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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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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.2012-01-12
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    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.2012-01-12
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    @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)2012-01-12
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    @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 !2012-01-12
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    @Andre Nicolas: I have added certain calculations in http://math.stackexchange.com/questions/98671/non-routine-application-of-inclusion-exclusion . You might like to have a look at it.2012-01-13
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    @André: You've counted the number of hands void in *at least* one suit, since you only subtracted the overcount of the multiply void ones, whereas to get the number with exactly one suit you'd have to subtract the entire count of the multiply void ones.2012-01-13
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    @joriki: Thanks, I had lost track of what was being counted.2012-01-13
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    @true blue anil: Thank you for your persistence. Inclusion-Exclusion worked just fine, with no adjustment. But it was being used correctly to count the wrong thing.2012-01-13