I'd like to ask for help with a task that I've been given.
The task goes like this: In a poker game (52 cards), 5 players are given 5 cards each. What is the probability that both the first and the third player will get a full-house?
It's apparent that the probability of the first player getting a full house is $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}}$
Then the second player gets 5 random cards, so $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5}$
Now the third player is supposed to get a full house, I'm assuming it should look like this: $\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}}$
The fourth and the fifth player both receive 5 random cards, so the final equation should be:$\frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{52}{5}} * \binom{47}{5} * \frac{\binom{13}{1} * \binom{4}{3} * \binom{12}{1} * \binom{4}{2}}{\binom{42}{5}} * \binom{37}{5}* \binom{32}{5}$
This is how I would solve the task, but I'm not sure that my solution for the third player is correct. Could anyone more experienced than me have a look at my solution?
Also I'm sorry if there is a solution for this type of task on this site, I tried looking for it and I found a lot of poker probability questions, most of them were however focused on only one player and not multiple players, which is my case.
You don't have to worry about any players but the first and third. (Imagine if you like, that the first player picks up his hand first, and the third player picks up his hand second).
Your calculation for the first player is correct. Your calculation for the third player is not. There are now $11$ ranks possible for the three of a kind in the third player's hand, because there aren't three cards available in either of the ranks in the first player's hand. For the pair in the third player's hand we have two possibilities, it can be one of the ten ranks not used yet, or it can be the same rank as the pair in the first player's hand. We have to take the sum of the two probabilities. For the first, we have $$ \frac{11\binom{4}{3}10\binom{4}{2}}{\binom{47}{5}}$$ For the second, have we $$\frac{11\binom{4}{3}}{\binom{47}{5}}$$ since there is only one way to choose the same pair as the first player.
Add these up and multiply by the probability for the first player.
P.S. Multiplying by $\binom{47}{5}$ for the second player is just wrong. This is the number of hands he can draw, not a probability.