Deck Odds, Explained
After reading this you will be able to compute the exact chance of drawing a combo piece in your opening hand, extend that chance turn by turn, and combine several requirements ("piece A and piece B by turn 4") without fooling yourself with a naive multiplication.
What deck odds actually measure
Every card game where you shuffle and draw is a sampling problem. You have a deck of known size, some cards you care about, and a hand drawn without putting cards back. The question is always the same shape: given the deck, what is the probability that the cards you drew satisfy some condition?
Here is the hook. A four-of in a 60-card deck feels like it should show up in your seven-card opening hand about 47% of the time, because 7 × 4/60 = 0.4667. That reasoning is wrong. The true chance is 39.9%. The gap comes from drawing without replacement: each card you draw changes what remains, and the naive product double counts the hands that hold two, three or four copies. The Deck Odds Calculator computes the exact figure so you never have to guess.
The tool describes a deck as groups of cards. You set the deck size, the hand size, and one or more groups such as "24 lands" or "4 combo pieces." It then answers questions like "at least one," "exactly two," or several of those at once.
When to use it, and when not to
Use it whenever the answer depends on counting cards in a shuffled deck. Good questions: the chance of a keepable opening hand, the odds of hitting your third land by turn three, the probability of assembling a two-card combo by a given turn, and how a mulligan shifts those numbers.
Do not reach for it when the outcome depends on choices made during play rather than on the shuffle. Whether an opponent has removal in hand, whether you sequence your plays correctly, or whether a scry lets you filter: those are decision and information problems, not pure draw problems. The calculator gives you the raw draw probability, which is the floor your play sits on, not the whole game.
Draw odds tell you how often the cards cooperate. They do not tell you how often you win. A 39.9% chance to see a combo piece by turn one is a fact about your deck; converting that into wins is a fact about your deck's plan and your play.
The formula and the intuition behind it
The engine is the hypergeometric distribution. You have a population of N cards, of which K are the ones you want. You draw n cards. The chance of getting exactly k of the wanted cards is:
Here N is the deck size, K the number of copies you want to draw from, n the number of cards drawn, and k the number of wanted copies in that draw. The symbol \binom{a}{b} is "a choose b," the number of ways to pick b items from a without regard to order.
The intuition: the denominator counts every possible hand. The numerator counts only the hands that hold exactly k good cards and n-k other cards. Divide, and you get the fraction of hands that qualify. For "at least one" you compute the complement: one minus the chance of zero copies, which is almost always the cleaner arithmetic.
Never multiply a per-card probability by the number of cards drawn. 7 × 4/60 gives 46.7% for a four-of in seven cards, but the real answer is 39.9%. The error grows as the wanted count grows, because you are counting the same card twice in different draw slots.
Worked example: the demo deck
A four-of in 60 cards, opening seven
The demo loads a 60-card deck, a 7-card opening hand, and a single group of 4 combo pieces. The question is: what is the chance the opening hand holds at least one of them?
- Set N = 60, K = 4, n = 7.
- Compute the chance of zero copies: P(X = 0) = \frac{\binom{4}{0}\binom{56}{7}}{\binom{60}{7}}.
- Numerically, \binom{56}{7} = 231{,}917{,}400 and \binom{60}{7} = 386{,}206{,}920.
- So P(X = 0) = 231{,}917{,}400 / 386{,}206{,}920 = 0.6006.
- The chance of at least one is
1 - 0.6006 = 0.3994, or 39.9%.
Compare that to the naive 0.4667. The difference of nearly seven percentage points is exactly the mass of hands with two or more copies, counted twice by the shortcut.
Odds across turns and the shape of the curve
Drawing does not stop at the opening hand. Each turn you draw one more card, so n grows: 7 on the opening hand, 8 by turn one on the draw, and so on. Feed the larger n into the same formula and you get a curve that climbs turn by turn.
