it was just free publicity.  Plus, Quicken Loans, which got a lot of personal data about all the contest entrants, paid Buffet an undisclosed fee.

It's fun to try and gauge in a rough sort of way why Buffet wasn't actually taking a rash gamble.  Let's  consider the difficulty of predicting correctly the outcome of every game in the first round of the tournament.  There are 32 first-round games, so if you just guessed, the chance of getting every game right is given by 1/2³².  This denominator is 4,294,967,296.  At least it's small enough so that we have recognizable words for it: just under 4.3 billion.  I think I read somewhere that only one entry per household was permitted.  The population of the country is a little more than 300,000,000, and I'm pretty sure there are considerably fewer than half that many households.  But let's just observe that 150,000,000/4,294,967,296 is about 0.035.  So now let's assume that everyone wants Warren Buffet to have to pay out and that all hundred and fifty million competitors collude to cover as many different outcomes as possible--no duplicate brackets, in other words.  They still cannot cover even 4% of the possible outcomes.  If there was collusion among the maximum number of permitted entrants, how many of the 96-plus percent of uncovered outcomes would have included Mercer over Duke? 

All of them, or for sure almost all.  And, remember, we are talking about the difficulty of submitting a bracket that's perfect for the first round only.

Car Talk Puzzler: dollars and envelopes

I'm a fan of the NPR show Car Talk, which, as I have mentioned before, includes a segment known as The Puzzler.  This week's problem is of a purely mathematical nature, so, as is not the case when the subject matter is automotive, I have a chance.  Here's the puzzle:

Suppose you have a thousand 1-dollar bills and ten envelopes.  The challenge is to put the bills in the envelopes in such a way that, if someone asks for any number of dollars from one to a thousand, you can turn over exactly the sum requested by handing out  envelopes.

I was in the car when I heard it, and the only thing I could figure out before I got home is that it's probably advisable to worry first about being able to comply with small requests.  After all, someone might ask for just a dollar, in which case you need (obviously!) one envelope with a single dollar bill in it.  And if they ask for two dollars, you need either an envelope with two dollars in it, or two envelopes with one dollar.  At first, it seems like it's going to be pretty hard to comply with these small requests and those in the 800- or 900-dollar range. 

But at home, with the aid of pencil and paper, it soon became apparent that you can make use of the fact that the expression 2ⁿ gets quite large quite fast as n rises through the smallest integers.  I had my 1-dollar and 2-dollar envelopes.  I surmised that a 3-dollar one was unnecessary, since 1 + 2 = 3.  But a 4-dollar envelope is necessary.  The next one that's necessary has eight dollars.  Perhaps this is enough to spot a pattern, since at this point the number of bills in the four used envelopes is given by 2⁰, 2¹, 2², and 2³, respectively.  In the fifth envelope, you need 2⁴, or sixteen, dollar bills.  And you proceed in this fashion until, in the ninth envelope, you put 2⁸, or 256, dollar bills.  Now put all the remaining bills--489 of them--in the tenth envelope.  It's pretty clear that you can now comply with any request, right?  The first nine envelopes cover any request up to $511.  All ten envelopes cover the request for all 1000 dollars.  And if the ten envelopes contain a total of $1000, and the first nine can be made to yield any sum from 1 to 511, it is clear that, to cover the numbers from 512 to 999, you can subtract an envelope, or some combination of them, from all ten to get the required amount. 

Probability and the outcome of games

My sister, a professional geek, meaning in her case that it's her job to teach chemistry to young people, recently let drop that a recent test included the following question:

In the "game" usually called "52 pick-up," what is the probability that 26 of the cards will land face down and 26 face up?

Since I'm an actual geek, meaning I have an interest in mathy problems that is purely recreational, I worked out what I think is the solution.  (It's not 1/2, the popular wrong answer, according to sis.)

It's maybe not the way a professional would think of it, but imagine cards falling one after another, face up or face down: up, down, down, down, up, down, up, up, up, up, etc., etc.--whatever the result is till there have been 52 landings, and therefore 52 results.  How many possible 52-string results are there? This is given by 2⁵².  If you don't believe it, you can try it out with smaller samples that can easily be counted, and perhaps persuade yourself that the pattern holds: for example, if there are two landings, the number of possible results is 2², or 4--up, up; up, down; down, up; and down, down.

So the problem becomes, of the 2⁵² possibilities, how many have 26 ups and 26 downs?  This is the number of different ways that you can choose 26 items out of a group of 52, which is written 52!/26! 26!  In other words, the number of ways you can "choose 26 from 52" is given by:

52x51x50x49x48. . . x3x2 / (26x25x24x23x22. . . x3x2)²

Again, skeptics may possibly persuade themselves by experimenting with smaller numbers.  The number of ways you can "choose 2 from 4" is given by 4!/2! 2! = 4x3/2 = 6.  Does this check?  Well, you could take (1,2), (1,3), (1,4), (2,3), (2,4), or (3,4).  Those are all the ways there are to do it, and there are 6.

The solution to the "52 pickup" problem, then, is the ratio of 52!/26! 26! (the number of results that have 26 ups and 26 downs) to 2⁵² (the number of all possible results).  This involves some number grinding but turns out to be almost exactly 0.11--an eleven per cent chance.

The solution to this problem brings into view others that are more interesting.  You know how, in baseball's World Series, the champion is the first team to win four times.  So suppose that one team is better than the other and has a 3/5 chance of winning any game.  What is the chance this better team will win the Series?  Well, their chance of a sweep is (3/5)⁴.  Their chance of winning in five games is (3/5)⁴ x (2/5) x 4.  Their chance of winning in six games is (3/5)⁴ x (2/5)² x 10.  And their chance of winning in seven games is (3/5)⁴ x (2/5)³ x 20.  (The terms 4, 10, and 20 in the above expressions come, respectively, from the number of ways you can choose 1 from 4, the number of ways you can choose 2 from 5, and the number of ways you can choose 3 from 6.)  Grinding the numbers yields the result that the superior team has about a 71% chance of winning the Series. 

Does that seem counterintuitive?  A team that is plainly inferior still has almost a 30% chance of prevailing. In general, romantic legends of improbable events--upsets made possible only by the hand of God--don't stand up to close scrutiny.  On the other hand, the longer the series, the better the chance that the superior team will win.  In our example, the inferior team's chance of winning goes from 29% to 40% if the championship is decided by the outcome of a single game instead of by a best-of-7 series.  That is why, at this time of year, when #2 seeds are taking on #15 seeds in the NCAA basketball tournament, you hear the commentators talking about how one team is going to try to "shorten the game"--slow it down, use all of the shot clock, hold down the number of times each team gets the ball.  It's like winning a single game instead of a long series.

