[math-fun] Bridge probability puzzle
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? --Dan
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?
210823 / 1675400449149, perhaps? -- Don Reble djr@nk.ca -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.
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
I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. --Joshua Zucker
I'll try an alternative counting method and see if I get the same answer as one of these other folks. Their two answers agree to about 7 significant figures, 0.00000012583437773342733107186438969809670960643871... vs 0.00000012583439386511151359617331968028508948288906.... Anyway, here's my approach. First of all, you could have only one suit: the probability of that happening is 4 possible hands out of (52 C 13). If that's the case, then the probability that partner has only two of the other three suits is, let's see, three ways to choose which two suits, times (26 C 13) ways of choosing the cards, out of (39 C 13) possible hands. But that overcounts the ways where partner has only one suit, counting each twice (say, if you have only clubs and partner has only spades, that's counted as one of the ways with hearts and spades and as one of the ways with diamonds and spades). So subtract the ways that partner has only one suit, 3 ways. On the other hand, you might have two suits: (4 C 2) ways to choose which two suits, times (26 C 13) ways to choose your cards, minus the two ways where you only have one suit, out of (52 C 13). Then partner has to have the other two suits, (26 C 13) out of (39 C 13) being that probability. So my final calculation looks like 4 / (52 C 13) * (3 * (26 C 13) - 3)/(39 C 13) + (4 C 2) * ((26 C 13) - 2) / (52 C 13) * (26 C 13) / (39 C 13) 54086240179999/429820857815004205200 is then my answer. This agrees with Gareth's answer, so that seems promising! --Joshua Zucker
participants (4)
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dasimov＠earthlink.net -
Don Reble -
Gareth McCaughan -
Joshua Zucker