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That is not a surprising statement. The surprise is just how large that effect is—and how difficult it can be to calculate. For example, suppose you give a 10-question true-or-false quiz to your class of 25 sixth-graders. Let’s do an accounting of the results a particular student might achieve: she could answer all questions correctly; she could miss 1 question—that can happen in 10 ways because there are 10 questions she could miss; she could miss a pair of questions—that can happen in 45 ways because there are 45 distinct pairs of questions; and so on. As a result, on average in a collection of students who are randomly guessing, for every student scoring 100 percent, you’ll find about 10 scoring 90 percent and 45 scoring 80 percent. The chances of getting a grade near 50 percent are of course higher still, but in a class of 25 the probability that at least one student will get a B (80 percent) or better if all the students are guessing is about 75 percent. So if you are a veteran teacher, it is likely that among all the students over the years who have shown up unprepared and more or less guessed at your quizzes, some were rewarded with an A or a B.

A few years ago Canadian lottery officials learned the importance of careful counting the hard way when they decided to give back some unclaimed prize money that had accumulated.3 They purchased 500 automobiles as bonus prizes and programmed a computer to determine the winners by randomly selecting 500 numbers from their list of 2.4 million subscriber numbers. The officials published the unsorted list of 500 winning numbers, promising an automobile for each number listed. To their embarrassment, one individual claimed (rightly) that he had won two cars. The officials were flabbergasted—with over 2 million numbers to choose from, how could the computer have randomly chosen the same number twice? Was there a fault in their program?

The counting problem the lottery officials ran into is equivalent to a problem called the birthday problem: how many people must a group contain in order for there to be a better than even chance that two members of the group will share the same birthday (assuming all birth dates are equally probable)? Most people think the answer is half the number of days in a year, or about 183. But that is the correct answer to a different question: how many people do you need to have at a party for there to be a better than even chance that one of them will share your birthday? If there is no restriction on which two people will share a birthday, the fact that there are many possible pairs of individuals who might have shared birthdays changes the answer drastically. In fact, the answer is astonishingly low: just 23. When pulling from a pool of 2.4 million, as in the case of the Canadian lottery, it takes many more than 500 numbers to have an even chance of a repeat. But still that possibility should not have been ignored. The chances of a match come out, in fact, to about 5 percent. Not huge, but it could have been accounted for by having the computer cross each number off the list as it was chosen. For the record, the Canadian lottery requested the lucky fellow to forgo the second car, but he refused.

Another lottery mystery that raised many eyebrows occurred in Germany on June 21, 1995.4 The freak event happened in a lottery called Lotto 6/49, which means that the winning six numbers are drawn from the numbers 1 to 49. On the day in question the winning numbers were 15-25-27-30-42-48. The very same sequence had been drawn previously, on December 20, 1986. It was the first time in 3,016 drawings that a winning sequence had been repeated. What were the chances of that? Not as bad as you’d think. When you do the math, the chance of a repeat at some point over the years comes out to around 28 percent.

Since in a random process the number of ways in which an outcome can occur is a key to determining how probable it is, the key question is, how do you calculate the number of ways in which something can occur? Galileo seems to have missed the significance of