ONE OBSERVATION BEHIND TWO PUZZLES

In two famous and popular puzzles a participant is required to compare two numbers of which she is shown only one. We show that there is one simple principle behind these puzzles. In particular this principle sheds new light on the paradoxical nature of the first puzzle. According to this principle the ranking of several random variables must depend on at least one of them, except for the trivial case where the ranking is constant. Thus, in the non-trivial case, there must be at least one variable the observation of which conveys information about the ranking. A variant of the first puzzle goes back to the mathematician Littlewood (1986) who attributed it to the physicist Schrodinger. See Nalebuff (1989), Brams and Kilgour (1995), and Blackwell (1951) for more detail on the historical background and for further elaboration on this puzzle. Below is the common version of the puzzle as first appeared in Kraitchik (1953). To switch or not to switch? There are two envelopes with money in them. The sum of money in one of the envelopes is twice as large as the other sum. Each of the envelopes is equally likely to hold the larger sum. You are assigned at random one of the envelopes and may take the money inside. However, before you open your envelope you are offered the possibility of switching the envelopes and taking the money inside the other one. It seems obvious that there is no point in switching: the situation is completely symmetric with respect to the two envelopes. The argument for switching is also simple. Suppose you open the envelope and find a sum x. Then, in the other envelope the sum is either 2x or x/2 with equal probabilities. Thus, the expected sum is (1/2)2x+(1/2)x/2 = 1.25x. This is true for any x, and therefore you should switch even before opening the envelope. Should you or should you not switch? Solutions to this paradox are discussed in numerous articles. The simplest and most common solution is this. In order to carry out the computation of the expected value, there must be some probability distribution over the two sums. But no probability distribution can have the property that for any envelope, and any given sum x in it, the sum in the other is equally likely to be 2x and (1/2)x.