The Problem of Points

Origin

Florence Nightingale David (1962) traced the problem of points to Fra Luca Paccioli (1445-1509) who published the problem in Summa de arithmetica, geometria, proportioni e proportionalità (1494). It is unknown, however, where Fra Luca obtained this problem. Because of Luca's publication, Gerolamo Cardano (1501-1576) found the problem of points and attempted to solve it. Then, in 1654, the Chevalier de Méré (1607-1684) posed the problem to Blaise Pascal who consulted with Pierre de Fermat, and both found different ways to solve it. Their correspondence led to the discovery of more probability rules.

Problem and Solutions

There are many variations of the problem of points. In his 1494 work, Paccioli posed the problem,

$A$ and $B$ are playing a fair game [meaning that the probability that either wins a round is $1/2$]... They agree to continue until one has won six rounds. The game actually stops when $A$ has won five and $B$ three. How should the stakes be divided? (as cited in David, 1962, p. 37)

In modern terms, a variation could be posed: Suppose two of your friends each contributed to a pool of money to play a game where a fair coin is tossed until either a heads is tossed $4$ times (not necessarily consecutively), in which case Friend A wins, or until a tails is tossed $4$ times, in which case Friend $B$ wins. Say that $3$ heads and no tails have been tossed when the game is stopped abruptly. How much of the pool should each friend take?

In the case proposed by Paccioli, he argued that the solution was a ratio of $5:3$, meaning that player $A$ should take $5/8$ of the pool and player $B$ should take $3/8$, because that was the ratio of the scores (David, 1962). However, the appropriate split is that player $A$ should take $7/8$ and player $B$ should take $1/8$. This is because from this point, player $A$ has a probability of $7/8$ to win, and player $B$ has a $1/8$ chance. One way to think about this is that the only way player $A$ does not win is if there are 3 consecutive rounds in favor of player $B$, a probability of $(1/2)^3$. Thus, player B has a $1/8$ probability and player $A$ has a $7/8$ probability of winning the game overall.

In the second case, the reasoning is similar. The only way Friend $B$ can win is if the coin is tails 4 times in a row. The probability of $4$ tails in a row is $(1/2)^4$. Then Friend B should win $1/16$ of the pool, so Friend $A$ should win $15/16$ of the pool.

Pierre de Fermat wrote a more general case of the problem in the second surviving letter he wrote to Pascal, dated 29 July 1654. In this case, each round still has a $1/2$ chance to be in favor of either player, but it is agreed to play until a player has won $n$ rounds, at which time, the winning player gets the whole pot. The game is interrupted when player $A$ needs $a$ more wins and player $B$ needs $b$ more wins. How much of the stake should each player take? To solve this problem, consider that the maximum number of rounds left to play is $a+b-1$. Then it is a matter of finding how many ways player $A$ can get at least a wins in the remaining $a+b-1$ rounds. This is represented in the following formula: $$\sum_{r=a}^{a+b-1}{\binom{a+b-1}{r}/2^{a+b-1}}$$

Pierre de Fermat and Blaise Pascal discussed the general form of the problem of points and independently arrived at the solution above. Their correspondence led to their discussion of other probability problems and topics, and motivated others to participate in discussions of these problems. Such discussion laid the groundwork for the later development of formal probability theory.