Probability of Drawing a Higher Card Than the Dealer

Probability · Easy · Free problem

You and a dealer are each dealt one card from a standard shuffled deck of 52 cards (dealt without replacement). You win if your card has a strictly higher rank than the dealer's card. What is the probability that you win?

(Assume ranks are

$ through
3$, with four cards of each rank.)

Hints

  1. There are three outcomes: you are higher, you tie, or the dealer is higher. What does symmetry tell you about two of these?
  2. Compute the probability of a tie. Once your card is dealt, how many of the remaining 51 cards share its rank?
  3. Use $P(\text{win}) + P(\text{tie}) + P(\text{lose}) = 1$ with $P(\text{win}) = P(\text{lose})$ to solve for $P(\text{win}) = (1 - 1/17)/2$.

Worked Solution

How to Think About It: There are three possible outcomes: your rank is higher, your rank is lower, or you tie. By symmetry, the probability of your rank being higher equals the probability of it being lower. So you just need to find the probability of a tie, and the rest follows from the fact that all three probabilities sum to 1.

Quick Estimate: If all 52 cards had distinct values, symmetry would give you exactly

/2$. But there are ties (same rank), and ties count as losses. So the answer should be a bit below
/2$. How often do you tie? After you draw a card, 3 of the remaining 51 cards match your rank, so $P(\text{tie}) = 3/51 = 1/17 \approx 5.9\%$. So $P(\text{win}) \approx (1 - 1/17)/2 = 16/34 = 8/17 \approx 0.471$.

Approach: Use symmetry to relate $P(\text{win})$ and $P(\text{lose})$, then subtract the tie probability.

Formal Solution: Let $E_1$ be the event that your card has strictly higher rank, $E_2$ be a tie, and $E_3$ be that the dealer's card has strictly higher rank.

By symmetry (you and the dealer are treated identically in the deal):

$P(E_1) = P(E_3)$

The tie probability: after your card is dealt, 51 cards remain, of which exactly 3 share your rank:

$P(E_2) = \frac{3}{51} = \frac{1}{17}$

Since $P(E_1) + P(E_2) + P(E_3) = 1$ and $P(E_1) = P(E_3)$:

$2P(E_1) + \frac{1}{17} = 1$

$P(E_1) = \frac{1}{2}\left(1 - \frac{1}{17}\right) = \frac{1}{2} \cdot \frac{16}{17} = \frac{8}{17}$

Answer: $P(\text{you win}) = \dfrac{8}{17} \approx 0.471$.

Intuition

This problem is a clean illustration of the symmetry-plus-complement technique. Instead of enumerating all $52 \times 51$ ordered pairs, you recognize that "you win" and "dealer wins" are symmetric events, so you only need to compute the easy one -- the tie probability -- and the answer drops out from $P(\text{win}) = (1 - P(\text{tie}))/2$.

The deeper lesson is about how duplicates break pure symmetry. With distinct values, the win probability would be exactly

/2$. The existence of ties (four cards per rank) steals probability from both "win" and "lose" equally, pushing each below
/2$. This same structure appears in many trading contexts: if two market participants have symmetric information but there's a chance of no trade (tied views), the probability of a favorable outcome for either side drops below the naive
/2$.

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