Probability your card rank beats mine

Question

You and an opponent each draw a card rank uniformly at random from $1$ to $13$ (Ace low through King), from independent well-shuffled decks. You win if your rank is strictly higher.

What is the probability that you win?

Accepted Answer

Let X be your rank and Y your opponent's, each uniform on \{1, \dots, 13\} and independent. The events \{X Y\}, \{X < Y\}, \{X = Y\} partition the outcomes, and by symmetry P(X Y) = P(X < Y). Ties: P(X = Y) = \frac{13}{169} = \frac{1}{13} (13 matching pairs out of 13 	imes 13). So P(X Y) = \frac{1 - P(X = Y)}{2} = \frac{1 - \frac{1}{13}}{2} = \frac{12}{26} = \frac{6}{13} \approx 0.4615. Direct count: of 169 ordered pairs, 13 tie and the remaining 156 split into 78 wins and 78 losses, giving \frac{78}{169} = \frac{6}{13}. Takeaway. The symmetry shortcut P(X Y) = \frac{1 - P(X = Y)}{2} works f…

Probability your card rank beats mine

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