You are given a choice between two games, each with a $1 payout, which do you prefer? Game 1: you are given 4 roles of a single die, and you win if you roll at least one six. Game 2: you are given 24 roles of a pair of dice, and you win if you roll a pair of sixes at least once.


a) Game 1: $1-P(no six) = 1- (5/6)^4 = 0.5177469$

b) Game 2: P(single roll no pair six) = $1-1/36$

So 1- P(no pair six for 24 rolls) = $1- (1-1/36)^{24}$

  • 1
    $\begingroup$ That is correct. Now, plug that into a calculator and see which value is larger. $\endgroup$ – JMoravitz Feb 19 '17 at 4:00

References for this question

This is a classical question called the "paradoxe du chevalier de Méré", which dates back to the 17th century. It was solved by Pascal and Fermat and constitutes one of the first attempts at probabilistic proof

I tried to find references in english :

Ref 1

Ref 2

Here is another one, in french, with much more historical background :

Ref 3


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