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 2^{52}. 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 2^{2}, or 4--up, up; up, down; down, up; and down, down.

So the problem becomes, of the 2^{52} 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 2^{52} (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.

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