Probability you don't lose the dice duel

Question

You and an opponent each roll one fair six-sided die. This time the rule favors you: you win if your number is at least as high, so ties go to you.

What is the probability that you win (that is, $X \ge Y$ )?

Accepted Answer

Let X be your roll and Y your opponent's. "At least as high" is the strictly-higher event plus the tie event, which are disjoint: P(X \ge Y) = P(X Y) + P(X = Y). From the symmetric argument, P(X Y) = \frac{1 - P(X = Y)}{2} = \frac{1 - \frac16}{2} = \frac{5}{12}, and P(X = Y) = \frac16 = \frac{2}{12}. So P(X \ge Y) = \frac{5}{12} + \frac{2}{12} = \frac{7}{12} \approx 0.5833. Sanity check by symmetry directly: awarding ties to you means P(X \ge Y) + P(Y X) = 1 with P(X \ge Y) = P(X Y) + P(	ext{tie}) and P(Y X) = P(X Y), so P(X \ge Y) = \frac{1 + P(	ext{tie})}{2} = \frac{1 + \frac16}{2} = \fra…

Probability you don't lose the dice duel

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