Gaming

I have been playing a lot of Yahtzee lately with Rianna, our 8-year-old.  If you've been exposed, you know that you play the game with five dice, and that the players in turn get to roll up to three times, setting aside after each roll the results they want to keep.  A "Yahtzee" is the same result on all five dice.  A pretty big part of the game is getting to three of a kind for all six numbers within the thirteen rounds that make a game.  Getting only two of a kind, especially for the higher numbers, will kill your score.  The first roll of a turn frequently yields two of one number together with three "singletons."  Rianna and I feel that it is an extraordinary run of bad luck not to pick up the needed third on either your second or third roll, yet it happens with sufficient regularity that I began to wonder just what the probability is.  The solution must be given by (5/6)⁶: the 5/6 representing the probablity that a single throw does not yield the desired result, and the exponent standing for the six (three dice times two rolls) straight undesired results.  This number turns out to be just about exactly 1/3.  In other words, the chance of failing to "triple up" after an initial pair is about the same as the chance of a major league batting champion getting a hit in his next at-bat.   He's more likely to make an out, but in a game in which he comes to bat several times you have to expect that he'll get at least one hit.  In fact, if he bats four times the chance of four outs is (2/3)⁴, which is less than 1/5.

Speaking of gaming and probability, isn't it sometimes in the interests of a football team to allow an immediate touchdown?  In today's Vikings game, for instance, Denver intercepted a Minnesota pass deep in Minnesota territory with the score tied and around 90 seconds left.  To me it seems obvious that the Vikings should have let Denver score a touchdown on the first play.  Instead, Denver ran a few plays, the last one being a quarterback keeper to position the ball immediately in front of the uprights, and kicked a winning field goal on the last play of the game.  If after the interception Minnesota had let Denver score an immediate touchdown, there would have been more than a minute in which to score a tying touchdown.  Isn't there a better chance of that than there is of the other team missing a chip-shot field goal?  But I've never seen a team pursue this strategy.  The commentators in today's game didn't mention it as a possibility.  It's almost as if it's against the ethics, but I don't see why doing what has to be done to increase the chances of victory should be frowned upon.  A team trying to win should pursue the optimal strategy.

Recreational math

Scraping around for a birthday present for one of my nephews, a college student majoring in computer science, I picked out a copy of Martin Gardner's My Best Mathematical and Logic Puzzles.  And since I'm pretty geeky myself, I couldn't help scanning it before the time came to wrap it in paper.  The first problem didn't interest me much.  Here's the second:

Two men play a game of draw poker in the following curious manner.  They spread a deck of 52 cards face up on the table so that they can see all the cards.  The first player draws a hand by picking any five cards he chooses.  The second player does the same.  The first player may now keep his original hand or draw up to five cards.  His discards are put aside out of the game.  The second player may now draw likewise.  The person with the higher hand then wins.  Suits have equal value, so that two flushes tie unless one is made of higher cards.  After a while the players discover that the first player can always win if he draws correctly.  What hand must this be?

That's down my alley.  I think I have an idea of how the first few games might go.  Since an ace-high straight flush is the highest hand in 5-card draw, the first player might in the first game thoughtlessly begin by choosing the ace, king, queen, jack and ten of a certain suit.  You can't lose with that.  But, according to the rules of this eccentric poker game, it quickly becomes evident that you can't win with it, either.  The second player need only choose the same  five cards of one of the other three suits.  It's easy to see how that game ends.  It's a tie.

The problem with the first player's first strategy is that it left the highest possible hand open to the other player.  The first player might prevent that in the second game by choosing, for his first hand, all four aces and one king.  Obviously the ace-high flush is now possible only for player one.  It turns out, however, that the first player's second strategy is worse than his first.  The second player should choose all four queens.  The highest possible hand available to player one is now a jack-high straight flush.  Meanwhile, player two is set up to fill in a queen-high straight flush on his second move.  The only way to prevent it is a defensive move that leaves player one with a hand lower than four queens.  Check mate.  Player two wins.

The way that hand works should, however, suggest a winning strategy to player one.  Picking four tens prevents player one from achieving an ace-high straight flush just as surely as picking four aces did.  Player two can now not make a hand higher than a 9-high straight flush, whereas player one is set up to fill in a 10-high straight flush on his second move.  Trying to block him by choosing four nines doesn't work (because player one, discarding three of his four tens, fills in an ace-high straight flush on his second move; player two cannot achieve a tie as none of the tens are available to him).  Of course, trying to block him by choosing four aces, or four kings or queens or jacks, is also useless (because player one uses his second move to complete a 10-high straight flush, and, as before, the highest hand availabe to player two is a 9-high straight flush). 

Is there something a little off about the wording of the problem?  "What hand must this be?"--makes it sound as if there is only one hand that guarantees victory when actually four tens and any fifth card will do it.  And you know what?  The first player can guarantee victory by choosing just three tens on his first play, but in that case he has to exercise some care in the choice of the other two cards--the jack and the six, for example, from the suit in which he lacks the ten will work.  I'll leave it to my dweeby readers to  work out all the details.

If you find these kinds of problems difficult, here is an interesting conversation that should make you feel better.  Almost everyones' instincts are wrong when it comes to puzzlers of the kind that interest recreational mathematicians and, increasingly, neuroscientists.  Our minds, the product of evolution by natural selection, are built to cope with other challenges.   

Car Talk and Puzzlers

On Saturdays, I like to lay about in the forenoon, drinking coffee, maybe roughhousing with the girls, and then, at 11:00, tuning into "Car Talk" on NPR.  Not that I care about cars.  I'm the kind of coffee-shop loitering, Obama-voting, public transportation-using urban liberal who, in my bachelor days, contemplated going car-less.  But then how would I get myself and my golf clubs to whatever course my friends were playing this weekend?  Plus, I had to travel forty miles each way on the big holidays.  So year in and year out I would drive 6000 miles, first in a Dodge Omni, then in another Dodge Omni, then in a Toyota Tercel that one of the regulars in our golf foursomes, the owner of a Suburban, nicknamed "Big Blue." 

I remember the first time I tuned into "Car Talk."  It was an accident.  I'd been listening to something else, and by the time this crazy car call-in show came on I was washing dishes and so didn't click off the radio.  The first caller was a woman with some car trouble, I can't remember what, that she described in considerable detail.  When she finally came to the end of the symptoms, one of the hosts asked: "What color is the car?"  I kept listening and soon was hooked.

Like other fans of the show I remember favorite calls.  One day Senator Tom Harkin called in with a car question.  I like the genre of calls involving women seeking an impartial adjudication of a car argument they've been having with the husband or boyfriend.  It's pretty clear to me that men, since they feel obliged to have an opinion on a car question, are more apt to be wrong than women, who are free to have no opinion and therefore only express one when they have an actual reason for believing as they do.  My favorite call of all time was probably the one in which the fellow's dog had gotten car sick and ralphed onto the dash, so that the puke seeped into the heat register, with the result that every time he ran the heater the car smelled of dog vomit.  They had a lot fun with that one, and finally advised the caller to sell the car--in the summertime.  They have a lot of fun with almost everyone before finally dispensing what I assume is consistently good advice.

