In 1654, the Chevalier de Mere, a French gambler, asked a question that earned the attention of mathematicians Pascal and Fermat:

Suppose two people, A and B, agree to play a series of fair games until one person has won six games. They each have wagered the same amount of money, the intention being that the winner will be awarded the entire pot. But suppose, for whatever reason, the series is prematurely terminated, at which point A has won five games and B three. How should the stakes be divided? (Stone Analytics, Inc, 2003)

This problem is generally regarded as the birthplace of mathematical probability. Though the study of probability today is deeply intertwined with engineering, marketing, insurance, and countless other professions, we have gambling to thank for its inception.

Solution:

The desired solution will reward both players in proportion to their respective chances of winning the tournament, had they continued playing. It is not acceptable to give A all of the money, since B was not fully defeated. On the other hand, the money should not be divided evenly, because A does indeed have the upper hand and is therefore more likely to win. The probabilistic solution will involve the number of possible outcomes and the proportion of those which are won by A.

Knowing that the contest will end when one player or the other reaches six victories, it is easily found that three more games would be sufficient to determine a winner. However, three games are not necessary; if A wins one more game, he is victorious.

This leads us to the list of possible conclusions for the tournament, had A and B continued playing:

This list shows us that the tournament can conclude in four different ways. Since A wins in three of the four possibilities, A should receive ¾ of the pot, and B should receive ¼, to reflect their chances of winning, had play continued.

However, this is problematic, as the cases listed above are not equally likely. For example, A and B are equally likely to win the first game of the conclusion, but the table shows B winning the first game in three of the four cases. Suppose instead that A and B play exactly three games, though the tournament may or may not be concluded by the time the third game is played. Now we get this table:

All eight outcomes are equally likely. The first four outcomes show A reaching six wins after the first game. The fifth and sixth cases show A winning the pot after the second game, and the seventh shows A winning after the third game. Only in case eight does B win. This means that P(A)=7/8 and P(B)=1/8. Therefore, this (correct) approach shows that A should receive 7/8 of the pot, and B the remaining 1/8.

Reference: Stone Analytics, Inc. 2003. www.secondmoment.org