November 16, 2000

Presidential election with 48% of the US popular vote does not come close to the extreme possibility, even with just two candidates, of winning with less than about 22% of the popular vote. See Math Chat of November 8, 1996.

**Old Challenge** (reappeared on recent "Green Chicken" math competition between Middlebury, Simon's Rock, and Williams College).

1. What is the expected (average) length of the shorter piece of a meter stick with a random cut?

2. What is the expected length of the shortest of 3 pieces from 2 random cuts?

3. What is the expected length of the shortest of 4 pieces from 3 random cuts?

4. What is the expected length of the shortest of n pieces from n-1 random cuts?

**Answers** (Joe DeVincentis). We may assume that the first piece is the shortest.

1. The cut can be anywhere from 0 to 1/2. The expected length is 1/4.

2. The cuts at must satisfy . Averaging x over this region of the unit square yields 1/9.

3. The cuts at must satisfy . Averaging x over this region of the unit cube yields 1/16.

4. 1/n^{2}. The n-1 conditions on the unit (n-1)-cube, all meeting at (1/n,..., 1/n), yield a pyramid of base b and height 1/n. Hence the expected value of x is

Substituting u = 1 - nx yields

Math Chat notes that similarly, 1/n^{2} is the expected fraction of a year between the closest of n birthdays. For 20 people, it is less than one day. A small variation on the above proof shows that with probability 1/2, it is less than (ln 2)/n(n-1). This is less than 12 hours (so they probably have the same birthday) when (ln 2)/ n^{2} 1/730, n^{2} 510, n 23.

**Incredible bridge hands** (Math Chat of September 21) apparently resulted from dealing out unshuffled new decks, typically boxed as Ace-2-3-...-King of Spades, similarly A-2-3-...-K of Diamonds, then K-Q-...-2-A of Clubs, and similarly K-Q-...-2-A of Hearts, according to analysis by Bart Bramley and Nick Straguzzi in the November 2000 ACBL Bridge Bulletin (p 19).

**New Challenge** (Joe Shipman). On the West Palm Beach ballot of the US Presidential election, it apparently turned out to be a big advantage for the Republican candidate to be listed first. Can you come up with a fair procedure for allocating ballot order from year to year between the two major parties, if you assume that being first is an advantage?

Send answers, comments, and new questions by email to Frank.Morgan@williams.edu, to be eligible for* Flatland *and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan's homepage is at www.williams.edu/Mathematics/fmorgan.

THE MATH CHAT BOOK, including a $1000 Math Chat Book QUEST, questions and answers, and a list of past challenge winners, is now available from the MAA (800-331-1622).

Copyright 2000, Frank Morgan.