# Thread: Bridge Probability Question/Inclusion, Exclusion

1. ## Bridge Probability Question/Inclusion, Exclusion

Hello:

How do we find the number of bridge hands that are weak in a suite (that is, lacking in either an ace, King or Queen) in a suite?

Would we want to do the following:

(1) find the number of hands that are lacking in a suite ->
Say, the suite is hearts
Since you have 52 cards, would you have then 52 - 3 cards to choose from.
So you would have (49 choose 13) ways of choosing a hand that is lacking in hearts. Then, since there are (4 choose 1) ways to choose which suite to be lacking in, total you have 4 x (49 choose 13) ways for a hand to be lacking in a suite.
(2) But then you have overcounted, and you must substract out the ones that are lacking in 2 suites, and then you have (46 choose 13) for those.
(3) Then you must add back in those that are lacking in 3 suites:
(2 choose 0) x (43 choose 13)
(4) Then you must substract out the number that is weak in 4 suites, which is (3 choose 0)*(40 choose 13)
(5) The next thing to do is to add back in the number that is weak in 5 suites, but that is impossible, and hence you are done.

2. Originally Posted by bodensee9
The next thing to do is to add back in the number that is weak in 5 suites, but that is impossible, and hence you are done.
Try this out: $\displaystyle \sum\limits_{k = 1}^4 {\left( { - 1} \right)^{k+1} \binom{4}{k} \binom{52-3k}{13}}$.

The last I played bridge there were only four suits.