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Chevalier de Méré’s first game involved taking a chance that at least one 6 would appear during a total of four rolls of one die.

Solution:  To find the probability of getting at least one 6 in four rolls, one would have to calculate the probability of getting one 6, two 6’s, three 6’s, and four 6’s.  Actually, it would be quicker to find the opposite of this event, called the complement.  This is the probability of not getting any 6’s.  Since the probability of not rolling a 6 is five out of six for each of the four rolls, this probability is (5/6)⁴.  To solve the game you need to subtract this probability from 1, which is 51.8 percent:  

1 - (5/6)⁴

or 51.8 percent

Chevalier de Méré’s second game involved taking a chance that he would roll a total of 12, or a double 6, once in 24 rolls of two dice.

Solution:  Instead of finding the probability of getting a total of 12, or a double 6, in 24 rolls of two dice, Pascal found the probability of de Méré losing, or not rolling a double 6.  Since there are 36 possible rolls of two dice and only one way to roll a double 6, then the probability of not rolling a double six is 35/36.  He rolled 24 times.  The probability that he loses the game is:

(35/36)²⁴

or 50.9 percent

To compare to the first game, one must calculate the probability that he wins in this game.  This is:

1 - (35/36)²⁴

or 49.1 percent