The gambler's two bets

Question

Two bets from the 1650s: - Bet A: roll one die four times; you win if at least one six appears. - Bet B: roll two dice twenty-four times; you win if at least one double six appears. The Chevalier de Méré reasoned that a six is 1/6 likely per roll and four rolls give 4/6; a double six is 1/36 likely per roll and 24 rolls give 24/36, the same. Yet he found that bet B lost him money over time and asked Pascal why. Compute the probability of winning each bet. Which is better, and what was wrong with the gambler's reasoning?

Accepted Answer

Bet A wins with probability about 51.8 per cent, bet B about 49.1 per cent. Bet A is better. Adding probabilities of overlapping events was the error. Bet A. The chance of no six in one roll is 5/6. In four independent rolls, no six at all has probability (5/6) to the 4 = 625/1296, about 0.482. So at least one six: 1 minus 0.482 = 0.518. Bet B. The chance of no double six in one roll of two dice is 35/36. In 24 rolls: (35/36) to the 24, about 0.509. So at least one double six: 0.491. The gambler's error. Adding 1/6 four times, or 1/36 twenty-four times, counts the outcomes where the event hap…

The gambler's two bets

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