Math Chat: It's All in the Name Old challenge
What is the probability that two people will randomly choose the same letter of the alphabet? What is the probability that the names of two random American presidents begin with the same letter? Estimate the probability that at least two of the names of 10 random Americans will start with the same letter.
Answer (Erik Randolph)
The likelihood that a second person will choose the same letter as the first is 1 in 26 or about 3.8 percent. As for the 42 presidents, since three have last names beginning with A (Adams, Adams, and Arthur), the probability of two A's is (3/42)(2/41). Adding up such possibilities for all 26 letters yields a total of about 5.3 percent (or 5.1 percent if you count Cleveland just once). This is higher than 3.8 percent because some letters are more common than others, and that makes duplication easier.
As for the 10 random Americans, first assume that all letters are equally likely and compute the chance they are all different. The first letter can be any of the 26, the second any one of the remaining 25, the third any one of the remaining 24, and so on. The probability they are all different is (26/26)(25/26)(24/26)(23/26) (22/26)(21/26)(20/26)(19/26) (18/26)(17/26), or about 13.7 percent. Therefore the probability that at least two are the same is about 86.3 percent. Because some letters are more likely than others, the actual figure would be even higher.
Eric Brahinsky used the last names of "Entertainment Personalities" in the 1997 World Almanac, of which a whopping 10.6 percent begin with an S. In 1,609 random computer trials he found that about 96 percent had duplicated first letters, much higher than the earlier 86.3 percent when all letters were equally likely.
In such an experiment, statistics suggests you can be confident that the error is less than 1/1,609 or about 2.5 percent. I do not know a nice way to compute by hand the exact probability of duplication given the relative abundance of the initial letters.
New challenge (Howard Sheldon)
5! = 5 x 4 x 3 x 2 x 1 = 120, which ends in a 0. Determine the number of zeros at the end of 25! (What about 510!)
* To be eligible for book awards, send answers, comments, and new questions to:
Princeton, NJ 08540
or e-mail: Frank.Morgan@williams.edu
(Professor Morgan is spending the year at Princeton University as its 250th-Anniversary Distinguished Teaching Professor.)