I also like "the puzzler," a regular feature that is as often mathematical as auto-related.  (An ongoing good-natured joke concerns the utility of math and science over and against "the humanities"; whenever someone calls in and says they are a student, inquiries concerning their field of study are sure to follow, and an answer like "art history" or "comparative literature" elicits loud groans.  Of course they have  no clue what to do about their problematic vehicles!)  Since I haven't placed a post in my "Math" category for awhile, let me give an example of a recent Car Talk Puzzler.

There are three scraps of paper, each with a different number written on it.  You have to pick the one that has the biggest number.  If you want, you can look at one and then decide whether or not to choose it.  If you decide not to choose it, that one is out: you have to choose one of the other two.  If you look at one of the remaining ones, you must either choose it or the last one.  The question is: What's the best strategy?  It's obvious that just guessing at the start of the game gives you a one-third chance of winning.  Can you do better than that?

Yes, you can.  The best strategy, it turns out, is to look at one paper but never choose it.  Then, if the second paper has a higher number than the first, keep it; but if it has a lower number reject it also and make the third your choice.  Here is the way to analyse it.  There are only six orders in which the three papers may be chosen.  Let us denote the papers 1, 2, and 3, where 1 stands for the paper with the smallest number and 3 for the one with the biggest.  The six possibilities are:

i. 1, 2, 3
ii. 1, 3, 2
iii. 2, 1, 3
iv. 2, 3, 1
v. 3, 1, 2
vi. 3, 2, 1

Each sequence is as likely as another to represent the order in which you choose them.  In (i.) you lose, because you end up with (2), not (3).  In (ii.), you win.  Scenarios (v.) and (vi.) are both losers but (iii.) and (iv.) are both winners.  That's three out of six winners, which is a fifty percent improvement over just guessing at the start of the game.  (Uncovering first one and then another but always settling on number three is the same as just guessing that three is the highest without turning any over.)

An archive of "Car Talk" puzzlers is here.  The one I've described was for February 12 of this year.

General theory of UConn

The women's basketball team at the University of Connecticut (the UConn Huskies) won its 89th consecutive game last night.  The victory, over 22nd-ranked Florida State, by a lopsided score of 93-62, broke the previous record held by the UCLA men's team, which won 88 games in a row back in the 1970s.  UConn's record-tying 88th consecutive win, over 12th-ranked Ohio State, was by a score of 81-60.  Last year UConn had its second straight undefeated season and, naturally, its second straight national championship.  The Huskies won the championship game of their regional tournament by forty points and their national semifinal by twenty points before prevailing in the championship game by just six points over Stanford.

There's been a lot of talk about just why UConn is so good.  No doubt they have a good coach, who is probably an even better recruiter, and their past success makes it easier to attract the best players.  But I wonder, too, whether there isn't a kind of pointy-headed, mathematical explanation akin to Stephen Jay Gould's account of why no baseball batters hit .400 anymore. 

It used to be fairly common for major-league baseball batters to hit for an average above .400, but no one has now done it since 1941, when Ted Williams averaged .406.  It probably isn't the case that Williams, Hornsby, Cobb and others were better hitters than the best contemporary batters.  But they were playing in a day when baseball was still a comparatively young game, the best strategies had yet to be devised (or discovered), and, consequently, there was more variation within the system of all batting averages.  And more variation meant the outliers were "outier," farther from the mean, possibly even north of .400.

The UConn women's team, in this view of things, is not just a great team--it is a great team playing in the (relatively) early development of the women's game, when the distance between the best team and most teams is greater than it is in the men's game now and will be in the women's game sometime in the future. More variation within the system of women's collegiate basketball means that the very best team is less apt to suffer an upset, because the gap between the best and the typical is greater.  A long winning streak is more likely to occur.

If you are skeptical, you probably won't be persuaded by the fact that the general theory would predict that UCLA's record, if ever surpassed, would be surpassed by a women's team.  After all, there is around a fifty-fifty chance of that without introducing any other considerations.  But what if the average margin of victory in all collegiate women's games was somewhat greater than the average margin of victory in all men's games? I think that would be strong evidence of greater variation within women's basketball. 

I don't have the energy to track the results of all collegiate basketball games.  If you'd like to, please tell me what you discover, as I'd be very interested.  Meanwhile, I think I will track the score differential in all men's and women's games played this season in the Big Ten Conference.  Then I will do the same for the NCAA tournament games.  That should give some idea.  I'll report in on what I find out.

POSTSCRIPT: For what it is worth, I've looked at the scores in last year's NCAA Division One Basketball Tournaments.  In the women's games, the average margin of victory was 16.4 points (14.5 points in games not involving UConn).  Whereas in the men's tournament, the average margin of victory was just 10.4 points.  It's not a lot of data--just 63 outcomes in each tournament--but the women's games were on average considerably more one-sided, suggesting more variation among the entrants.

Some more calculus problems

Here's a problem from Amanda's most recent calculus quiz.  A print shop is to make a poster with a four-inch margin at the top and bottom, a two-inch margin on the sides, and a print area of 50 square inches.  What are the dimensions of the smallest poster--least area--that can be made.

Well, if the horizontal print area is x, and the vertical print area y, then xy = 50 and y = 50/x.  The area of the entire poster might then be expressed by the equation (x + 4) (50/x + 8).  Expand this equation and you get 50 + 8x + 200/x + 32.  Differentiating, we arrive at 8 - 200/x ².  So an extreme value exists where 8 = 200/x ², which would be when x = 5.  And when x = 5, y = 10.  Alack, this is where I quit when working the problem, which did not ask anything about the print area but rather about the dimensions of the whole poster.  That would be (5 + 4) x (10 + 8), or 9 x 18 inches.

Martin Gardner, in Calculus Made Easy, is fond of similar problems involving various shapes or solids inside of bigger ones--for example, find the volume of the largest cylinder inscribed within a cone the height of which is the same as the radius of its base.  It would help to draw a picture but I am not clever enough for that--even if it is possible.  You will have to imagine a cone--the often orange thing that kids use to mark a goal for their pick-up soccer games, or that the highway department uses to keep cars off a patch of road they are working on.  The cone the problem describes has a height equal to the radius of its base and it also has a cylinder--a solid shaped like a can of anything at the grocery store--inside of it.  If the cylinder is tall, its radius will be small; and as the radius gets bigger and bigger, finally approaching that of the cone, its height shrinks and the shape becomes "squat."  Its radius can't be greater than that of the cone or else it won't "fit."

