I'm making a tabletop roleplaying game which uses a handmade deck of cards as a chance generator. All we need to know is that there are 16 cards and four suits, but:

  • Each card is of two suits;
  • Six cards in the deck are of two different suits;
  • And one card in the deck is of the same suit twice.

Thus, our cards look like this (with Hearts highlighted as an example):

Card Suits
1 Hearts and Clubs
2 Hearts and Spades
3 Hearts and Diamonds
4 Hearts and Hearts
5 Clubs and Hearts
6 Clubs and Spades
7 Clubs and Diamonds
8 Clubs and Clubs
9 Spades and Hearts
10 Spades and Spades
11 Spades and Diamonds
12 Spades and Clubs
13 Diamonds and Spades
14 Diamonds and Diamonds
15 Diamonds and Hearts
16 Diamonds and Clubs

If I choose Hearts as my suit, and turn over a Hearts card, that's called a success. If I turn over the card with both suits as Hearts, that's two successes.

I need to calculate the probability of getting at least n successes when I draw x number of cards without replacement. For instance, it is quite common to need two successes and to draw three cards. The following sets of cards would all qualify:

  • One Hearts card, one Hearts card, and one other card (2 successes)
  • One double Hearts card and two other cards (2 successes)
  • One double Hearts card, one Hearts card, and one other card (3 successes)
  • Three Hearts cards (3 or 4 successes)

My knowledge of probability maths is basic at best, and I've set up what feels like quite a challenging system, so I'm getting nowhere calculating it. I'd be grateful for the specific probability of getting 2 successes on 3 cards, but a general formula for me to calculate successes would be even better!

  • $\begingroup$ In your example of qualifying draws, you've missed the possibility of 4 successes in 3 cards, if you draw the double Heart and two more Hearts. Also, I assume we draw without replacement here? Writing out the closed-form exact solution of no-replacement problems can be tricky since there are many branches based on the sequence of what you pick - it may be easier to do a Monte Carlo approximation and just simulate it empirically if you know a little programming. $\endgroup$ Feb 24, 2021 at 17:07
  • $\begingroup$ Yes, we draw without replacement (sorry, I mention that in the question, but in the middle of a bunch of other stuff!) And yes, you're right, I'll edit that! $\endgroup$ Feb 24, 2021 at 17:15

2 Answers 2


There are $6$ single-Heart cards and $1$ double-Heart card, so to get at least $n$ successes in $x$ draws, we must either

  • Draw at least $n$ single-Heart cards and no double-Heart card, or
  • Draw the double-Heart card and get at least $n-2$ single-Heart cards in the other $x-1$ draws.

In the first case, if we are to draw $k$ single-Heart cards, there are $\binom 6k$ ways to choose them, and there are $\binom 9{x-k}$ ways to choose the remaining $x-k$ cards from among the $9$ non-Heart cards, so there are $$\sum_{k=n}^x{\binom 6k\binom 9{x-k}}$$ ways in all.

We can do the second case in much the same way. The double-Heart cards must be chosen, and then we have $\binom 6k\binom 9{x-1-k}$ ways to choose $k$ single-Hearts and $x-1-k$ non-Hearts, giving $$\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}$$

There are $\binom{16}{x}$ ways to draw $x$ cards, so the probability is $$\frac1{\binom{16}{x}}\left(\sum_{k=n}^x{\binom 6k\binom 9{x-k}}+\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}\right)$$


In your specific case, it is probably easier to find the probability of drawing one or zero hearts.


  • zero hearts: $\dfrac{9 \choose 3}{16 \choose 3}=0.15$

  • one heart: $\dfrac{{6 \choose 1}{9 \choose 2}}{16 \choose 3}\approx0.386$

and then subtract both from $1$ to give $\frac{13}{28}\approx 0.464$

  • $\begingroup$ This was very helpful to get a quick calculation, thank you! $\endgroup$ Feb 25, 2021 at 13:57

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