At least one six in four rolls

Question

You roll a fair die four times. What is the probability that you roll at least one six? Would you take an even-money bet on it?

Accepted Answer

Work with the complement: no six in four rolls. The rolls are independent, and each avoids a six with probability 	frac{5}{6}: P(	ext{no six}) = \left(\frac{5}{6}ight)^4 = \frac{625}{1296}. Therefore P(	ext{at least one six}) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.5177. Since this exceeds 	frac{1}{2}, the even-money bet is (slightly) favorable, this is exactly the bet the Chevalier de Méré profitably made in the 1650s, and his losses on the "double six in 24 rolls" variant (1 - (35/36)^{24} \approx 0.4914 < 	frac12) prompted the Pascal–Fermat correspondence that founded pr…

At least one six in four rolls

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