Gaming · Card game odds
Card combo draw probability calculator.
Enter your deck size, how many copies of each combo piece you run, and which pieces each combo needs; the calculator gives the exact chance you hold at least one complete combo by your chosen turn. It counts every possible hand once using the multivariate hypergeometric formula, so overlapping combos are not double counted. Example: in a 60-card deck, a combo needing one of four A cards and one of four B cards is complete 26.5% of the time by turn 4 on the play (10 cards seen); adding a second combo from two 3-of pieces lifts the chance of at least one combo to 40.2%.
Deck and combos
Result
Working out the result.
How the combo probability is calculated
A shuffled deck of N cards has C(N, n) equally likely sets of n cards. The multivariate hypergeometric formula from Wikipedia gives the chance of seeing exactly k1 copies of piece 1, k2 of piece 2 and so on: the product of C(Ki, ki) over every kind, divided by C(N, n), where Ki is how many copies of kind i are in the deck. Every card that is not a named piece counts as one more kind. The tool walks every split of your n cards across the pieces, keeps the splits that satisfy at least one combo, and adds their probabilities. Because each distinct split is counted once, two combos that can be completed together are not double counted.
Cards seen is the opening hand plus one card per turn. The player on the play skips the draw on turn one, so on the play by turn 4 you have seen 7 + 3 = 10 cards, and on the draw 7 + 4 = 11.
Worked examples with real numbers
- One piece, four copies, opening 7 of 60. Exact: 39.9% chance of at least one. The shortcut 4 × 7 ÷ 60 gives 46.7%, which overcounts because hands holding two copies get counted twice.
- Combo 1 = one A and one B (4 copies each), turn 4 on the play, 10 cards seen. 26.5%.
- Add combo 2 = one C and one D (3 copies each). Combo 2 alone is 17.2%. Added naively the two give 43.7%, but 3.5% of hands finish both, so the exact chance of at least one combo is 40.2%.
Chance of at least one of those two combos, turn by turn
| Turn | Cards seen (play) | Chance (play) | Cards seen (draw) | Chance (draw) |
|---|---|---|---|---|
| Turn 1 | 7 | 22.7% | 8 | 28.5% |
| Turn 2 | 8 | 28.5% | 9 | 34.3% |
| Turn 3 | 9 | 34.3% | 10 | 40.2% |
| Turn 4 | 10 | 40.2% | 11 | 45.9% |
| Turn 5 | 11 | 45.9% | 12 | 51.5% |
| Turn 6 | 12 | 51.5% | 13 | 56.8% |
| Turn 8 | 14 | 61.8% | 15 | 66.4% |
| Turn 10 | 16 | 70.7% | 17 | 74.6% |
What this calculator leaves out
It assumes a fully shuffled deck and no card that searches, draws extra, or mulligans you into a new hand. It treats all copies of a piece as identical, and it does not model order, so it cannot say whether the pieces arrive in time to be played, only whether you have seen them. The result is a chance of having seen the cards, not of being able to cast them. For the hand-keeping decision use the mulligan probability calculator, and for a single card the card draw probability calculator.
Common mix-ups
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Often assumedTo get the chance of at least one of 4 copies in a 7-card hand from a 60-card deck, multiply 4 copies by 7 cards and divide by 60.
What the source saysThat shortcut gives 46.7%. The hypergeometric formula, which counts each distinct hand once, gives 39.9%, because hands with two or more copies are counted once rather than once per copy. [1]
-
Often assumedThe chance of finishing either of two combos is the two chances added together.
What the source saysAdding gives 43.7% for the example, but the formula counts each hand once, and 3.5% of hands finish both combos, so the exact chance is 40.2%. [1]
Checked against these sources on .
Frequently asked questions
How do I calculate the probability of drawing a combo?
Work out how many cards you will have seen, then count every hand that contains all the pieces. The multivariate hypergeometric formula does that exactly: multiply the ways to choose each piece, divide by the ways to choose the whole hand. This tool does it for you.
Why not just multiply the odds of each piece?
The draws are not independent: each card drawn changes what is left. Multiplying separate odds ignores that and overcounts or undercounts depending on the deck. The hypergeometric method accounts for it.
How many cards have I seen by turn 4?
In a typical 7-card opening hand game, 10 cards on the play (one draw each on turns 2, 3 and 4) and 11 cards on the draw. Change the hand size and turn order above to match your game.
Can a combo need two copies of the same card?
Yes. In a combo row enter 2 under that piece. The tool then only counts hands with at least two copies of it.
What if two combos share a piece?
Enter that piece with a number in both combo rows. Shared cards are counted once per hand, so a hand completing both combos is not counted twice.
Does this account for mulligans or draw effects?
No. It assumes a random sample of cards from the top of a shuffled deck. Extra draws can be modelled by raising the opening hand size; the mulligan calculator covers the hand-keeping step.
Sources
Retrieved
- Wikipedia: Multivariate hypergeometric distribution — probability mass function: product of C(Ki, ki) over kinds divided by C(N, n) (read 2026-10-26)
Odds of having seen the cards, not of casting them
The probabilities are exact for a randomly shuffled deck and the draws you describe, as computed on 26 October 2026. They do not model play order, costs or card effects. Nothing you type is sent or stored.