I see that the truth is the same in Toulouse as in Paris.
Je vois bien que la vérité est la même à Toulouse et à Paris.
Pascal begins joyfully: he and Fermat agree on the dice problem. Then he puts another problem to Fermat. Two players play a match to three wins; each has staked 32 pistoles (gold coins), 64 in total. The game is interrupted. Depending on how far they got, one of three scenarios arises.
- Game to three winseach staked 32 pistoles, 64 in total
- 1.2–1Pascal needs 1 more win, Fermat 248:16
- 2.2–0Pascal needs 1 more win, Fermat 356:8
- 3.1–0Pascal needs 2 more wins, Fermat 344:20
Here Pascal and Fermat play the two players themselves; in the letters the players have no names. Each scenario builds on the result of the previous one.
1. Scenario 2–1
- Score 2–1next game
- ½Bingo! Takes all 64
- ½score 2–2, even chances
- ½takes 64
- ½takes 64
If Pascal wins, it is over and he takes all 64. If Fermat wins, the score is 2–2 and the stake is divided equally. So Pascal already holds 32 pistoles for certain, and the remaining 32 are played for at even odds.
Chances ¾:¼, that is, 48:16
2. Scenario 2–0
- Score 2–0next game
- ½takes all 64
- ½score 2–1, see scenario 1: Pascal holds 48
Pascal has 48 for certain, because even if he lost the next game he would end up in scenario 2–1. For the remaining 16, fortune is even.
56:8, because ½ · 64 + ½ · 48 = 56
3. Scenario 1–0
- Score 1–0next game
- ½score 2–0, see scenario 2: Pascal holds 56
- ½score 1–1, even chances: Pascal holds 32
Pascal has 32 for certain. The difference 56 − 32 = 24 is played for at even odds, which makes 12.
44:20, because ½ · 56 + ½ · 32 = 44