You versus the house

A casino offers a simple card game. You draw a card from a standard 52-card deck. The dealer draws another card from the remaining 51. If your card's value is higher, you win. If it's equal or lower, the house wins.

As always, the house has the edge. But how much?

What do you think?

What is your probability of winning this card game?

Play a few rounds:

Card Game

After enough draws, your win rate settles near 47.1%. Let's see why.

The brute-force approach

There are 13 card values (2 through A), each with 4 cards. You could enumerate all cases:

Your card	P(win)	Logic
2	0/51	No card is lower
3	4/51	Only the four 2s are lower
4	8/51	Four 2s and four 3s
...	...	...
A	48/51	Everything except the other three Aces

Each value appears with probability 1/13, so:

$P(\text{win}) = \frac{1}{13} \times \frac{4}{51} \times (0 + 1 + 2 + \cdots + 12) = \frac{1}{13} \times \frac{4}{51} \times \frac{12 \times 13}{2} = \frac{24}{51} = \frac{8}{17}$

That works, but there's a faster way.

The symmetry shortcut

Just like the coin toss game, this problem yields to symmetry. Consider three events:

$E_1$ : Your card is higher than the dealer's

$E_2$ : Both cards have the same value

$E_3$ : Your card is lower than the dealer's

Symmetry Solution

P(E_1) = P(E_3)

By symmetry. Nothing distinguishes 'your' draw from the 'dealer's' draw. You each have one random card from the deck.

Step 1 of 5

The symmetry approach reduces the problem to one simple calculation: what's the probability of a tie? Once you know that, the win probability falls out immediately. No summation needed.

This is the same pattern from the coin toss problem. Identify the asymmetry (here, ties favor the house), compute just that probability, and let symmetry handle the rest.

The house edge

The house wins with probability $P(E_2) + P(E_3) = 1/17 + 8/17 = 9/17 \approx 52.9\%$ . Their edge is $9/17 - 8/17 = 1/17 \approx 5.9\%$ .

What do you think?

If the rules changed so ties result in a redraw (no one wins), what would your win probability become?

decimal like 0.50

Playing with fewer values

The pattern becomes clearer if you simplify. With a deck that has $k$ distinct values, each with $m$ cards ( $km$ cards total):

$P(\text{tie}) = \frac{m - 1}{km - 1}$

$P(\text{you win}) = \frac{1 - \frac{m-1}{km-1}}{2}$

Deck	Values (k)	Cards per value (m)	P(tie)	P(you win)
Standard	13	4	3/51 = 5.9%	8/17 = 47.1%
2 suits only	13	2	1/25 = 4.0%	12/25 = 48.0%
No duplicates	13	1	0%	50.0%
4 values x 4	4	4	3/15 = 20%	2/5 = 40%

More distinct values means smaller tie probability, which means a fairer game. If every card were unique (like numbering 1 through 52), you'd have P(win) = exactly 50%.

In a deck with 10 distinct values and 4 copies each (40 cards), if you draw and the dealer draws, what's P(tie)? (decimal like 0.08)

1/3

The takeaway

This problem and the coin toss game use the same technique: decompose into win/tie/lose, observe that win and lose are symmetric, then compute only the tie probability. The symmetry argument works because mathematical structure doesn't care which player is "you" and which is "the dealer." The only asymmetry is the rule about who wins ties, and that's easy to handle separately.