Dice and Coins

The Chevalier de Mere Dice Problem

The 17th-century dice puzzle behind early probability theory: one six in four rolls versus a double six in 24 rolls of two dice, with the exact numbers.

On this page
  1. The historical puzzle
  2. The first bet
  3. The second bet
  4. Why the ratio argument fails
  5. Why it matters
  6. A note on gambling

The historical puzzle

In seventeenth-century France, a gambler and writer known as the Chevalier de Mere (a title; his name was Antoine Gombaud) is said to have posed a puzzle to the mathematician Blaise Pascal. The details come to us through later accounts, so it is best to hold the colourful parts loosely. What is well established is that questions of this kind, discussed in letters between Pascal and Pierre de Fermat in 1654, are often cited as an early step in the history of probability.

The puzzle compares two bets involving dice:

  1. Roll one die four times. Is there at least one six?
  2. Roll two dice 24 times. Is there at least one double six (both dice showing six on the same throw)?

According to the usual telling, de Mere believed the two bets should be equally good. His reasoning was a ratio. A six on one die has chance 1/6, and a double six on two dice has chance 1/36, which is six times smaller. If four rolls are enough for the first bet, then six times as many, 24 rolls, should be enough for the second. The numbers 4 to 6 and 24 to 36 are in the same proportion.

The first bet

Use the complement rule: find the chance of no six and subtract from 1. On one roll the chance of no six is 5/6, so for four independent rolls it is (5/6)^4 = 625/1296.

P(at least one six) = 1 - 625/1296 = 671/1296, about 0.5177, or 51.8 percent.

This is slightly above one half. Over many repeated sets of four rolls, a six would show up in a bit more than half of them, though any one set is uncertain.

The second bet

Two dice give 36 equally likely outcomes, and exactly one of them is a double six. The chance of a double six on a throw is 1/36, and the chance of not getting one is 35/36. Over 24 independent throws:

P(no double six) = (35/36)^24, which is about 0.5086.

P(at least one double six) = 1 - 0.5086 = about 0.4914, or 49.1 percent.

That is slightly below one half. If you extend to 25 throws, the chance is about 50.55 percent, which tips just above half. So 25 throws is the smallest number for which the double-six bet is better than even.

Bet Number of tries Chance of success
At least one six, one die 4 about 51.8%
At least one double six, two dice 24 about 49.1%
At least one double six, two dice 25 about 50.6%

Why the ratio argument fails

De Mere’s argument treated probability as if it scaled in direct proportion to the number of tries. If you add the chances (4 x 1/6 and 24 x 1/36) you do get the same figure, 2/3, for both. But adding probabilities is not the right operation for repeated tries, because it double counts cases where success happens more than once. That is the same error as thinking four rolls give 4/6 for a six.

The right operation raises the “no success” chance to a power. The ratio 4 : 6 matches 24 : 36, but (5/6)^4 and (35/36)^24 are different numbers, since the two bases are different. The two bets are close, which makes the gap hard to see by intuition. The difference is under three percentage points, so a gambler would have needed a lot of repeated play to feel the difference, even though over the long run it matters.

Why it matters

The puzzle shows how reasonable-sounding shortcuts can mislead and how careful counting resolves them. Pascal and Fermat’s correspondence on this and a related problem about dividing stakes in an unfinished game is often presented as the start of formal probability theory, though ideas about chance existed earlier. The episode also shows why a theory of chance developed from questions about games.

It also illustrates a modern point: a bet that looks slightly worse than even is not a bet that always loses. In 24 throws you will get at least one double six in just under half of repeated sets of 24, so the “worse” bet still often succeeds. A small edge shows itself only over many repetitions. This is the reasoning behind the house edge in casino games, where a small average disadvantage per bet compounds over many bets.

A note on gambling

This article is a history and maths story, not a suggestion to bet. The probabilities here describe fair dice with fair payouts, and real casino games are set so the venue keeps an advantage. Past results never change the chance on the next roll. Gambling can harm people; if it is affecting you or someone you know, a local support service can help.

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