A bridge hand void in one suit

Question

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 ?

Answer 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.

Answer 2

Here, I use another way (differing from 1 and 2) for getting the answer. The solution method can be simply used to answer similar questions. I also compare the problem with other related problems.

For each $i=1,2,3,4$, let $A_i$ be the event of that a specific hand has no card of the $i\text{th}$ suit (i.e., one of $\spadesuit$, $\diamondsuit$, $\clubsuit$, and $\heartsuit$).

Then, the probability that one hand is void in exactly one hand is 4 times of the following:

$$ P\left (A_1\cap A'_2\cap A'_3\cap A'_4 \right )= P(A_1)-P\left (A_1\cap(A_2\cup A_3 \cup A_4) \right )=P(A_1)-{3 \choose 1}P(A_1\cap A_2)-{3 \choose 2}P(A_1\cap A_2\cap A_3)+{3\choose 3} P(A_1\cap A_2 \cap A_3\cap A_4) . $$

This follows from the relation $P(CD')=P(C)-P(CD)$, the De Morgan's law, and the inclusion-exclusion principle.

The probabilities can be computed as follows:

$$ P(A_1)= \dfrac{ \binom{39}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} }= \dfrac{ \binom{39}{13}}{ \binom{52}{13} } $$

$$P(A_1\cap A_2)=\dfrac{ \binom{26}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} }= \dfrac{ \binom{26}{13}}{ \binom{52}{13} }$$

$$P(A_1\cap A_2 \cap A_3)=\dfrac{ \binom{13}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} }= \dfrac{ 1}{ \binom{52}{13} }$$

$$P(A_1\cap A_2 \cap A_3\cap A_4)=\dfrac{ 0}{ \binom{52}{13} }.$$

Hence, the final answer is

$$\binom{4}{1}\times P\left (A_1\cap A'_2\cap A'_3\cap A'_4 \right )=4\frac{\binom{39}{13}-3\binom{26}{13}+3}{\binom{52}{13}}=0.05096725.$$

This solution method can be simply used to answer similar questions, e.g., a specific hand is void in exactly $k$ suits. For $k=2$, we have

$$\binom{4}{2}\times P\left (A_1\cap A_2\cap A'_3\cap A'_4 \right)$$

where

$$ P\left (A_1\cap A_2\cap A'_3\cap A'_4 \right )= P(A_1\cap A_2)-{2 \choose 1}P(A_1\cap A_2\cap A_3)+{2 \choose 2}P(A_1\cap A_2\cap A_3\cap A_4). $$

For a better comparison with other variants of this problem, also consider the related probability that a specific hand is void in at least one suit, computed below:

$$ P\left ( \cup_{i=1}^4A_i \right )={4 \choose 1}P(A_1)-{4 \choose 2}P(A_1\cap A_2)+{4\choose 3} P(A_1\cap A_2 \cap A_3)-{4\choose 4} P(A_1\cap A_2 \cap A_3\cap A_4)=0.051065521. $$

The answer of this problem is the sum of the answers to the previous problem where $k$ is 1, 2, or 3.

$$\binom{4}{1} P\left (A_1\cap A'_2\cap A'_3\cap A'_4 \right)+\binom{4}{2} P\left (A_1\cap A_2\cap A'_3\cap A'_4 \right)+\binom{4}{3} P\left (A_1\cap A_2\cap A_3\cap A'_4 \right)$$

Note that the void probability given in Wikipedia 3, i.e. 0.0512, is not accurate.

Another related and challenging probability that at least one hand is void in at least one suit is studied here 4, which is 0.18376632718730682.