Drunkard's Walk - Leonard Mlodinow [38]
In order for us to use Pascal’s triangle, let’s say for now that your guest list consists of just 10 guests. Then the relevant row is the one at the bottom of the triangle I provided, labeled 10. The numbers in that row represent the number of distinct tables of 0, 1, 2, and so on, that can be formed from a collection of 10 people. You may recognize these numbers from the sixth-grade quiz example—the number of ways in which a student can get a given number of problems wrong on a 10-question true-or-false test is the same as the number of ways in which you can choose guests from a group of 10. That is one of the reasons for the power of Pascal’s triangle: the same mathematics can be applied to many different situations. For the Yankees-Braves World Series example, in which we tediously counted all the possibilities for the remaining 5 games, we can now read the number of ways in which the Yankees can win 0, 1, 2, 3, 4, or 5 games directly from row 5 of the triangle:
1 5 10 10 5 1
We can see at a glance that the Yankees’ chance of winning 2 games (10 ways) was twice as high as their chance of winning 1 game (5 ways).
Once you learn the method, applications of Pascal’s triangle crop up everywhere. A friend of mine once worked for a start-up computer-games company. She would often relate how, although the marketing director conceded that small focus groups were suited for “qualitative conclusions only,” she nevertheless sometimes reported an “overwhelming” 4-to-2 or 5-to-1 agreement among the members of the group as if it were meaningful. But suppose you hold a focus group in which 6 people will examine and comment on a new product you are developing. Suppose that in actuality the product appeals to half the population. How accurately will this preference be reflected in your focus group? Now the relevant line of the triangle is the one labeled 6, representing the number of possible subgroups of 0, 1, 2, 3, 4, 5, or 6 whose members might like (or dislike) your product:
1 6 15 20 15 6 1
From these numbers we see that there are 20 ways in which the group members could split 50/50, accurately reflecting the views of the populace at large. But there are also 1 + 6 + 15 + 15 + 6 + 1 = 44 ways in which you might find an unrepresentative consensus, either for or against. So if you are not careful, the chances of being misled are 44 out of 64, or about two-thirds. This example does not prove that if agreement is achieved, it is random. But neither should you assume that it is significant.
Pascal and Fermat