A bridge hand void in one suit

A bridge hand consists of 13 cards from a standard deck of 52 cards.

What is the probability of getting a hand that is void in exactly one suit, ie consisting of exactly 3 suits ?

Solution 1:

Corrected: Thanks to the OP for querying my previous answer, and to joriki for pointing out that I was counting the wrong thing.

There are $\binom{52}{13}$ ways of choosing $13$ cards, all equally likely. We will be finished if we count the number of hands that have exactly one void. This number is $4$ times the number of hands void in $\spadesuit$ only.We now proceed to count these.

The number of hands void in $\spadesuit$ is $\binom{39}{13}$. This overcounts the hands void in $\spadesuit$ alone. To adjust, we use the
Inclusion-Exclusion Principle.

How many hands are void in both $\spadesuit$ and $\heartsuit$? Clearly $\binom{26}{13}$. The same is true for $\spadesuit$ and $\diamondsuit$, and for $\spadesuit$ and $\clubsuit$. So from our first estimate of $\binom{39}{13}$ we subtract $3\binom{26}{13}$.

But we have subtracted too much. We need to add back the number of hands that are void in all but one of $\heartsuit$, $\diamondsuit$, or $\clubsuit$. There are $3$ of these. Thus the number of hands with exactly one void is $$4\left(\binom{39}{13}-3\binom{26}{13}+3\right).$$

Comment: From the "practical" point of view, we could have stopped with the first term, since in a well-shuffled deck multiple voids have negligibly small probability compared to single voids.