Classic Problems of Probability

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 = {A₁, B₁A₂, B₁B₂A₃, B₁B₂B₃}. Here B₁A₂, 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 A₁ occurs if any one of the following four equally likely events occurs: A₁A₂A₃, A₁A₂B₃, A₁B₂A₃, and A₁B₂B₃. Event B₁A₂ occurs if any one of the following two equally likely events occurs: B₁A₂A₃ and B₁A₂B₃. In terms of equally likely sample points, the sample space is thus

There are in all eight equally likely outcomes, only one of which (B₁B₂B₃) results in B hypothetically winning the game. Player ...