Probability of Void Hand in Card Game of BridgeDate: 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 52! ------ = 635,013,559,600 13!39! 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 39! ------ = 8,122,425,444. 13!26! Well, that was easy! So we divide and get approximately 8,122,425,444 --------------- = 0.01279 635,013,559,600 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? 26! ------ = 10,400,600 13!13! 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 13! ----- = 1 13!0! 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 up: The number of hands void in at least one suit is 39! 26! 4 ------ - 6 ------ + 4. 13!26! 13!13! And this divided by 52! ------ 13!39! 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/