Euler's Eyes
2026-10-11
Fermat and Pascal: fairly dividing an interrupted game
Two players bet the same amount on a game of several rounds: whoever first wins a certain number of rounds takes everything. Midway through, the game has to be stopped. How should the money be divided fairly?
This is the problem of points. It looks simple, but it defeated mathematicians for more than 150 years.
The wrong answer
In 1494 the Italian friar Luca Pacioli proposed dividing the prize in proportion to the rounds already won. If A has won 2 rounds and B has won 1, A takes $\frac23$ of the prize.
It seems reasonable, but it is not. Imagine a game to 100 wins, stopped with A leading 1–0. By Pacioli’s rule, A takes everything. But the game had barely started, and B had almost the same chance of winning. What matters is not what has happened, but what can still happen.
The letters of 1654
The problem reached Blaise Pascal through the Chevalier de Méré, a French nobleman who loved games of chance. Pascal wrote to Fermat in Toulouse, and in the summer of 1654 the two exchanged a series of letters. Each found the solution by a different path: Pascal by step-by-step reasoning, working backwards; Fermat by counting all the possibilities.
When Pascal saw that the two methods gave the same answer, he wrote:
I see that the truth is the same in Toulouse as in Paris.
— Blaise Pascal, letter to Fermat (29 July 1654)
Fermat’s method: counting the futures
Suppose each round is like tossing a coin: each player has a chance of $\frac12$ of winning it. The game is stopped when A needs 1 more win and B needs 2.
Fermat reasoned like this: within at most two rounds, someone will certainly have won. So we imagine both rounds are played, even if the first already decides the game. The possible outcomes, all equally likely, are:
| Rounds | Winner |
|---|---|
| A A | A |
| A B | A |
| B A | A |
| B B | B |
A wins in 3 of the 4 cases; B in 1. The probabilities are
$$P(A)=\frac34,\qquad P(B)=\frac14.$$If the prize is 80 coins, the fair division is 60 to A and 20 to B.
But wait: in “A A” and “A B” the game would end in the first round. Why count the second? Because playing one more round after the end does not change the winner, and the trick makes every case the same length and equally likely. That makes counting easy.
The general rule
If A needs $a$ wins and B needs $b$, the game would be decided in at most $a+b-1$ rounds. Fermat lists all $2^{a+b-1}$ possible sequences and counts those in which A wins at least $a$ rounds. This count uses the numbers of the arithmetic triangle, which Pascal was studying at the same time. See the Pascal trail for his method.
The new idea
Before 1654 nobody had treated chance with serious mathematics. The letters of Pascal and Fermat introduced the idea of expected value: what each player “deserves” is the prize multiplied by their chance of winning. The same idea underlies insurance, investment and any decision involving risk today.
Exercise: each player needs one more win. How should 80 coins be divided?
The next round decides everything, and each has a chance of $\frac12$. The fair division is 40 each.
Exercise: A needs 1 win and B needs 3. How should 80 coins be divided?
Within 3 rounds the game is decided. Of the $2^3=8$ sequences, B wins only with BBB. So $P(A)=\frac78$ and $P(B)=\frac18$: 70 coins to A and 10 to B.
Sources: Fermat biography, MacTutor · Pascal biography, MacTutor.
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