The volume of a cylinder is the procuct of pi, its height, and its radius squared.  The one we are talking about is limited in the respect that it must be inscribed within the cone, which is to say that the sum of its height and radius is equal to the radius (which is the same as the height) of the cone. The state of affairs suggests the following algebraic expression

h + r = R

where h refers to the height of the cylinder, r to the radius of the cylinder, and R to the radius of the cone.  It follows that

h = R - r

in which case the volume of the cylinder may be rewritten thus to eliminate one variable:

A = (pi) r² (R - r) = (pi) R r² - (pi) r ³. 

Now we differentiate and set the derivative equal to zero to determine the extreme value:

A' = 2 (pi) R r - 3 (pi) r ²
2 (pi) R r = 3 (pi) r ²
2 R = 3 r
r = 2R/3

To maximize the volume of this inscribed cylinder, make its radius two-thirds that of the cone. 

It may seem strange that the solutions to both problems are so regular and involve 2/3 or the ratio 2-to-1 instead of some more complicated fraction.  But if you work a lot of these problems you begin to notice that this is almost always the case.  The poster with the smallest area has one dimension exactly twice that of the other.  The cylinder with the greatest volume has a radius exactly two-thirds that of the cone in which it is inscribed.  You can verify, if you are a dweeb, that a cylinder of a given volume will have the maximum surface area if its height is exactly twice the length of its radius. 

Pleasures of calculus

I'm still enjoying Amanda's calculus course.  She uses her book, so I read up the topics on the syllabus in my old textbook, or in Martin Gardner's Calculus Made Easy, and we then work the problems from her homework together.  The couple that calculates together, sticks together. 

So here's a recent problem in "related rates." A ten foot ladder stands straight against a wall.  Then someone starts moving its bottom end away from the wall at the rate of one foot per second.  How fast is the top of the ladder going down the side of the wall at the moment the bottom of the ladder is six feet out from the wall?  Eight feet out from the wall?

Maybe your intuition is that the top of the ladder moves, if not at the same steady rate as the bottom, at least at some constant rate.  But it is relatively easy to verify that that is not so.  After six seconds, for example, the base of the ladder will be six feet from the wall.  How far will the top of the ladder have gone down?  Since the ladder is always ten feet long, we are considering a right triangle with a base of six feet and a hypotenuse of ten feet.  From the Pythagorean Theorem, it follows that the height is eight feet (8² + 6² = 10²), and that in the first six seconds the ladder has gone down two feet.  Two seconds later, however, the base is eight feet and the height is only six feet, so the ladder went down as far in those two seconds as it did in the first six. Determining instantaneous velocity at certain points is a problem in calculus.

The key to solving the problem is to analyze this constantly changing right triangle.  If we call the height of the ladder on the wall h, and the distance from the bottom of the ladder to the wall b, then

h² + b² = 10². 

What we are looking for is the ratio of the change in height to the change in time--dh/dt, in the language of calculus.  Differentiating implicitly, we get

2h (dh/dt) + 2b (db/dt) = 0.

Now, let us recall that we are looking for dh/dt when b is six and eight, respectively; also, that db/dt, the ratio of the change in the length of the base to the change in time, is a constant one foot per second.  So all that really remains is to make the necessary substitutions and then do the algebra:

2h (dh/dt) + 2b = 0
h (dh/dt) = -b
dh/dt = -b/h

When b is six, h is eight; and when b is eight, h is six--remember our right triangle with the hypotenuse a constant ten.  So at the instant the bottom of the ladder is six feet from the wall, the top of the ladder is going down the wall at the rate of 9 inches per second.  When the bottom is eight feet from the wall, the top is going down at the rate of 16 inches per second.

Wherein I prove I am a dweeb

Amanda's calculus course is progressing.  I am progressing, too--through Martin Gardner's update of Silvanus Thompson's Calculus Made Easy, and it's making me nostalgic.  I remember how, after first learning to differentiate functions, we had our first word problems, and how cool I thought it was to be able to solve, so easily, very practical problems that would be impossible were it not for the calculus. For example, I can still recall one of the problems from the second exam in Calculus I:

You want to build a storage shed that is to have a square base, a flat roof, and a volume of 1800 cubic feet.  Find the dimensions of the least expensive shed that can be built if material for the floor costs $1.20 per square foot, for the sides $3.00 per square foot, and for the roof $2.00 per square foot.

So there is a cost function, which we want to minimize.  The cost is the sum of certain areas multiplied by the cost of the material for that area, or

C = 1.2x² + (3.00)(4)xh + 2.00x²

The two variables are a potential road block but, as the shed is to have a volume of 1800 cubic feet, the height (h) can  be expressed in terms of the dimension of the square base (x), or 1800/x².  Now, combining like terms and substituting, we have

C = 3.2x² + (12x)(1800/x²) = 3.2x² + 21,600/x

This expression will have a minimum value when its derivative--here comes the calculus--is equal to zero.  So, differentiating, we get

dC/dx = 6.4x - 21,600/x²,

which is equal to zero when

6.4x = 21,600/x².

Doing the algebra, we get

6.4x³ = 21,600: x³ = 21,600/6.4:  x = 3375^1/3 = 15.

If the square base is 15 feet on a side, and the volume 1800 cubic feet, then the height has to be 1800/225, or 8.  The least expensive shed that can be built is 15 feet wide by 15 feet deep by 8 feet high.

Now calculus, like, say, "Shakespeare," has an undeserved mystical aura among subjects in the school curriculum.  We just got the right answer to what might have seemed to be quite a difficult problem.  All that was needed, besides a bit of algebra, was the power rule for derivatives--more like a straightforward technique than anything needing deep, mystic knowledge. 

If you are as big a dweeb as I you might enjoy this and this, among others.

Humor in a calculus text

Amanda is enrolled in a calculus course.  In order to brush up and be of occasional use to her, I've pulled from the shelf Martin Gardner's update of Silvanus Thompson's Calculus Made Easy.  I laughed out loud this morning when, reading along in chapter two, on limits, Gardner elucidates with a joke Zeno's paradox.  You know Zeno's paradox.  To complete a race, the runner must first get half way to the finish line, then must cover half of the remaining distance, and then half of that, and on and on. . . he can't ever make it to the finish, because it is guarded by an infinite number of "half ways."  The elucidation concerns a teacher who, to make the paradox vivid, told the prettiest girl in class to stand against one wall and one of the boys against the opposite wall.  The boy then advances across the room till he is half way to the girl.  After a pause, he advances again, covering half the remaining distance.  After another couple advances, the girl teases, "You'll never get here!" 

"True," answers the boy, "but I'll get close enough for all practical purposes."

DiMaggio and some baseball math

Publicity for this photograph, currently on display in a San Francisco museum, set me to noodling around with numbers connected with hitting safely in 56 consecutive games, a feat performed by DiMaggio during the 1941 baseball season.  Many say it is the most amazing and improbable record in sports.  So I wondered what the probability might actually be.

