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!
Best Answer
There are $6$ single-Heart cards and $1$ double-Heart card, so to get at least $n$ successes in $x$ draws, we must either
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 $$\frac1{\binom{16}{x}}\left(\sum_{k=n}^x{\binom 6k\binom 9{x-k}}+\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}\right)$$