At least one six when rolling three dice

Question

You roll three fair dice at once. What is the probability of at least one six? Would you take an even-money bet on it?

Accepted Answer

Use the complement: no six on any of the three dice. The rolls are independent, and each avoids a six with probability 	frac{5}{6}: P(	ext{no six}) = \left(\frac{5}{6}ight)^3 = \frac{125}{216}. Therefore P(	ext{at least one six}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421. Since this is below 	frac12, an even-money bet on "at least one six" is a loser: you would win only about 42\% of the time. Note that the tempting shortcut 3 	imes 	frac{1}{6} = 	frac12 overcounts, it double-counts the outcomes with two or three sixes, which is exactly why probabilities of overlapping eve…

At least one six when rolling three dice

Your answer

More Probability questions