Of course you have to make assumptions.  They don't have to be perfect in order to give a good idea of what magnitude of improbability we are looking at.  Let us assume a batting average of .400 and four at-bats each game (over the 56-game streak, DiMaggio went 91-for-223, essentially four at-bats per game and an average of .409).  It's obviously very difficult to bat for an average higher than that.  In the same year that DiMaggio hit in 56 consecutive games, Ted Williams batted .406 for the season--and it hasn't been done since then.  I remember reading somewhere that over the time that DiMaggio was compiling his legendary streak--it began on May 15 and ended on July 17--Williams hit for a higher average.  That's not too surprising.  His average for the whole season was only three points lower than DiMaggio's during the streak.

Anyway, if you have a 0.4 chance of getting a hit in each at-bat, the likelihood of getting no hits in a game in which you have four official at-bats is given by 0.6⁴, which is 0.1296.  The likelihood of getting at least one hit is therefore 1 - 0.1296 = 0.8704.  If there is a 0.8704 chance of something happening in each independent trial, what is the chance of it happening 56 times in a row?  That is given by (0.8704)⁵⁶, which is quite a small number: 0.00042--or about 1-in-2500.

Now, that is not a rough answer to the question: what is the probability of a .400 hitter ever duplicating DiMaggio's streak?  DiMaggio hit in 56 straight games played from May 15 to July 16, but we'd be no less impressed if he had done it in 56 consecutive games played from late July to sometime in September.  The season is now 162 games long, so a player who plays in almost every game has right around 100 different groupings of 56 consecutive games (games 1 to 56, 2 to 57, 3 to 58, and so on).  Consequently it seems reasonable to expect that the chance of our .400 hitter getting at least one hit in 56 straight games played within the same season is something like 1 - (2499/2500)¹⁰⁰, which is about 0.039--a little less than a 4% chance.

I think the above calculations are sufficient to demystify DiMaggio's streak.  Yes, it was improbable, but not freakishly so.  A batter who could stay healthy and consistently hit .400 over the course of ten seasons would have a roughly 1-in-20 chance of equalling DiMaggio's record sometime over the course of the ten years.  As averages fall away from .400, however, probabilities become vanishingly small.  The highest lifetime batting average among active players is Ichiro Suzuki's .333.  His chance of getting a hit in a game is about 0.8.  His chance of getting a hit in 56 consecutive games somewhere over the course of 1500 games is about one-half of one per cent, or 1-in-200, which is just one-tenth the chance a .400 hitter has.  Ichiro's longest career consecutive game batting streak is 27.

Contemplating the percentages

I recently referred to a bit of innumeracy in a Hendrik Hertzberg politcal column and challenged readers to spot it.  Since my mailbox has not been overflowing with responses from my thousands of alert readers, it is time to reveal the solution myself. 

Hertzberg himself directed attention to the error on his blog, which is how I came to know about it.  He wrote in the March 9 issue of The New Yorker that a 140 million dollar federal outlay for monitoring volcanoes, mocked by Louisiana's Governor Bobby Jindal in his lame response to the President's televised speech to a joint session of Congress, amounted to just 0.0002 percent of the total stimulus package.  Actually, it is 0.0002 ("two-ten thousandths") of the total package, which is 0.02 percent. 

I've noticed people have a hard time with the concept of percent.  Maybe they just have a hard time  figuring with numbers but the examples of innumeracy that cross my line of sight frequently have to do with percent problems.  For example, a friend, by no means dumb, charged with overseeing the United Way drive at her office, calls to ask how to determine whether someone's pledge is at least ten percent more than what they gave last year (in which case they get a prize).  I launch into an explanation and she interrupts, "Just tell me what buttons to push."

Hertzberg's error is rather common.  People understand (I think) that "percent" signifies "what part of a hundred" and would have no trouble telling you that 8 is 20% of 40.  But when the "part of a hundred" is somewhat more than 100 or somewhat less than 1, it's harder to think straight about it.  In fact I think it is a little disreputable to speak of percent in certain cases.  It seems like, instead of telling the truth, you are going for a "Wow!" with "a 100% increase" (sounds like it increased a hundredfold when it only doubled) or "one-tenth of one percent" (which may sound more impressively small than "a thousandth").

Of course you can achieve even more wonderful effects if you are willing to dispense altogether with the truth of the matter.  The New Yorker recently ran an essay--abstract only, the whole thing is behind a subscriber wall--by John McPhee concerning the Olympian efforts of fact checkers who, before an article can appear in print, verify every statement of fact, no matter how trivial or tangential to the subject at hand.  So I guess first Hertzberg and then Homer nodded.  My own theory of the case is that those toiling as fact checkers for The New Yorker are probably "literary types" and numbers are not their strength. 

But it is indeed odd, as Hertzberg says, that the Governor of Louisiana should be the one to go on TV to complain, on behalf of the Republican party, that President Obama appears too interested in preparing for natural disasters.

Contemplating e

The number e, sometimes called Euler's number, may be contemplated by recreational mathematicians in the way that others gaze fondly at a painting without knowing how much time has passed.  It is, like pi, irrational, and its importance in calculus is roughly analogous to that of pi in geometry.  But you don't meet up with pi anywhere except in geometry, whereas e keeps popping up in the strangest places.

The most common definition of e is the limit, as x increases without bound, of (1 + 1/x)^x.  The first time I had to think about this, I thought the answer must be either 1 or infinity.  In favor of the former, the expression in the parenthesis goes to 1 when x is very large, and 1 to any power, no matter how huge, is 1.  On the other hand, the expression in the parenthesis is always greater than 1, and, as exponents are powerful, it seems possible the expression "blows up" when x is very large.  But it is as if my two arguments to myself fight to a draw, neither able to win in a gazillion rounds, with the result that the limit turns out to be "e," or 2.7182818284. . . .  If you are skeptical, get out a calculator and perform, successively, (1.1)¹⁰, (1.01)¹⁰⁰, (1.001)¹⁰⁰⁰, and so on.  The display on the calculator will get closer and closer to 2.7182818284. . . .

The physicist Richard Feynman created a stir with his pastime of cracking safes while working on the Manhattan Project.  His methods, however, were not magical.  For example, he got access to some top secret papers when, trying combinations that he thought would occur to a physicist, a safe swung open after he dialed 27-18-28.

Suppose you were asked the sum of the following infinite series:

1/0! + 1/1! + 1/2! + 1/3! + 1/4! + . . . .

Would you believe that this, too, is e?  That two expressions, one exponential in nature and the other a sum involving factorials, should both equal the same transcendental number--pretty weird, I think.  Possibly the diverse means of deriving e is connected to its all-around mathematical versatility.  Do you have a problem involving interest compounding continuously?  You need e.  In calculus, the derivative of the function

f(x) = e^x

is e^x, a vital fact that turns out to be exceedingly useful.  It follows that the slope of a line drawn tangent to the graph of y =e^x will be equal to e^x at every point.  Calculus teachers around the world have made use of these qualities to invent clever, marginally tricky problems.  Q: What is the slope of the line tangent to f(x) = e^x at (0,1)?  A: 1.  Q: What are the x- and y-intercepts of a line tangent to f(x) = e^x at (1,e)?  A: 0, 0.

