Bridge Hand Probability
Date: 12/02/2008 at 21:31:40 From: Chandramouli Subject: Bridge hand Your right hand opponent opens 1 No Trump; you have 9 cards in two suits [5 in one and 4 in the other]. What is the probability of your partner having 4 cards in one of your two suits? There are bidding conventions that show that you have 2 suits. I want to know the utility of such conventions. How often these two suit advertisement yield positive results. One view is that they show off more than they yield.
Date: 12/09/2008 at 00:46:16 From: Doctor Vogler Subject: Re: Bridge hand Hi Chandramouli, Thanks for writing to Dr. Math. You were probably just trying to give all of the relevant information, but actually you made the problem much harder when you gave the opening bid. You see, it's a simple matter of combinatorics and probability to answer the question, "If I have exactly 5 cards in one suit and 4 cards in another suit, what is the probability that my partner has at least 4 cards in one of those two suits?" But when you also give the opponent's bid, that changes the probabilities, probably only very slightly, but in a very complicated way! What I'm going to do is count the number of ways to distribute the cards such that your partner has enough cards in the right suits. But if you want to account for the opponent's bid, then you have to decide, first of all, under what circumstances (what his hand contains) would he bid that way, and, secondly, how many of the ways to distribute the cards such that your partner has enough cards in the right suits also have a bidding hand in the opponent's hand. I won't account for that, and I expect that the difference would be very small anyway. Since the particular suits don't matter, just fer terminology, we'll say that you have 5 spades and 4 hearts. Then there are 8 other spades and 9 other hearts among three other hands. We don't really care what the opponents have, so how many ways are there to make up your partner's hand from the remaining 39 cards? The answer is 39-choose-13, or 39!/(13!*26!) = 8,122,425,444. How many of those ways leave your partner with 4 or more spades or hearts? Well, there are 8-choose-4 times 31-choose-9, or (8!31!)/(4!4!9!22!), ways for your partner to have exactly 4 spades. There are (8!31!)/(4!4!9!22!) + (8!31!)/(3!5!8!23!) + (8!31!)/(2!6!7!24!) + (8!31!)/(1!7!6!25!) + (31!)/(5!26!) = 1411205250 + 441768600 + 73628100 + 5890248 + 169911 = 1932662109 ways for him to have 4 or more spades. And since 1932662109/8122425444 = 0.2379, that means he has a 23.79% chance of having 4 or more spades. Similarly, there are (9!30!)/(4!5!9!21!) ways for your partner to have exactly 4 hearts, and (9!30!)/(4!5!9!21!) + (9!30!)/(5!4!8!22!) + (9!30!)/(6!3!7!23!) + (9!30!)/(7!2!6!24!) + (9!30!)/(8!1!5!25!) + (30!)/(4!26!) = 1802700900 + 737468550 + 171007200 + 21375900 + 1282554 + 27405 = 2733862509 ways for him to have 4 or more hearts, which gives a 33.66% chance. But some of these intersect, so we should also count the number of ways for him to have 4 or more hearts AND 4 or more spades. For example, the number of ways that he has EXACTLY 4 of each is (8!9!22!)/(4!4!4!5!5!17!). There are lots of other combinations, too, and you should add them all up. Then since this would be counted in both of the two above sums, the probability that your partner has 4 spades OR 4 hearts is equal to the probability that he has 4 hearts PLUS the probability that he has 4 spades MINUS the probability that he has both 4 hearts and 4 spades. I'll leave that last part up to you. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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