What is the probability of drawing a four of a kind when 20 cards are drawn from a deck of 52? Yesterday my housemates and I were playing card games and someone popped this question. We tried to solve the problem, but we couldn't figure it out. This morning I woke up and I am still wondering how to solve it. Will you please help me out?    
 A: To draw at least $k$ specified four-of-a-kinds, we must draw all $4k$ required cards. This is a hypergeometric distribution, where we must draw all $4k$ successes from population of size $52.$ There are $\binom{13}{k}$ such sets of four-of-a-kinds. Therefore, the chance of getting at least $k$ four-of-a-kinds is
$\binom{13}{k} \frac{\binom{4k}{4k}\binom{52-4k}{20-4k}}{\binom{52}{20}} = \binom{52}{20}^{-1} \binom{13}{k} \binom{52-4k}{20-4k} ,$ for $0\leq k\leq5.$
By the inclusion-exclusion principle, the probability of drawing at least one four-of-a-kind is therefore equal to 
$\binom{52}{20}^{-1} \sum_{k=1}^5 (-1)^{k+1} \binom{13}{k} \binom{52-4k}{20-4k} = -\binom{52}{20}^{-1} \sum_{k=1}^5 (-1)^k \binom{13}{k} \binom{4(13-k)}{4\times 8} .$
This can be calculated numerically to be about $0.2197706.$
The above sum has the form $\sum_{k=0}^n (-1)^k \binom{n}{k} \binom{r(n-k)}{rm},$ if we subtract the $k=0$ term afterwards, since the terms for $5<k\leq 13$ are equal to zero. I wonder if there's a way to simplify that kind of sum.
A: There are 13 kinds, so we can solve the problem for a single kind and then move forward from there.
The question then is, what is the probability of drawing 4 successes (like kings) in 20 samples from the same distribution of 4 successes (kings) and 48 failures without replacement?
The hypergeometric distribution (wikipedia) gives us the answer to this question, and it is 1.8%.
If one friend bets on getting 4 kings, and another bets on getting four queens, they both have 1.8% chance of winning. We need to know how much the two bets overlap in order to say what the probability is of at least one of them winning.
The overlap of both winning is similar to the first question, namely: what is the probability of drawing 8 successes (kings and queens) in 20 samples from a distribution of 8 successes (kings and queens) and 44 failures, without replacement?
The answer is again hypegeometric, and by my calculation it's 0.017%.
So the probability of at least one of the two friends winning is 1.8% + 1.8% - 0.017% = 3.6%
In continuing this line of reasoning, the easy part is summing the probabilities for individual kinds (13*1.8%=23.4%), and the difficult part is to figure out how much all of these 13 scenarios overlap.
The probability of getting either 4 kings or 4 queens or 4 aces is the sum of getting each four-of-a-kind minus the overlap of them. The overlap consists of getting 4 kings and 4 queens (but not 4 aces), getting 4 kings and 4 aces (but not 4 queens), getting 4 queens and 4 aces (but not 4 kings) and of getting 4 kings and 4 queens and 4 aces.
This is where it gets too hairy for me to continue, but proceeding this way with the hypergeometric formula on wikipedia, you can go ahead and write it all out.
Maybe somebody can help us reduce the problem? 