My favorite uxexpected appearance of e is in certain probability problems.  If there is a 1/n chance of winning a casino game, and you play n times, what's the chance you never win?  Notice that if n is 2, the answer comes in at (1/2)²  = 1/4, or 0.25; if it's 3, it rises to (2/3)³ = 8/27 = 0.296; and it it's 4 it rises again to (3/4)⁴ = 0.316.  What if it's 100?  That would be given by (0.99)¹⁰⁰, which is very close to 0.366, which is getting very close to. . . (1/e).  There is a famous problem involving a butler checking hats at a party.  Each guest has a space reserved for his hat, but the butler is mute and doesn't know anyone's name, so can only guess.  If there are n guests and thus n spots for hats, what is the chance that the butler does not put a single hat in its proper place?  For a real big party, that is 1/e, too!

A little more poker math

Since a flush is just below a full house in the hierarchy of poker hands, let's see how the probabilities compare.  We determined, here, that the probability of being dealt a full house is 3744/2,598,690, or about 0.00144--slightly better than 1-in-700.  Also, the probability of drawing to a full house from two pair is 4/47, about 8.5%.

You can think of the probability of being dealt a flush in the following manner. You're dealt five cards.  The first one can be anything.  The second one has to be the same suit as the first.  There are twelve of that suit left, and 51 unknown cards, for a probability of 12/51.  And, if that chance is realized, you're down to 11/50 for the third card, then 10/49 on the fourth, and 9/48 on the last card.  The probability of a flush is the product of the probabilities of these separate events, which comes out to 11,880/5,997,600, or very close to 0.00198, which is very close to 1-in-500.  (This represents the probability of all flushes, including "straight flushes," such as the 4, 5, 6, 7 and 8 of spades.)  Since a full house beats a flush (except for the very rare straight flush), you'd expect the former to be less likely, and indeed it is: out of every 3500 hands, you'd expect about seven flushes on the deal, and about five full houses.

If you have any doubts about  my method of calculating the probability of a flush, you can check the result against the number of possible combinations that make a flush divided by all possible combinations.  The number of different ways of having a flush is given by "13 choose 5" (the number of ways of choosing five cards from thirteen) times "4 choose 1" (since you can choose five from thirteen in any one of the four suits).  The number of possible five-card hands is given by "52 choose 5."  Grinding the numbers, you get 5148/2,598,960, which is the same as 11,880/5,997,600.

Remember how, starting from two pair, you have about an 8.5% chance of drawing to a full house?  It raises the question: what if you are dealt four cards of the same suit? what is the chance of drawing to a flush?  Well, if you have, say, four hearts and one of something else, then there are 47 unknown cards, of which nine are hearts, and 9/47 is just over 0.19--more than twice the chance of drawing to a full house from two pair.  Of course, if you have two pair, and the 8.5% chance of improving to a full house fails, you still have two pair, which will beat four hearts most of the time.

My wife's geeky husband

Her geeky husband looking over her shoulder, my wife Amanda prepares for her final exam in Discrete Mathematics.  More than one of the exercises my eye falls on seem to rebut the familiar complaint, "This is so impractical and will never help me in real life!"  For example:  What is the probability of being dealt a full house in five-card poker?

Well, that would be the ratio of the number of different ways there are to have a full house to the number of unique hands one might be dealt in five-card poker.  Since there are 52 cards in the deck, the latter is the number of different ways of selecting five from 52.  "Fifty-two choose five" is given by 52 factorial divided by the product of 47 factorial and 5 factorial, denoted

52!/(47!)(5!) = 52 x 51 x 50 x 49 x 48/5 x 4 x 3 x 2 x 1 = 2,598,960.

Now, how many ways are there of having a full house?  In one denomination, you need three of the four cards--three of the four aces, say.  There are four ways to choose three from four.  You need two of the four cards in a second denomination, and there are six ways of choosing two from four (4!/2! x 2! equals 6).  So the product of four and six--24--represents the number of ways there are of having a particular full house, like "tens over queens."  Since there are thirteen denominations (ace through king), there are thirteen choices for one denomination and only twelve for the other--obviously, if you have three tens, ten is not a possibility for the pair.  Putting this all together, we have 24 x 13 x 12 ways of having a full house: 3744.

And 3744 divided by 2,598,960 is about 0.00144, or 1-in-694.

A full house is two pair plus a good draw, so a poker player should know the probabilities along this line.  Turns out you are 33 times more likely to be dealt two pair than a full house.  If you are dealt two pair, the probability of drawing to a full house is given by 4 (the number of remaining cards that will improve your hand)/47 (the number of unknown cards).  Four divided by 47 is 0.0851, or between 8 and 9 percent.

We geeks have a hard time understanding why more people aren't interested in this kind of stuff.  Surely they don't think poker is remote from "real life"?

The Monty Hall problem

My wife, Amanda, came home from her course in "Discrete Mathematics" the other night eager to talk about a problem the professor had discussed in class.  Turns out I had already posted on the very same problem, here.  But in honor of her interest let's elaborate.

The problem concerns a television game show, popular in my youth, called "Let's Make a Deal." In one popular game, the host, Monty Hall, would show a contestant three doors.  You win the prize behind the door you choose.  There's always a goat behind two of the doors, but behind the third there'd be a sports car, something like that.  After the contestant chose a door, Monty would direct that one of the unchosen doors be opened. Invariably a goat was revealed--Monty knew where the car was, and this was a way of building up the suspense.  Here's the interesting part: Monty would also allow the contestant to switch doors.  In other words, once it was known that a certain unchosen door held a goat, you could accept an offer to trade your first choice for the remaining unopened door.

Should you switch?  Does it matter?

Despite what almost everyone thinks, it matters a lot.  You double your chances of winning the sports car by accepting Monty's offer to switch.  Let's see if I can persuade skeptics.  Suppose the car is behind door 1 and you choose a door at random.  If you choose door 1, Monty opens, say, door 2, revealing a goat.  You then accept the offer to switch to door 3 and win--a goat.  If however you choose door 2, Monty opens door 3, revealing a goat.  You accept the offer to switch to door 1 and win--a car.  In the third scenario, you choose door 3, Monty shows the goat behind door 2, you switch to door 1 and win--the car, again.

At the start of the game, your chance of winning is 1/3. By switching doors in the middle of the game, the probability of winning rises to 2/3--you lose if you began with the winning door (1/3 probability), but win if you began with a goat (2/3 probability).

