What is the fair division of stakes in a game which is interrupted before its conclusion? This was the problem posed in 1654 by French gambler Chevalier de Mere to philosopher and mathematician, Blaise Pascal, of Pascal’s Triangle and Pascal’s Wager fame, who shared it in now-famous correspondence with mathematician Pierre de Fermat (best known these days perhaps for Fermat’s Last Theorem). It has come to be known as the Problem of Points.

The question had first been formally addressed by Franciscan Friar and Leonardo Da Vinci collaborator, Luca Bartolomeo de Pacioli, father of the double-entry system of bookkeeping. Pacioli’s method was to divide the stakes in proportion to the number of rounds won by each player to that point. There is an obvious problem with this method, however. What happens, for example, if only a single round of many has been played. Should the entire pot be allocated to the winner of that single round? In the mid-1500s, Venetian mathematician and founder of the theory of ballistics, Niccolo Fontana Tartaglia, proposed basing the division on the ratio between the size of the lead and the length of the game. But this method is not without its own problems. For example, it would split the stakes in the same proportion whether one player was ahead by 40-30 or by 99-89, although the latter situation is hugely more advantageous than the former.

The solution adopted by Pascal and Fermat defied prevailing intuition by basing the division of the stakes not on the history of the interrupted game to that point as on the possible ways the game might have continued were it not interrupted. In this method, a player leading by 6-4 in a game to 10 would have the same chance of winning as a player leading by 16-14 in a game to 20, so that an interruption at either point should lead to the same division of stakes. As such, what is important in the Pascal-Fermat solution is not the number of rounds each player has yet won but the number of rounds each player still needs to win.

Take another example. Suppose that two players agree to play a game of coin-tossing repeatedly to won £32, and the winner is the first player to win four times.

If the game is interrupted when one of the players is ahead by two games to one, how should the  £32 be divided fairly between the players?

In Fermat’s method, imagine playing another four games. Outcomes of each coin-tossing game are equally likely and are P (won by Player 1) and Q (won by Player 2).

The possible outcomes of the next four games are as follows:

PPPP; PPPQ; PPQP; PPQQ

PQPP; PQPQ; PQQP; PQQQ

PQQQ; PQQP; PQPQ; PQPP

PPQQ; PPQP; PQQP; QQQQ

The probability that Player 1 would have won is 11/16 (in bold) = 68.75%.

The probability that Player 2 would have won is 5/16 = 31.25%.

The method can be generalised to any game of chance which ends before the game is complete.

Appendix

Pascal proposed an alternative method which dispenses with the need to consider possible steps after the game had already been won. In doing so, he was able to devise a relatively simple formula which would solve all possible Problems of Points, without needing to go beyond the point at which the game resolves in favour of one or other of the players, based on Pascal’s Triangle, demonstrated below.

1                                     = 1

1       1                                 = 2

1    2      1                            = 4

1   3    3     1                        = 8

1   4   6   4    1                   = 16

1  5  10  10  5   1                = 32

1 6 15  20 15  6   1            = 64

1 7 21 35 35 21 7 1           = 128

Each of the numbers in Pascal’s Triangle is the sum of the adjacent numbers immediately above it.

If the game is interrupted, as above, 2-1, after three games, in a first to four match, the resolution is 1+4+6 / 16 to 4+1 / 16, i.e. 11/16 to Player 1 and 5/16 to Player 2.

More generally, Pascal’s method establishes the modern method of expected value when reasoning about probability. To show this, consider the probability that Player 1 would win if leading 3-2 in a game in which the first player to win four games is the outright winner. If Player 1 wins the next coin toss, he goes ahead 4-2 and wins outright (value = 1). There is a 50% chance of this. There is a 50% chance of player 2 winning the coin toss, however, in which case the game is level (3-3). If the game is level, there is a 50% chance of player 1 winning (and a 50% chance of player 2 winning).

So the expected chance of player 1 winning when leading 3-2 = 50% x 1 + 50% x 50% = 50% + 25% = 75%. Expected chance of player 2 winning = 25%.

Now consider the probability that Player 1 would win if leading 3-1 in a game in which the first player to win four games is the outright winner. If Player 1 wins the next coin toss, he goes ahead 4-1 and wins outright (value = 1). There is a 50% chance of this. There is a 50% chance of player 2 winning the coin toss, however, in which case the game goes to 3-2. We know that the expected chance of player 1 winning if ahead 3-2 is 75% (derived above).

So the expected chance of player 1 winning when leading 3-1 = 50% x 1 + 50% x 75% = 50% + 25% = 87.5%. Expected chance of player 2 winning = 12.5%.

Now consider the question that we solved using Fermat’s method, i.e. the probability that Player 1 would win if leading 2-1 in a game in which the first player to win four games is the outright winner. If Player 1 wins the next coin toss, he goes ahead 3-1, and has an expected chance of winning of 87.5% and wins outright (value = 1). There is a 50% chance of this. There is a 50% chance of player 2 winning the coin toss, however, in which case the game goes to 2-2. We know that the expected chance of player 1 winning if tied is 50%. There is a 50% chance of this.

So the expected chance of player 1 winning when leading 2-1 = 50% x 87.5% + 50% x 50% = 43.75% + 25% = 68.75% (i.e. 11/16). Expected chance of player 2 winning = 31.25% (i.e. 5/16).

So both Fermat’s method and Pascal’s method yield the same solution, by different routes, and will always do so in determining the correct division of stakes in an interrupted game of this nature.

Exercise

Suppose that two players agree to play a game of coin-tossing repeatedly to won £32, and the winner is the first player to win four times.

If the game is interrupted when one of the players is ahead by two games to zero, determine how the  £32 should be divided fairly between the players.