# 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 ?

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If this is homework, please add the homework tag. – Dilip Sarwate Jan 11 '12 at 19:23
It is not homework. – true blue anil Jan 12 '12 at 4:49

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.

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If you write out those terms explicitly, you get $32489701776 - 62403600 + 4$, confirming the Comment. – Michael Lugo Jan 12 '12 at 6:02
Andre, this is what i thought the # of favorable ways would be initially, but there is an unexpected twist ! And, of course, we are seking the exact figure. – true blue anil Jan 12 '12 at 6:27
@Andre Nicolas: More careful analysis reveals that the # of ways should be C(4,1)*C(39,13) - 2*C(4,2)*C(26,13) + 3*C(4,3)*C(13,13) – true blue anil Jan 12 '12 at 7:32
@Andre Nicolas: Well, take some time. And while at it, can you (or anyone else who sees it) give some guidance as to cases where adjustments need to be made while applying inclusion-exclusion, which is why i posted the question ! – true blue anil Jan 12 '12 at 8:12
@Andre Nicolas: I have added certain calculations in math.stackexchange.com/questions/98671/… . You might like to have a look at it. – true blue anil Jan 13 '12 at 10:00