If you don't believe me, you can model the game with playing cards and tally the result.  Take, say, a joker (sports car) and the black twos (goats).  Shuffle, then give "the house" two cards and yourself one (the first choice).  Have "the house" discard a two--there will be at least one--and then trade your card for the house card.  Repeat many times.  See if you can't satisfy yourself that switching cards, or doors, is the best strategy. 

Why do most people have a really hard time accepting the solution to this problem?  I think in most cases it's because they feel intuitively that there is no reason to suppose the car is more likely to be behind one door than another--a supposition that is true at the start of the game only.  Since Monty has determined to reveal only goats before moving to the last stage of the game, there is a nonrandom reason that the two remaining doors are this one (the one the contestant chose) and that one (the one Monty chose).

If this kind of stuff interests you, check out the Wikipedia article on "the Monty Hall problem."  Turns out Amanda's math professor wrote it (at least in part)!  Here's his website.  

The compensations of algebra

Before he retired, my dad was a physicist.  When as a teenager I approached him with questions about my algebra homework, I got more than I bargained for.  I remember him casting an appraising eye on my textbook, the better I assumed to address my questions in an appropriate fashion, before remarking, "It is obvious that the author of this textbook knows nothing of mathematics.  I really must speak with your teacher."

Not what I had in mind. 

His subsequent interest in my mathematical education was the source of some tense moments.  He cared more for math than for my self esteem and sometimes had a hard time understanding how I could not understand.  His first language is Norwegian, and he worked hard to eliminate the inflections of that language from his spoken English, but when agitated, as he was by my trouble simplifying algebraic expressions by factoring, his voice would rise, and he would suddenly sound scary and foreign.  The sounds of our algebra sessions carried to other parts of the house and  I remember my mom peering into the room we were working in while nervously drying her hands on a dish towell.  I wished she would interrupt but she didn't.  How could I concentrate with her just standing there and him talking loud at me, his voice rising sharply in that weird Norwegian way at the end of his emphatic speech acts?

But I received a grounding in algebra.  It's its own reward, I think, but once in awhile you get to amaze the uninitiated with your mystical powers.  Today, in a meeting, someone wanted to know the product of 54 and 46.  While someone else began reaching for a calculator, I announced, "Two thousand four hundred and eighty four."  When a few seconds later the guy with the calculator verified my answer, jaws sagged, and I think my colleagues now regard me as a genius, or possibly as an "idiot savant" along the line of the Dustin Hoffman character in Rain Man.  But it is really just the simplest algebra.  It just happens that 54 is four more, and 46 four less, than fifty.  So you have an instance of (a + b)(a - b) where a = 50 and b = 4.  Since (a + b)(a - b) = a² - b², it follows that 54 x 46 = 50² - 4², which is not too hard to do in your head.  If the problem had been the product of 54 and 44, the fellow with the calculator would have gotten there first.

Well, maybe not: 54 x 44 = (44 x 50) + (44 x 4) = 1/2 (4400) + 176 = 2376.  It was from my dad's way of teaching algebra that I learned to think like this.  Before, algebra had been to me a kind of incomprehensible code, the whole object of which was to memorize certain abstruse manipulations that allowed you to "solve" meaningless "problems."  The realization that it is arithmetic generalized--that made all the difference.

Recreational mathematics

Being a big geek who works at math problems for recreation, I was excited when my wife enrolled in a math class at a local college.  My plan was to read up the material and work the same exercises as she was assigned.  It would be fun for both of us and I'd learn something, too.  So far, though, it hasn't been working.  I have a job, two kids, something resembling an obsession with American politics, an unambiguous obsession with baseball, especially the fortunes of the Minnesota Twins, and, lately, in my spare time, I've been progressing, at the rate of about ten pages a day, through Benny Morris's Righteous Victims: A History of the Zionist-Arab Conflict, 1881-2001.  No time for my wife's math.

Still, I'm dimly aware that the first lessons of the class involve the Gaussian elimination and the solution for systems of linear equations.  The name comes from the inventor, or discoverer, of the method, Carl Friedrich Gauss, one of the great mathematicians of all time and a child prodigy.  Once, a math instructor related to a class in which I was enrolled the following anecdote.  In what we call elementary school, the very young Gauss, bored during an arithmetic lesson, was raising a disruptive ruckus.  Finally the exasperated teacher said, "Carl, I don't want to hear another peep until you've summed the first hundred integers and given me the answer."  Gauss, who had been loudly tapping his pencil on his desk, kept it up.  "Carl, I mean it!"  The boy answered loudly above the sound of his thumping pencil, "Five thousand and fifty!"

The teacher of course thought the boy would have to do 1 + 2, then add 3 to 3, then 4 to 6, then 5 to 10, which gets tedious fast.  Plus you'll probably get confused and make a mistake before you make it into the thirties.  But the young Gauss must have realized that 1 + 2 + 3 + 4 + ... + 99 + 100 may more simply be thought of as (1 + 100) + (2 + 99) + (3 + 98), etc.--that is, it's the same as a hundred and one fifty times.  A boy clever enough for such an insight is also going to be sufficiently clever to do the multiplication in his head: 101 x 50 = (100 x 50) + 50 = 5050.

Probability blindness

In the field of mind science, the inability of people to think straight about odds and probability is such a well known phenomenon that the term "probability blindness" has been coined to describe it.  Is it a simple matter of innumeracy?  In "How the Mind Works" Steven Pinker reports the results of some experimental problems in probability.  My favorite concerns a woman, "Linda," 31, bright and outspoken, a college philosophy major interested in social justice and the anti-nuke cause.  Given this description, test subjects are asked to rate the likelihood of the following statements: (1) Linda is a bankteller, and (2) Linda is a bankteller who is active in the feminist movement.  One is tempted to say that (1) is "obviously" more likely, but in tests there is considerable support for (2).  It seems that the mind forms an idea of Linda's character and scans the propositions, searching for "hits."  It is the binary world of computer science.  The first statement is a "miss," the second at least a partial "hit," and thus the tendency to rate it more likely than (1).

So some people are bad at math, others just slow all around: what is so interesting about that?  Try another problem described by Pinter, one almost identical to the coin problem at Mathproblems.info.  A certain invisible medical condition afflicts one person in a thousand.  It can be detected before onset by a test that yields 5% false positives.  If a randomly selected person tests positive, what is the probability she has the condition?

The answer is: just about exactly 2%.  You may satisfy yourself of this, I hope, by considering (as in the coin problem) what results you'd expect with more than a single trial--in this case, say, 100,000.  Since the condition afflicts one person in a thousand, you'd expect 100 to test positive on account of having the condition.  Of the remaining 99,900, five per cent, or 4995, would test positive.  So the ratio of afflicted to all positive results is given by 100/5095, or 0.0196.