For the four-of in 60, the at-least-one chance rises like this.
| Cards seen | Turn (on the draw) | P(at least one) |
|---|---|---|
| 7 | Opening | 0.3994 |
| 8 | 1 | 0.4416 |
| 10 | 3 | 0.5160 |
| 13 | 6 | 0.6079 |
| 17 | 10 | 0.7061 |
Combining requirements the right way
Single-card questions are easy. The hard ones ask for several things at once: "the combo piece and at least two lands," or "piece A and piece B by turn 4." You cannot just multiply the two separate probabilities, because the cards compete for the same slots in your hand. Drawing more lands leaves fewer slots for the combo piece.
The correct model is the multivariate hypergeometric. You split the deck into disjoint groups (piece A, piece B, everything else), then sum the probability over every combination of counts that satisfies your condition. The calculator does this exactly by convolution, so overlaps never distort the answer.
Two combo pieces by turn 4
Take a 60-card deck with two single copies, piece A and piece B, and 58 other cards. On the play you have seen 10 cards by turn 4. What is the chance you hold at least one A and at least one B?
- P(no A) among 10 seen: \binom{59}{10} / \binom{60}{60} style ratio equals
50/60 = 0.8333, so P(at least one A) is0.1667. - By symmetry P(at least one B) is also
0.1667. - Multiplying naively gives
0.1667 × 0.1667 = 0.0278. - The exact multivariate answer is P = \binom{58}{8} / \binom{60}{10} subtracted correctly, which gives
0.0254.
The exact 2.54% is lower than the naive 2.78% because the two pieces compete for the same ten slots. Whenever requirements share the draw, the naive product runs high.
Reading the results and trusting them
The tool shows two numbers side by side: the exact hypergeometric figure and an empirical rate from the Monte Carlo dealer, which shuffles and deals thousands of sample hands in front of you. Watch the empirical rate wobble at first, then settle toward the exact value. After 10,000 deals the empirical rate for the demo typically sits within about 0.5 points of 39.9%.
Interpret the exact number as a long-run frequency, not a promise about your next game. A 39.9% opener means that across many shuffles you will see the piece in roughly two openings out of five. Any single game lands on one side or the other.
When two methods should agree and do not, trust the exact formula and suspect the simulation setup. A Monte Carlo rate that stalls at the wrong value almost always means the sampling rule does not match the question you meant to ask.
Common mistakes
- Multiplying per-card odds by cards drawn
- This gives 46.7% instead of 39.9% for a four-of in seven. It ignores that copies share your hand.
- Multiplying separate requirements
- P(A) times P(B) overcounts because A and B compete for slots. Use the multivariate model.
- Forgetting the mulligan
- A London mulligan lets you draw a fresh seven and bottom cards. It raises your effective chance to keep a good hand well above the single-draw figure, and the tool's mulligan mode shows by how much.
- Counting cards seen wrong across turns
- On the play you draw no card turn one, so by turn 4 you have seen 10 cards, not 11. On the draw you have seen 11. Off-by-one here shifts every number.
Related tools
Deck odds is a counting problem where order does not matter. If you enjoy the exact-combinatorics style of thinking applied to physical objects, the Rubik's Cube Solver & Timer works the other side of the same coin: a finite state space you navigate with a deterministic solver rather than a random draw.
Frequently asked questions
Why is a four-of only 39.9% in the opening hand and not almost half?
Because you draw without replacement. The naive 7 × 4/60 counts hands with two or more copies multiple times. The exact hypergeometric answer removes that double counting and lands at 39.9%.
How many copies do I need for a roughly 50% opener in 60 cards?
Five or six copies. Five copies give about 44.9% in the opening seven, and six copies give about 49.2%. Seven copies cross 53%.
Does the calculator handle Commander and Hearthstone deck sizes?
Yes. Presets cover Magic (60/7), Commander (99/7), Hearthstone (30/3 to 4), Yu-Gi-Oh (40/5) and Pokemon (60/7). Change the deck and hand size and the same formula applies.
How many simulated hands should I watch before I trust the empirical rate?
The standard error of a proportion shrinks with the square root of the sample. At 10,000 deals the error on a 40% event is about 0.5 percentage points, which is close enough to confirm the exact figure.
Can it tell me the chance of exactly two lands rather than at least two?
Yes. "Exactly two" uses a single term of the formula with k = 2, while "at least two" sums the terms for k = 2, 3, \dots. The tool reports whichever you request.
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