Probability your d20 beats mine

Question

You and an opponent each roll one fair twenty-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 events \{X Y\}, \{X < Y\}, \{X = Y\} partition the outcomes, and by symmetry (identically distributed, independent rolls) P(X Y) = P(X < Y). Ties: P(X = Y) = \frac{20}{400} = \frac{1}{20} (any of 20 matching pairs out of 20 	imes 20). So P(X Y) = \frac{1 - P(X = Y)}{2} = \frac{1 - \frac{1}{20}}{2} = \frac{19}{40} = 0.475. The count confirms it: of 400 ordered outcomes, 20 are ties and the remaining 380 split evenly, 190 wins and 190 losses, giving \frac{190}{400} = \frac{19}{40}. Takeaway. For i.i.d. quantities, P(X Y) = \frac{1 - P(X = Y)}{2}. A…

Probability your d20 beats mine

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