Problem 4

The Chevalier de Méré Problem II: The Problem of Points (1654)

Problem Two players A and B play a fair game such that the player who wins a total of 6 rounds first wins a prize. Suppose the game unexpectedly stops when A has won a total of 5 rounds and B has won a total of 3 rounds. How should the prize be divided between A and B?

Solution. The division of the prize is determined by the relative probabilities of A and B winning the prize, had they continued the game. Player A is one round short, and player B three rounds short, of winning the prize. The maximum number of hypothetical remaining rounds is (1 + 3) − 1 = 3, each of which could be equally won by A or B. The sample space for the game is Ω = {A_{1}, B_{1}A_{2}, B_{1}B_{2}A_{3}, B_{1}B_{2}B_{3}}. Here B_{1}A_{2}, for example, denotes the event that B would win the first remaining round and A would win the second (and then the game would have to stop since A is only one round short). However, the four sample points in Ω are not equally likely. Event A_{1} occurs if any one of the following four equally likely events occurs: A_{1}A_{2}A_{3}, A_{1}A_{2}B_{3}, A_{1}B_{2}A_{3}, and A_{1}B_{2}B_{3}. Event B_{1}A_{2} occurs if any one of the following two equally likely events occurs: B_{1}A_{2}A_{3} and B_{1}A_{2}B_{3}. In terms of equally likely sample points, the sample space is thus

There are in all eight equally likely outcomes, only one of which (B_{1}B_{2}B_{3}) results in B hypothetically winning the game. Player ...