On Tuesday 30 May 2006 12:59, dasimov@earthlink.net wrote:
What is the probability there is some pair of suits your bridge hand does not include, and your partner's hand does not include the other pair?
1. Consider a particular pair: you have no S,H and partner has no D,C. Number of ways to arrange this: (26 C 13) ways to choose your hand, (26 C 13) independently to choose partner's, and then (26 C 13) for dividing the remaining cards between your opponents. Total number of such deals: (26 C 13)^3. 2. There are (4 C 2) = 6 pairs of suits, so naively 6.(26 C 13)^3 deals with the claimed property. 3. This counts some deals more than once. Which ones? You can't have no S,H and also no D,C. So multiple counting arises from situations like this: you have no S,H and no H,D; partner has no D,C and no S,C. Thus: you have only one suit and partner has another. For any given pair of suits there are (26 C 13) ways to do this; each such deal has been counted (it turns out after a little scribbling) twice instead of once under (1,2) above. 4. So the total number of deals is 6.(26 C 13)^3 - 12.(26 C 13), out of (52 C 13;13;13;13), or 54086240179999 / 429820857815004205200, or about 1.26 * 10^-7. Thus, if you play an average of one hand of bridge a day for 50 years, the chance that you'll see this at least once is about 0.0023. (There may be any number of mistakes in the foregoing.) -- g