What's the difference between a hand and drawing one at a time?

A hand dealt at once only depends on final composition and uses combinations, while drawing one at a time in a specific order uses a chain of shrinking dependent probabilities.

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Card Probability Calculator

Covers single-card draws, drawing several cards at once (like a poker hand), and drawing cards one at a time without putting them back, the case that needs a different formula than dice or coins because the deck itself changes after every draw.

Choose a scenario

MATCHING CARDS IN THE DECK

TOTAL CARDS IN THE DECK

Example: 13 hearts out of a standard 52-card deck.

Your step-by-step card probability will appear here once you calculate.

Why cards need a different formula than dice or coins

Every dice roll and coin flip is independent: the deck of possibilities resets completely between attempts. Cards almost never work this way. Once a card is drawn and not returned, the deck is permanently smaller and has a different mix of remaining cards. This single fact is why card probability problems generally use the hypergeometric distribution rather than the binomial distribution used for dice and coins, even though the two formulas look structurally similar.

The formulas for each scenario

Single Card Probability

P(match) = matching cards / deck size

The simplest case. 13 hearts in a 52-card deck gives 13/52, which simplifies to 1/4.

Hypergeometric Distribution

P(X=k) = C(K,k) × C(N−K, n−k) / C(N,n)

For drawing n cards at once from a deck of N, where K cards in the deck match your target. C(K,k) counts ways to pick the matches, C(N−K, n−k) counts ways to fill the rest of the hand, and C(N,n) is every possible hand.

Sequential Draws Without Replacement

P(all match) = (K/N) × ((K−1)/(N−1)) × ...

Each successive draw uses a shrinking deck and, if the previous draws matched, one fewer matching card too. The denominator and numerator both shrink by 1 with each draw.

A worked example: 2 hearts in a 5-card hand

N = 52 (deck size), K = 13 (hearts in the deck), n = 5 (cards drawn), k = 2 (target hearts).

C(13,2) = 78, the number of ways to choose 2 hearts from the 13 available. C(39,3) = 9,139, the number of ways to fill the remaining 3 hand slots from the 39 non-hearts. C(52,5) = 2,598,960, the total number of possible 5-card hands.

P(exactly 2 hearts) = (78 × 9,139) / 2,598,960 ≈ 0.274, about 27.4%. Notice how different this is from treating each card as an independent 13/52 draw and applying the binomial formula, which would overstate the deck's consistency across draws.

A worked example: three hearts in a row

Drawing 3 hearts in a row, one at a time, without replacement: the first draw has 13/52 hearts available. If it's a heart, the second draw has 12 hearts left out of 51 total cards. If that's also a heart, the third draw has 11 out of 50.

Multiply the three fractions: (13/52) × (12/51) × (11/50) ≈ 0.0129, about 1.3%. Each successive fraction shrinks both top and bottom, which is exactly what "without replacement" means in practice.

Mistakes people make with card probability

Using the binomial formula for a fixed deck. Binomial assumes the same probability on every draw, which only holds if cards are replaced and reshuffled between draws. For a single hand or a shrinking deck, hypergeometric is the correct tool. The two give similar answers when the deck is very large relative to the number of draws, but diverge noticeably for smaller decks or larger hands.

Forgetting that a "hand" is not the same as "one at a time." A 5-card poker hand dealt all at once is a hypergeometric problem about combinations. Asking for the probability of specific cards arriving in a specific order is a sequential, dependent-draw problem instead, even though both start from the same deck.

Treating suits, ranks, and colours as independent categories. A card belongs to one suit and one rank simultaneously. "Probability of a red card or a king" needs the addition rule with the overlap subtracted (the two red kings), not simple addition.

Questions people actually ask about this

Why can't I just use the binomial formula for card hands?

Binomial assumes every draw has the same probability, which requires replacing and reshuffling the card each time. Dealing a hand removes cards permanently, changing the odds for every subsequent card, which is exactly what the hypergeometric distribution is built to handle.

How different are binomial and hypergeometric results in practice?

For a small number of draws from a very large deck, the two get close, since removing one card barely changes the remaining proportions. For a standard 52-card deck and a 5-card hand, the difference is large enough to matter, which is why poker odds calculations always use hypergeometric-style counting.

What's the difference between "a hand" and "one at a time"?

A hand dealt all at once only cares about the final composition, not the order cards arrived in, so it's a combinations-based hypergeometric problem. Drawing one at a time and asking about a specific sequence cares about order, so it's a chain of shrinking dependent probabilities instead.

Does shuffling between hands reset the calculation?

Yes. Each freshly shuffled, full deck restarts the probabilities from scratch. The shrinking-deck math only applies within a single deal from a single shuffle.