Antoine Gombaud, who styled himself the Chevalier de Méré, was a writer at the French court and a passionate gambler. Around 1654 he brought Pascal two questions about bets. Pascal found them so good that he wrote to Pierre de Fermat, and probability theory was born from those letters.

About his friend, Pascal remarked to Fermat:

He has a very good mind, but he is not a geometer; and that, as you know, is a great defect.

— Blaise Pascal, letter to Fermat (29 July 1654)

First problem: the six and the double six

Méré knew a profitable bet: roll one die 4 times and bet that at least one six comes up. He won more often than he lost.

So he reasoned: with two dice, a double six is 6 times rarer than a six on one die. To make up for it, just roll 6 times as often: $4\cdot6=24$ rolls. The bet “at least one double six in 24 rolls of two dice” should be as good as the first.

It was not. Méré lost money and could not understand why.

The right calculation goes through the opposite event. The chance of not rolling a six with one die is $\frac56$. In 4 independent rolls, the chance of never rolling a six is $\left(\frac56\right)^4$. So

$$P(\text{at least one six in 4})=1-\left(\frac56\right)^4=1-\frac{625}{1296}\approx0.518.$$

With two dice there are $36$ outcomes, and only one is a double six. The chance of not rolling a double six in one throw is $\frac{35}{36}$:

$$P(\text{at least one double six in 24})=1-\left(\frac{35}{36}\right)^{24}\approx0.491.$$

The first bet wins 51.8% of the time; the second, only 49.1%. Méré’s proportion rule looked logical, but chances do not add up that way. With 25 throws the chance goes over one half: $1-\left(\frac{35}{36}\right)^{25}\approx0.506$.

Second problem: the interrupted game

Two players each stake 32 pistoles, 64 in total. Whoever first wins 3 rounds takes everything. The game is interrupted when A has 2 wins and B has 1. How should the 64 pistoles be split fairly?

Fermat solved it by counting every possible sequence. Pascal did something different: he looked only at the next round.

  • If A wins the next round, he reaches 3 and takes all 64.
  • If B wins, it is 2 all, and then the two are level: each deserves 32.

Both players have the same chance in the next round. So A is guaranteed 32 pistoles whatever happens, and the other 32 are uncertain, with an even chance for each. Pascal concluded that A should receive

$$32+\frac{32}{2}=48\ \text{pistoles},$$

and B the remaining 16. These are exactly the numbers in his letter.

Working backwards

The idea works for any score: solve the easy case and go back one step at a time. Suppose now that A has 2 wins and B has none. If A wins the next round, he takes 64; if he loses, the score becomes 2 to 1, which we already know is worth 48 to A. The average is

$$\frac{64+48}{2}=56.$$

And if A has 1 win and B none? If A wins, it goes to 2 to 0, worth 56; if he loses, it is 1 all, worth 32. The average is

$$\frac{56+32}{2}=44.$$

Pascal sent Fermat exactly this sequence: 48, 56 and 44 pistoles.

The general rule

Let $P(a,b)$ be A’s chance of winning the game when he still needs $a$ wins and B still needs $b$. If $a=0$, A has already won: $P(0,b)=1$. If $b=0$, B has already won: $P(a,0)=0$. In every other case, the next round gives

$$P(a,b)=\frac12P(a-1,b)+\frac12P(a,b-1).$$

A’s fair share is the prize times $P(a,b)$. In the example of 2 to 1 in a game to 3, A needs $a=1$ and B needs $b=2$, and $P(1,2)=\frac34$; $\frac34$ of 64 gives the 48 pistoles.

Notice that the rule has the same shape as the arithmetic triangle: each value is the average of two neighbours. That is no coincidence, and Pascal used the triangle to solve the problem in one go for any score.

Choose how many wins each player still needs. The control computes A’s chance with Pascal’s rule and splits a prize of 80 coins.

A new mathematics

Before 1654, chance was a matter for gamblers, not mathematicians. Pascal’s letter introduces the idea we now call expected value: what each player deserves is the prize multiplied by the chance of winning it. In 1657 the Dutchman Christiaan Huygens, who had heard about the letters in Paris, published the first book on the subject. Insurers, lotteries and banks began doing this calculation, and they still do.

Exercise: in a game to 3 wins with 64 pistoles, the score is 1 to 1. What if it is 0 to 0?

At 1 to 1 both players are in the same position: 32 each. At 0 to 0 as well: nobody has an advantage, and each gets back his own 32 pistoles.

Exercise: what is P(2,1)?

$P(2,1)=\frac12P(1,1)+\frac12P(2,0)=\frac12\cdot\frac12+\frac12\cdot0=\frac14$. With a prize of 80 coins, A gets 20 and B gets 60.

Sources: the 1654 letters of Pascal and Fermat, in English translation · Pascal biography, MacTutor.

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