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A deck has n = 52.
Must not be larger than n.
52!
52! / 47!
P(n, k) / 5! = 311,875,200 / 120
Drawn without replacement: no card appears twice.
The same seed and settings always reproduce exactly the same results.
Which cards count as “targets” in the question.
Whole number from 1 to 52.
5 × 13/52
| Target cards drawn (k) | Ways | Probability |
|---|---|---|
| 0 | 575,757 | 22.1534% |
| 1 | 1,069,263 | 41.1420% |
| 2 | 712,842 | 27.4280% |
| 3 | 211,926 | 8.1543% |
| 4 | 27,885 | 1.0729% |
| 5 | 1,287 | 0.0495% |
Ways = C(13, k) × C(39, 5 − k), out of C(52, 5) = 2,598,960 equally likely draws.
Permutations count ordered selections: P(n, k) = n! / (n − k)!. Combinations ignore order: C(n, k) = n! / (k! (n − k)!). All values are computed with exact big-integer arithmetic, so even 52! is shown without rounding.
The draw-count tool uses the hypergeometric distribution, which describes sampling without replacement. If K of the 52 cards are targets and you draw n cards, the probability of exactly k targets is C(K, k) × C(52 − K, n − k) / C(52, n). Its mean is n K / 52. The random generator uses a seeded shuffle, so you can share a seed and another person can reproduce your draws exactly. See the methodology page for details.