According to Pinker, the average wrong answer submitted by Ivy-league test subjects is on the order of 0.5, which is around 25 times the actual risk.  I know that I myself, barred from doing the number grinding, would have guessed a number much higher than 0.02.  Something about the way the problem presents itself to the mind seduces us toward the conclusion that the subject is much more likely to be afflicted than she actually is.  And the error, it seems, is similar in all the problems I have discussed in the last two posts--a failure to recognize, or to weigh, different bits of information.  In the coin problem, people tend to consider only that a "head" result does not eliminate the possibility that the conventional coin has been selected; in the "Linda" problem, people are so sure she is a feminist that any proposition including "feminist" is deemed more likely than one containing only dissonant information; and in the problem relating to the test for an invisible medical condition, people tend to discount the low incidence of the condition in favor of elevating the significance of a quite unreliable test.

Pinker thinks that these tendencies are evidence that our minds, the products of evolution by natural selection, are essentially for information processing of the binary variety.  That lower function still asserts itself, tempting us, at least  on first consideration, toward false conclusions on problems that require subtler reasoning, the deployment of a higher order of processing.  I think it is exciting, in a humbling sort of way, to consider that our lizard minds, unelected monarchs in certain conditions, such as the presence of physical danger, may also explain our fondness for casino gambling and state lotteries.

Geeky probability

Okay, I majored in English, so probably do not qualify for the the geekiness competitions that sometimes break out over at the science blogs.  But my wife will tell you that I work math problems for recreation, more or less the way others might do sodoku or crosswords.  Lately I've been using for a source Mathproblems.info.  Problem 16 posits a box with two coins, one "fair" and the other two-headed, and asks what is the probability that, if a coin is drawn that displays a head, the reverse side of that coin is a head as well.

I can easily satisfy myself that the answer is 2/3 by observing that a single trial can have four equally likely results: one side of the two-headed coin, the other side of the two-headed coin, a head on the "fair" coin, and a tail on the "fair" coin.  So you get a head displayed three times, and, of those three, you are holding the two-headed coin twice.  Hence, 2/3. 

The problem is probably less interesting than the host's hilarious account of the mail generated by this problem and his claim that the answer is 2/3.  It seems like very many people, observing that a result of heads does not preclude the possibility that the "fair" coin has been drawn, insist that the answer must be 1/2.  Of course, 2/3 acknowledges that the "fair" coin may have been drawn, and the host patiently illustrates with an example concerning mislabeled soup cans.  Suppose there are 2 million cans of tomato soup, half correctly labeled but the other half labeled "chicken noodle," and that one store receives 999,999 correctly labeled cans and one bad one, while another store receives the other 1 million, all but one incorrectly labeled.  If you went to one of the stores, opened a can, and found it to be correctly labeled, would you ignore that piece of information when trying to decide which store you most likely were in?  So why ignore an analogous piece of information when trying to decide what coin you are most likely holding?

Nevertheless, one of the host's correspondents, a high-school biology teacher, could not persuade his principal, still less many of his students, that the correct answer to the coin problem is 2/3.  He recruited the host to remonstrate with the principal, which the host did, finally persuading the principal to relent, though in grudging fashion festooned with weasel-words.  Check it out.

In my next post I plan on describing what Steven Pinker, author of How the Mind Works, says about episodes such as this one.  It turns out that the mind scientist, though up-to-speed on the math, has considerable sympathy for the principal.

The 12 ball problem

Over at Time to Blog Abhay Parekh poses a problem:

The Oddball Problem: You are given n identical looking balls. n-1 of them weigh the same, but one of them is either heavier or lighter than the others (you don't know which). Given a two pan weighing machine what is the minimum number of weighings you need to do to be sure that you have identified that odd ball? You can only use the weighing pan as follows: put some balls in the left pan, some in the right pan and observe one of three possible outcomes: either the left pan is heavy or the right pan is heavy, or they are even.

Since you don't know if the odd ball is heavier or lighter things get a bit tricky. This problem is often posed with 12 balls. Here's a problem worth working on:

Show that for 12 balls you can always identify the oddball in 3 weighings!

It happens that, a few days ago, I was confronted with the 12-ball variety of the problem by the local Big Brothers Big Sisters organization, which every week, before I meet my "little" at a nearby school, sends out to the "bigs" a "suggested activity."  As I worked on the problem during a boring work meeting, and then some more over my lunch break, and yet more at a coffeeshop in the afternoon, I began to question how much Big Brothers Big Sisters knows about the mental development of sixth graders and what they are apt to enjoy.  I was hooked by the difficulty, however, and eventually worked out what I take to be a solution.  Since my "little brother" exhibited no interest, I shall set it out here, before Abhay posts his version of a solution.  If nothing else it is an interesting exercise in "technical" writing.

For ease of exposition, let us think of the balls as being numbered 1 through 12.  Begin by weighing 1, 2, 3 and 4 against 5, 6, 7 and 8.  Suppose they balance.  You now know that the "oddball" is 9, 10, 11 or 12.  Use the second trial to weigh, say, 1, 2 and 3 against 9, 10 and 11.  If the result is another balance, you know the "oddball" is 12, and it is a simple matter to use the third trial to determine whether it is heavier or lighter than the rest.  But suppose 1, 2 and 3 are, for example, lighter than 9, 10 and 11.  That could only mean that 9, 10 or 11 is the "oddball" and heavier than the rest.  On the last trial, weigh 9 against 10.  If they balance, 11 is odd (and heavy).  If they don't balance, the heavier one is the "oddball."

Now we need to consider the case where, on the first trial, 1, 2, 3 and 4 do not balance with 5, 6, 7 and 8.  If, for example, 1, 2, 3 and 4 are lighter than 5, 6, 7 and 8, we know that either 1, 2, 3 or 4 is the "oddball" (and light) or 5, 6, 7 or 8 is the "oddball" (and heavy).  On the second trial, weigh 1, 2, 3 and 5 against 4, 9, 10 and 11.  Suppose that 1, 2, 3 and 5 are lighter. Since 5 cannot be light and 4, 9, 10 and 11 cannot be heavy (refer to result of the first trial), this means that 1, 2 or 3 is the oddball (and light).  Weighing 1 against 2 on the third trial will reveal which of those three is the (light) "oddball."

We have yet to consider the case where 1, 2, 3 and 5 balance 4, 9, 10 and 11 and the case where 1, 2, 3 and 5 are heavier than 4, 9, 10 and 11.  In the former case, we learn that 6, 7 or 8 must be the "oddball" (and heavy).  Use the third trial to weigh 6 against 7, which will suffice to reveal which of those three is heavier than the other twelve.  In the latter case, we learn that either 5 is odd (and heavy) or 4 is odd (and light).  The solution is clear if on the third trial either 4 or 5 is weighed against any one of the ten that we know are not the "oddball": if, for example, 4 balances with 1, then 5 is odd (and heavy), whereas if 4 is lighter than 1 then 4 is the (light) "oddball."