Probability your die beats mine

Question

You and an opponent each roll one fair six-sided die. You win if your number is strictly higher. What is the probability that you win?

Accepted Answer

Let X be your roll and Y your opponent's. The three events \{X Y\}, \{X < Y\}, \{X = Y\} partition the sample space, and by symmetry (the dice are identically distributed and independent) P(X Y) = P(X < Y). Ties are easy: P(X = Y) = \frac{6}{36} = \frac{1}{6}. So P(X Y) = \frac{1 - P(X = Y)}{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12} \approx 0.4167. A direct count confirms it: of the 36 ordered outcomes, 6 are ties, and the remaining 30 split evenly, 15 wins and 15 losses, giving 	frac{15}{36} = 	frac{5}{12}. Takeaway. For i.i.d. quantities, P(X Y) = \frac{1 - P(X = Y)}{2}. Compute the t…

Probability your die beats mine

Your answer

More Probability questions