Card Draw Probability (Hypergeometric)
Probability of drawing N of M target cards from a deck of any size.
Overview
The Card Draw Probability tool applies the hypergeometric distribution to deck-drawing problems. Tell it the size of the deck, the number of "success" cards in that deck, the number of cards you draw, and the threshold for what counts as a win, and it returns the exact probability of meeting that threshold.
It works equally well for ordinary 52-card decks, custom card games, lottery setups, and Magic: the Gathering or other TCG opening-hand questions. Because it computes the closed-form hypergeometric probability — not a simulation — the answer is exact to floating-point precision regardless of deck size.
How it works
The hypergeometric distribution models sampling without replacement. The probability of drawing exactly k successes from a deck of N cards containing K success cards in n total draws is the ratio of two combinations: pick k successes from K and (n - k) failures from (N - K), divided by the total ways to pick n from N. The tool computes this for each k from the lower threshold up to the smaller of n and K, then sums to get cumulative probabilities like "at least 2" or "no more than 3".
Combinations are evaluated using log-gamma to avoid overflow on large factorials, then exponentiated back to the probability. This means decks of tens of thousands of cards work without numerical issues.
Examples
- Drawing at least one Ace in a 5-card poker hand from a standard 52-card deck: 1 - C(48,5)/C(52,5) = about 34.1 percent.
- An MTG deck of 60 cards with 24 lands: probability of opening with exactly 3 lands in 7 cards = C(24,3) * C(36,4) / C(60,7) = about 30.9 percent.
- A 40-card commander side deck containing 4 win conditions: probability of at least one in the top 10 = 1 - C(36,10)/C(40,10) = about 70.0 percent.
- Lottery-style: pick 6 from 49, all 6 must match — probability = 1/C(49,6) = about 1 in 13.98 million.
FAQ
What's the difference between hypergeometric and binomial?
Binomial assumes replacement (each draw is independent); hypergeometric assumes drawing without replacement, which is what card games actually do.
Why does removing a card from the deck change the odds for the rest?
Because the deck is finite, each draw shifts the composition of the remaining cards. Hypergeometric captures that conditional shift exactly.
Can I model mulligans or scrying?
This tool answers single-shot questions. Multi-step decisions need a custom Monte Carlo, but the per-step probabilities can be combined manually.
Does the order of cards matter?
No. Hypergeometric counts unordered combinations, which is the right model for "any time in the next n draws".
What if the deck size and success count are equal?
You'll always draw all successes; the probability of "at least K" is 1 and of "fewer than K" is 0.