Three of a kind and a pair on hands bigger than 5

Question

Given a standard deck and a random $n$-cards hand the probability of having two of one rank and three of another is given by:

for $n=5$: $$ \frac{{13 \choose 1} {4 \choose 3} {12 \choose 1} {4 \choose 2} }{{52 \choose 5}}$$

(In words, I choose one rank, for which there are 13 ways, then there are 4 ways to get three of a kind in this rank. Next choose any of the remaining ranks and count the ways you can get a pair in that rank ${4 \choose 2}$. Finally divide by the total number of possible hands.)

But what about $n=15$?

Using similiar (naive) logic, just adding that the 10 extra cards can be any combination of the remaining 47, I get a non-sensical value: $$ \frac{{13 \choose 1} {4 \choose 3} {12 \choose 1} {4 \choose 2} {47 \choose 10}}{{52 \choose 15}} > 1$$

My own simulation shows the actual answer is in the neighbourhood of $0.66$.

What is the failed step I have taken?

EDIT1: corrected a typo in the denominator $49 \choose 10$ to $47 \choose 10$...

Answer 1

You have overcounted hands with multiple pairs and three of a kinds, besides that the last factor in your logic should be $47 \choose 10$. If you think of a hand with two three of a kinds and two pairs plus five other cards, you could take either three of a kind to be the one you count and also either pair, so you will count this hand four times. You have also counted cases where there are four of a kind four times for the three of a kind.

Answer 2

To have exactly three of a kind once and a pair once with $n=15$, you need to pick the rank that has three of a kind, the rank that has two of a kind, the ranks that occur once, and then assign suits: $$ \frac{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}{10\choose 9}4^9}{52\choose 15}=\frac{9814671360}{4481381406320}\approx 0.0022.$$

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To have at least one rank that occurs with at least three cards and at least one other rank that occurs with at least two cards, we would instead first determine the complement: To have at most one multiple rank, or two multiple ranks that are bot pairs.

a) only one multiple rank, this being three-of-a-kind: $$ \frac{{13\choose 1}{4\choose 3}{12\choose 12}4^{12}}{{52\choose 15}}=\frac{872415232}{4481381406320}$$ b) only one multiple rank, this being four-of-a-kind: $$ \frac{{13\choose 1}{4\choose 4}{12\choose 11}4^{11}}{{52\choose 15}}=\frac{654311424}{4481381406320}$$ c) two pairs (and nothing else): $$ \frac{{13\choose 2}{4\choose 2}^2{11\choose 11}4^{11}}{{52\choose 15}}=\frac{11777605632}{4481381406320}$$ Thus the final answer is $$1-\frac{872415232+654311424+11777605632}{4481381406320}\approx 0.997 $$

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Possibly ither interpretations are possible, but their results are presumably somewhere between the above numbers ...