# Combinatorics for a 15 card deck. Not choosing two sets with 6 drawn?

I have a combinatorics problem for a game I am playing that I can't seem to wrap my head around. If anyone could take a stab on it that would be greatly appreciated!

I have a deck of 15 cards. The deck is composed of 5 sets of 3 cards:

AAA BBB CCC DDD EEE

I shuffle the deck and draw 6 cards. What are the odds that I do not draw all of A or all of B? With drawing all of A being a failure, all of B being a failure, and all of A and B being a failure.

I tried to follow combinations for 52 card decks, but I've been messing up when it comes to choosing how to set up the x choose y sections.

You have $$15$$ cards. There $${15\choose6}$$ ways to choose $$6$$ cards. There are $${12\choose3}$$ ways to choose all the A's, because after drawing all the A's, you still have to choose $$3$$ of the remaining $$12$$ cards.

Can you finish it now?

• Got it, thank you! The only thing I needed to add was removing the double-counted case, which I got from Austin's answer. Mar 11, 2019 at 19:09

How many 6-card hands contain all the A's? You're forced to use three of the cards in your hand to be the A's. From the remaining 12 cards in the deck, you may now freely select any 3, giving a total of $$\binom{12}{3}$$ possible hands.

The number of hands containing all the B's is also $$\binom{12}{3}$$ by an analogous argument.

If we simply add these outcomes together, there is exactly one that gets counted twice: AAABBB. Correcting for this overcount, there are $$\binom{12}{3} + \binom{12}{3} - 1 = 439$$ hands having all the A's or all the B's.

Since there are $$\binom{15}{6} = 5005$$ total 6-card hands, we get a probability of $$439 / 5005 \approx 8.77\%$$ of drawing all the A's or all the B's (and therefore a $$91.23\%$$ likelihood that this does not occur).

• Thank you! This makes a lot more sense now. The last piece after reading the other answer was the double-count. Mar 11, 2019 at 19:11