The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Probability of Void Hand in Card Game of Bridge

Date: 12/02/2004 at 03:10:48
From: naseer ahmad
Subject: (no subject)

What is the probability of being dealt a bridge hand void in a 
specified suit from an ordinary deck of 52 playing cards?

Date: 12/02/2004 at 10:00:56
From: Doctor Vogler
Subject: Re: (no subject)

Hi Naseer,

Thanks for writing to Dr. Math.  The probability of something is the
number of ways it works divided by the total number of ways possible.
There are a total of

  ------  =  635,013,559,600

different possible bridge hands.  So we just need to count how many
have a void in a specific suit.  Let's say spades.  There are 39
non-spade cards, so the number of ways to get a hand with 13 of those
39 cards is

  ------  =  8,122,425,444.

Well, that was easy!  So we divide and get approximately

  --------------- = 0.01279

or about a 1.279 percent chance.  Of course, if the person before you
bids "pass," then that makes it less probable that he has a
distributional hand, which makes it less probable that you have a
distributional hand, and therefore the probabilities change at that
point.  But it's 1.279% as the cards are being dealt.

If, instead, you asked what was the probability of a hand being void
in at least one suit, then that would be different.  There are four
ways to pick a suit, and each will have the same number of hands void
in it.  So we multiply the above by four and divide by the number of
total hands, right?  Well, almost.  If you do that, you will be off
only by a tiny fraction of a percent, but it won't be exactly right. 
You see, that would count the number of hands void in spades, and add
the number of hands void in hearts, and the number void in diamonds,
and the number void in clubs.  But it will count twice every hand
that's void in TWO suits!  So how many of those are there?

  ------  =  10,400,600

for any particular two suits, six times that in total (for all six
pairs of suits).  Now how many times have we counted hands void in
three suits?  Well, first we counted them three times (once for each
suit) and then we subtracted them off again for each of the three
pairs of suits they are void in, which means we haven't counted them
at all!  So we still need to count how many hands are void in three
suits.  Well, that's

  -----  =  1

for any particular three suits (obviously, it is only the hand that
has every card in the remaining suit), or a total of 4 hands for all
four suits.  There are no hands void in four or more suits.  So we sum

  The number of hands void in at least one suit is

        39!          26!
    4 ------  -  6 ------ + 4.
      13!26!       13!13!

  And this divided by

is the probability of being dealt a bridge hand being void in at least
one suit.

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 Probability

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.