Frank Morgan

July 15, 1999

The mass M of a universe is less than the sum m₁ + m₂ + . . . of its constituent masses, due primarily to negative potential energy when masses get near each other. The question is how much less. Classically, the total mass could actually be negative if the constituent masses get close enough together. In general relativity, masses cannot get small enough to get that close together: they become black holes and stay a certain size. In 1979 R. Schoen and S.-T. Yau proved that the total mass of a universe must be positive. Last year G. Huisken and T. Ilmanen proved that the total mass is at least as large as the largest constituent mass: M  m₁. Finally this year H. Bray of MIT proved that M  sqrt (m₁² + m₂² + …). These results are forms of the so-called Penrose Conjecture.

Note. Actually, mass is hard even to define in general relativity and requires much more care than we have been able to take here.

OLD CHALLENGE (Joe Shipman). Abel writes down two distinct positive integers from 1 to 4. Beta sees one of them at random and tries to guess whether it is the larger of the two. What are Beta’s chances? (What if Abel can choose from more numbers?)

ANSWER. Elliot Kearsley and Jean-Pierre Carmichael explain that if Abel picks the integers from 1 to 4 at random, Beta can win 5/6 of the time by guessing that a 3 or 4 is the larger (and that a 1 or 2 is not). He is always right about the 4 or 1, and right 2/3 of the time about the 3 or 2, for a 5/6 winning average. Similarly with integers from 1 to N (say N even), by guessing that any number larger than N/2 is the larger, Beta can win with probability (3N-2)/(4N-4), which checks out to be 5/6 when N = 4 and which approaches 3/4 as N increases.

Joseph DeVincentis further explains that Abel can do better by picking only consecutive integers at random. For example, from integers from 1 to 4, he could pick randomly from {1,2}, {2,3}, (3,4}. Now Beta’s strategy wins just 2/3 of the time. As N increases, Beta’s chances approach 1/2, no better than guessing at random.

In the final word, Arthur Pasternak points out that against this particular strategy of Beta’s, Abel could actually do better. For example, Abel could always pick {1,2} and Beta would be correct only the half of the time she saw the 1. To prevent this Beta must introduce some randomness. For general N, her best strategy is to state that the number K she sees is the higher number with probability (K-1)/(N-1). Now Abel can not improve on his strategy, and Beta will still win with probability N/2(N-1), which starts at 2/3 when N = 4 and approaches 1/2 as N gets large.

Pasternak proposes the following

NEW CHALLENGE. Abel and Beta are each given a box containing cash. It is known that one box contains 10 times the cash that is in the other box. They are give the opportunity to swap boxes. Abel reasons as follows:

Assume I have 10X in my box. Then the other box has either X or 100X with equal probability. If it has 100X, I will gain 90X by swapping; if it has X, I will lose 9X. Therefore I expect to gain 40.5X by swapping boxes.

This seems logical but, of course, Beta reasons the same way and concludes that she is better off swapping as well. Can they both be right?

Send answers, comments, and new questions by email to [email protected], 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.

Copyright 1999, Frank Morgan.