RE: [math-fun] Bridge probability puzzle
I get Joshua's and Mr. McCaughan's answer to 14 decimal places, but it differs beyond that. There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards. Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places) I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-) Bill C. -----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Joshua Zucker Sent: Tuesday, May 30, 2006 10:32 AM To: math-fun Cc: dasimov@earthlink.net Subject: Re: [math-fun] Bridge probability puzzle 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi Bill, dunno if you remember me from EPGY tutoring/ARML days ... I see Bob is winning all kinds of math awards out there these days! He's off to college this fall, is that right? Anyway, the correction you're missing is that if you have only ONE suit, then partner can have any two of the remaining three. So the probability is very slightly higher for that reason. And it's so slight because there are a LOT more two-suited hands than one-suited hands. --Joshua On 5/31/06, Cordwell, William R <wrcordw@sandia.gov> wrote:
I get Joshua's and Mr. McCaughan's answer to 14 decimal places, but it differs beyond that.
There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards.
Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places)
I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-)
Bill C.
-----Original Message----- From: math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+cordwell=sandia.gov@mailman.xmission.com]On Behalf Of Joshua Zucker Sent: Tuesday, May 30, 2006 10:32 AM To: math-fun Cc: dasimov@earthlink.net Subject: Re: [math-fun] Bridge probability puzzle
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
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William Cordwell wrote:
There are (4 C 2) = 6 ways to pick which two suits you have; this also fixes the suits of your partner. You have, from these two suits, (26 C 13) ways of getting your 13 cards; your partner also has (26 C 13) ways of getting his 13 cards from the other two suits. You then divide by (52 C 13) ways of getting your hand randomly, and divide also by (39 C 13) ways of your partner getting his hand randomly from the remaining 39 cards.
Thus, P = 6 * (26 C 13) * (26 C 13) / [ (52 C 13) * (39 C 13) ] = 21850/173640942909 = (acc'ding to Mathematica)1.25834377733429657622523386340107748418239154034425 x 10^(-7)(forty places)
I'm not convinced that this is right, but I'd like to know why, if not. What are the odds of getting the same answer to 14 digits, but differing otherwise? :-)
In a problem like this? Very high. :-) You've over-counted a few deals: the ones in which you and your partner have only *one* suit apiece. You say "pick which two suits you have", but in that situation there's more than one way to do that picking while still getting the same deal. You have therefore counted those deals multiple times (twice, as it turns out): once for each way of picking "which two suits you have". -- g
participants (3)
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Cordwell, William R -
Gareth McCaughan -
Joshua Zucker