Card Draw Probability Calculator

How to Calculate Card Draw Probability

Drawing cards from a deck can be modeled two different ways, depending on whether each card is put back before the next draw. A real deck of cards is drawn without replacement — every card removed stays out, so the deck's makeup keeps changing. That case uses the hypergeometric formula: $$P(\text{exactly } k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$ If cards were instead drawn with replacement — each card put back and the deck reshuffled before the next draw — every draw would have the same fixed probability, and the binomial formula applies instead: $$P(\text{exactly } k) = \binom{n}{k}\left(\frac{K}{N}\right)^k\left(1-\frac{K}{N}\right)^{n-k}$$

Why Without Replacement Gives a Different Answer

Each card removed from the deck changes what's available for the next draw. Pull an Ace on the first draw and only 3 Aces remain among 51 cards for the second draw — the odds for that second draw are no longer 4/52, they've shifted to 3/51. This is exactly what "without replacement" means, and it's why a real deck of cards is always modeled with the hypergeometric formula rather than the with-replacement binomial assumption, which incorrectly treats every draw as if the deck reset back to its full 4/52 odds each time. In practice, without-replacement probabilities tend to differ from with-replacement probabilities most noticeably when the deck is small relative to the number of cards drawn — the composition shifts more, proportionally, with each card removed.

With Replacement: The Simpler (but Less Realistic) Case

With replacement, every draw is an independent trial at the same fixed probability K/N, exactly like a repeated coin flip or dice roll. That's mathematically simpler — it's the same binomial math used elsewhere on this site — but it's rarely how real card games actually work, since almost nobody reshuffles a drawn card back into the deck before dealing the next one.

Worked Examples

1. Poker hand, 1 ace, without replacement. $N=52$, $K=4$, $n=5$, $k=1$: $\binom{4}{1}\binom{48}{4}/\binom{52}{5} = 4 \times 194580/2598960 \approx 29.95\%$.
2. One card, one whole suit, without replacement. $N=52$, $K=13$ (hearts), $n=1$, $k=1$: $\binom{13}{1}\binom{39}{0}/\binom{52}{1} = 13/52 = 25\%$ — the simplest possible case, matching plain single-event odds.
3. Poker hand, 1 ace, with replacement. Same setup as example 1, but with replacement: $\binom{5}{1}(4/52)^1(48/52)^4 \approx 27.92\%$ — noticeably lower than the without-replacement answer, since replacement never lets the "remaining aces" ratio improve after a near-miss.
4. Both cards are aces, without replacement. $N=52$, $K=4$, $n=2$, $k=2$: $\binom{4}{2}\binom{48}{0}/\binom{52}{2} = 6/1326 \approx 0.45\%$.
5. Both cards are aces, with replacement. Same setup, with replacement: $\binom{2}{2}(4/52)^2(48/52)^0 \approx 0.59\%$ — here with-replacement is actually higher, since drawing an ace without replacement makes the second ace harder to find, not easier.

Card Draw Terms You Should Know

Hypergeometric Distribution — the probability distribution for drawing a fixed number of items from a finite population without replacement, when that population splits into "matching" and "non-matching" groups.

Binomial Distribution — the probability distribution for a fixed number of independent yes/no trials at the same fixed probability each time; the correct model for card draws with replacement.

Without Replacement — a sampling method where each item drawn is not returned to the population, so the odds for every subsequent draw depend on what's already been drawn.

Population — the full set being drawn from; for a card draw, the deck itself, sized N with K matching cards inside it.