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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
Associated Topics:
College Probability
High School Permutations and Combinations
High School Probability

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