Here’s a probability problem which is causing me some trouble:
In a deck of $52$ cards, in how many ways can you draw $13$ cards such that at least one suit is missing.
What I've done at first was: $\binom{4}{1}\binom{13}{13}+ \binom{4}{2}\binom{26}{13} + \binom{4}{3}\binom{39}{13}$, but I think that in here, I'm counting some elements twice. I know about the Inclusion-Exclusion Principle but have no idea on how to apply it to this problem.
Thanks a lot for your help!
There are $\binom{4}{1}$ ways to exclude a suit and $\binom{39}{13}$ ways to choose the $13$ cards from the remaining three suits, giving an initial count of $$\binom{4}{1}\binom{39}{13}$$ However, we have counted those distributions in which two suits are missing twice, once for each way we could have designated one of the missing suits as the missing suit.
There are $\binom{4}{2}$ ways to exclude two suits and $\binom{26}{13}$ ways to choose the $13$ cards from the remaining two suits, so we must subtract $$\binom{4}{2}\binom{26}{13}$$ from our initial count.
However, doing so subtracts too much. We counted each case in which three suits are missing three times in our initial count, once for each way we could have designated one of the missing suits as the missing suit, and subtracted each case in which three suits are missing three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the missing suits as the missing pair of suits. Therefore, we have not counted those cases in which three suits are missing at all. We must add them to our total.
There are $\binom{4}{3}$ ways to exclude three of the four suits, and one way to choose all $13$ cards of that suit. Hence, there are $$\binom{4}{3}\binom{13}{13}$$ ways to exclude three of the suits.
By the Inclusion-Exclusion Principle, the number of $13$ card hands from which at least one suit is missing is $$\binom{4}{1}\binom{39}{13} - \binom{4}{2}\binom{26}{13} + \binom{4}{3}\binom{13}{13}$$ Therefore, the probability that at least one suit is missing when $13$ cards are drawn from a standard $52$-card deck is $$\frac{\dbinom{4}{1}\dbinom{39}{13} - \dbinom{4}{2}\dbinom{26}{13} + \dbinom{4}{3}\dbinom{13}{13}}{\dbinom{52}{13}}$$
Yes, there's some double counting going on. Certainly $${4\choose3}{39\choose13}$$ covers all the cases, but cases where only two suits occur have been counted twice. (For example, hands with only Hearts and Spades, have been counted as having only Clubs, Hearts and Spades, and also as having only Diamonds, Hearts, and Spades.) So we must subtract these, giving $${4\choose3}{39\choose13}-{4\choose2}{26\choose13}$$ Now, what about hands with only one suit? The have been added in ${3\choose2}$ times, and subtracted out ${3\choose1}$ times, so we haven't counted them at all, and we need to add them in. The final answer is $${4\choose3}{39\choose13}-{4\choose2}{26\choose13}+{4\choose1}{13\choose13}$$
