A bit of background (can skip this), I play this card game called Coup and in this game there are $ 6 $ types of characters with the Duke being the best card at the start of the game, there are $ 3 $ Dukes in the deck of $ 18 $. I usually play with $ 2 $ others and each person gets dealt $ 2 $ cards each, so $ 6 $ cards get dealt.
In the deck of $ 18 $ cards there are $ 6 $ types of cards (for simplicity just $ A,B,C,D,E,F $), for each type of card there are $ 3 $ identical cards. I want to know the probability of when dealt $ 6 $ cards at least one has an $ A $ on it. I tried to work this out with the binomial formula however this gave a very low chance of drawing $ 4 $ $ A $ cards which is impossible because there are only $ 3 $ $ A $ cards in the deck. Did I do something wrong? And how can prevent to make this mistake in the future?
Thanks a lot!
There are ${18 \choose 6}$ ways to choose 6 cards from a deck of 18, and ${15 \choose 6}$ ways to choose 6 cards from the 15 non-A cards. So the chance of drawing at least one A-card is $$1-\displaystyle\frac{{15 \choose 6}}{{18\choose 6}}.$$