You play a simple game against the house:

  • From a standard 52-card deck, you draw one card.
  • Then the dealer draws one card without replacement.
  • If your card’s rank is strictly higher, you win $1.
  • If it’s equal or lower, you lose.

Question: What’s the probability that your card beats the dealer’s?

Step 1: Understand the Deck and the Rules

  • Each rank (2 through Ace) appears 4 times.
  • Only the rank matters—not suit.
  • You and the dealer each get one card, no replacement.

You want: \[ P(\text{your card rank} > \text{dealer’s card rank}) \]

Step 2: Use Symmetry

Instead of brute-force enumeration, let’s use a key insight:

  • The game is symmetric—each pair of distinct ranks occurs equally often.
  • For example, there are:
    • 4 × 4 = 16 ways for you to draw a 7 and dealer a 5.
    • Also 16 ways to draw a 5 and 7 in the opposite order.

Step 3: Count Winning Pairs

There are a total of: \[ 52 × 51 = 2652 \text{ possible (ordered) card pairs} \]

Let’s compute how many of these result in your rank > dealer’s rank.

There are 13 ranks, and for every pair of distinct ranks \( (r_i, r_j) \), there are:

  • 4 × 4 = 16 ways to draw one card of each rank.

Among the \(\binom{13}{2} = 78\) unordered rank pairs, half of the 16 × 78 = 1248 ordered pairs favor you (i.e., you have the higher rank), the other half favor the dealer.

So:

  • Total wins: 1248
  • Total losses or ties: 2652 - 1248 = 1404

Hence, the probability of winning is:

\[ \frac{1248}{2652} = \frac{8}{17} \]

Final Answer

The probability that your card beats the dealer’s in this single-draw game is:

\[ \boxed{\frac{8}{17}} \]

Reference