I think the answer is :C(51,12)-C(38,12).Since one card is fixed in this 13 cards.I just need to choose 12 cards from 52-1=51 cards.And rather than listing all the possibilities of choosing at least 1 diamond,I choose it in the opposite way.That is ,assuming that no diamonds are in the 12 cards.In this case, I choose 12 cards from 51-13=38(exclude all the diamonds) and it gives C(38,12).At last, I just need to subtract C(38,12) from C(51,12) .Is that correct?
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2Nicely reasoned. – N. F. Taussig May 17 '15 at 16:50
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Check out this site or this site. IIRC they give sound looking data about bridge related distributions and their probabilities. – Jyrki Lahtonen May 18 '15 at 17:19