Probability everyone else has a club when I have two clubs?

Question

A problem in probability I was toying with:

Using the standard 52-card set, four cards are dealt to four players. Given that Player 1 has exactly two clubs, what is the probability that the other three players each have at least one club?

I'm imagining counting the number of permutations of the deck that would result in such hands, then dividing by $52!$. So far, I'm fairly sure that there are $$\underbrace{\binom{13}{2}}_{\text{club}}\cdot \underbrace{\binom{39}{2}}_{\text{non-club}} = \frac{13\cdot 12}{2}\cdot \frac{39\cdot 38}{2} = \frac{231192}{4}=57798$$ possible hands for Player 1. This leaves a 48-card deck with 11 clubs and 37 non-clubs from which to draw hands for the other 3 players. But from here, I'm not sure how to continue counting.

A Python simulation suggests the probability is around $26\%$.

Answer 1

Outline/Major Hints:

Without loss of generality, suppose that player $1$ was dealt specifically $A\clubsuit, 2\clubsuit, A\heartsuit, 2\heartsuit$.

You have then $11$ remaining clubs and $37$ remaining non-clubs to distribute. We consider all ways to distribute these $48$ cards among the second, third, and fourth players as our new restricted sample space.

Let $A,B,C$ represent the events that the second, third, and fourth players have no clubs respectively.

We are interested in calculating $P(A^c\cap B^c\cap C^c)$, i.e. the second, third, and fourth players all have at least one club (given that player one has exactly two clubs)

By De'Morgan's: $P(A^c\cap B^c\cap C^c)=P((A\cup B\cup C)^c) = 1-P(A\cup B\cup C)$

By inclusion-exclusion: $\dots = 1-P(A)-P(B)-P(C)+P(A\cap B)+P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)$

Continue by calculating each of $P(A),P(A\cap B),P(A\cap B\cap C)$ and plugging in to calculate the probability you are after.

$P(A)$ is the probability that player $2$ receives no clubs out of the $11$ clubs and $37$ non-clubs that are being distributed. There are $\binom{48}{4}$ possible hands that player $2$ could have gotten (given that player $1$ received the hand he did) and $\binom{37}{4}$ of which have no clubs. The probability $P(A)$ is therefore $\binom{37}{4}/\binom{48}{4}$. Continue similarly for the rest.

$~$

I calculate an exact probability of $\frac{37960141}{144841028} \approx 0.2620814110764251\dots$ seemingly confirming your simulation. I leave it to you to come up with the algebraic expression that I used to arrive at this result given my earlier hints.