# Calculating the probability of drawing two cards of the same suit from a deck of dual-suit cards

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
3 Hearts and Diamonds
4 Hearts and Hearts
5 Clubs and Hearts
7 Clubs and Diamonds
8 Clubs and Clubs
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!

• 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. Feb 24, 2021 at 17:07
• 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! Feb 24, 2021 at 17:15

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.

So

• 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 $$\frac{13}{28}\approx 0.464$$

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