Higher Card Wins: One Draw Against the Dealer
You and a dealer play a one-round card game with a standard, well-shuffled 52-card deck. You draw one card, then the dealer draws one card from the remaining 51 (so the draws are without replacement). Only ranks matter: $2 < 3 < \cdots < 10 < J < Q < K < A$, and suits are ignored.
You win if and only if your card's rank is strictly higher than the dealer's. A tie counts as a loss for you.
What is the probability that you win?
Hints
- Do not enumerate all 52 x 51 ordered pairs. Ask first: what is the probability that the two cards tie?
- If the two cards do not tie, is there any reason your card should be more likely to be the higher one than the dealer's card?
- $P(\text{tie}) = 3/51$. By symmetry $P(\text{win}) = P(\text{lose})$, so $P(\text{win}) = \tfrac{1}{2}(1 - P(\text{tie}))$.
Worked Solution
How to Think About It: The two cards are exchangeable random draws, so before we look at them there is no asymmetry between "my card" and "the dealer's card." The only outcome that is not a win for one of us is a tie. So find the tie probability, and split what is left evenly.
Quick Estimate: After you draw, 3 of the remaining 51 cards match your rank, so a tie happens about 6% of the time. Roughly 94% of games have a winner, and half of those go to you: about 47%.
Formal Solution:
*Step 1 -- Probability of a tie.* Whatever your card is, exactly 3 of the 51 remaining cards share its rank:
$$P(\text{tie}) = \frac{3}{51} = \frac{1}{17}.$$
*Step 2 -- Symmetry.* The ordered pair (your card, dealer's card) is a uniformly random ordered pair of distinct cards. Swapping the two cards is a bijection on this sample space that maps every "you win" outcome to a "you lose" outcome and vice versa, so
$$P(\text{win}) = P(\text{lose}).$$
*Step 3 -- Combine.* Since win, lose and tie partition the sample space,
$$P(\text{win}) = \frac{1 - P(\text{tie})}{2} = \frac{1 - \tfrac{1}{17}}{2} = \frac{16/17}{2} = \frac{8}{17} \approx 0.4706.$$
*Check by direct counting.* There are $52 \cdot 51 = 2652$ ordered outcomes. Ties: $52 \cdot 3 = 156$. Wins: for each of the 13 ranks with $k$ ranks below it, $4 \cdot 4k$ ordered pairs, giving $16 \cdot (0 + 1 + \cdots + 12) = 16 \cdot 78 = 1248$. So $P(\text{win}) = 1248/2652 = 8/17$.
Answer: $P(\text{win}) = \dfrac{8}{17} \approx 0.471$.
Intuition
Your card and the dealer's card are exchangeable: nothing about the draw order favors either player, so the only thing that breaks the 50/50 symmetry is the tie, and ties go to the house. The whole calculation collapses to "one minus the tie probability, halved." This symmetry-plus-tie decomposition is the fastest way to price almost any "who has the higher value" bet, and it is the same logic a trader uses when a contract pays on strict inequality: the house edge lives entirely in the tie mass.