Explanation for solution of a combinatorial problem

Question

The given problem is:

An ordinary deck of cards is dealt to four people: Joe, Bob, Jim, and Larry. If Larry has exactly one ace, what is the probability that Jim has all the remaining aces?

My solution: $\frac{\binom{4}{1}\binom{51}{12}\binom{36}{10}\binom{26}{13}}{\binom{52}{13}\binom{39}{13}\binom{26}{13}}$

This equates to $\frac{22}{703}$, which is apparently the right answer. However, is the solution right? It made so much sense to me when I was answering it, but now that I take a better look, it seems like the $\binom{51}{12}$ in the numerator should be replaced by $\binom{48}{12}$, since it's like setting aside $4$ of the aces, which means that there are $48$ cards left to choose from.

My solution looks weird. If it's wrong, please tell me it's wrong. If it's right, please explain why because I don't understand my own thoughts and need to be saved from them.

Answer 1

I think your answer $\frac{\binom{4}{1}\binom{51}{12}\binom{36}{10}\binom{26}{13}}{\binom{52}{13}\binom{39}{13}\binom{26}{13}}$ is wrong but you got the right result, by chance.

It looks like you've selected any of the $4$ aces, then $12$ other cards among the remaining $51$ for Larry.

Then, you gave the $3$ aces to Jim along with $10$ other cards among $36$.

After that, you gave $13$ cards among $26$ to any of the other players.

And you forgot to give the remaining $13$ cards to the last player.

In any case, this problem is equivalent to give $10$ cards among $36$ and $3$ aces to Jim, which is equivalent to $$\frac{\binom{36}{10}}{\binom{39}{13}}=\frac{22}{703}$$

Answer 2

The question is:

$P(\text{Jim 3 aces }| \text{Larry 1 ace})=\frac{P(\text{Jim 3 aces, Larry 1 ace})}{P(\text{Larry 1 ace})}=\frac{P(\text{Jim 3 aces, Larry 1 ace, Bob 0 aces, Joe 0 aces})}{P(\text{Larry 1 ace, Bob 0 aces, Joe 0 aces})}$.

So how can we give Jim 3 aces, Larry 1 ace, Bob 0 aces, Joe 0 aces. Start with Larry:

We have 4 ways to deal him an ace, so that gives the $\binom{4}{1}$. Then we have 48 cards left to deal to him (no aces) and pick 12 from them $\binom{48}{12}$. Then look at Jim. We can choose 3 aces from 3 cards, so $\binom{3}{3}$ and we have left then 10 cards from 36 cards.

Now look how to give Larry one ace, while we don't make any restriction for Jim. That is the option to choose again one ace, and twelve cards, hence $\binom{4}{1}$ and $\binom{48}{12}$. Now to choose 13 cards for Jim gives then $\binom{39}{13}$.

Note that we don't have to restrict to Joe and Bob.

So what we get is

$\frac{\binom{4}{1} \binom{48}{12} \binom{3}{3} \binom{36}{10}}{\binom{4}{1} \binom{48}{12} \binom{39}{13}}=\frac{22}{703}$.