We have 12 cards with four different suits: hearts, diamonds, spades, and clovers. Each suit has three cards with numbers from 1 to 3. We randomly deal these cards to three people, giving each person four cards. What is the probability that at least one person has four of a kind?
I came across this probability problem while playing 'Spoons' with my friend.
I tried to solve the problem and got the results, but I couldn't check if this is the right answer, so I'm asking here.
My answer is $$\frac{3\cdot 3{}\cdot \left({\binom{8}{4}-2}\right)+6}{\binom{12}{4}\cdot \binom{8}{4}}$$
We approach by Inclusion-Exclusion over the events "Player 1 has a 4 of a kind", "Player 2 has a 4 of a kind" and "Player 3 has a 4 of a kind", hereafter abbreviated as events $A,B,C$
The probability of event $A$ will be $\dfrac{3}{\binom{12}{4}}$. Yes, technically you could have called it $\dfrac{3\binom{8}{4}}{\binom{12}{4}\binom{8}{4}}$ but there is no reason to refer to the $\binom{8}{4}$'s as they will cancel out. The other players don't need to even really exist for this calculation.
Similarly we find the probabilities of $B$ and $C$.
We then turn our attention to $\Pr(A\cap B)$, the probability that both of the first two players simultaneously have a four of a kind. The only way this can happen is if all three have four of a kinds since the only remaining four cards for the third player will all be a match as well. We get a probability here of $\Pr(A\cap B)=\Pr(A\cap B\cap C)=\dfrac{3!}{\binom{12}{4}\binom{8}{4}}$
By inclusion-exclusion we have as a final answer:
$$\Pr(A\cup B\cup C)=\Pr(A)+\Pr(B)+\Pr(C)-\Pr(A\cap B)-\Pr(A\cap C)-\Pr(B\cap C)+\Pr(A\cap B\cap C)$$
which simplifies by symmetry to
$$=3\Pr(A)-2\Pr(A\cap B) = \dfrac{3\cdot 3\binom{8}{4}-2\cdot 3!}{\binom{12}{4}\binom{8}{4}}$$
or however else you decided to write or simplify the expression. Yes, your answer is correct (though it was not immediately obvious the line of logic you used to get there)