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bridge probability question (OT)

Posted By: eXtreme Gammon
Date: Thursday, 15 December 2011, at 9:53 p.m.

In Response To: bridge probability question (OT) (Chuck Bower)

(Finally) the probability question: Author Alder begins his analysis with the statement: "The odds slightly favor two aces being split between the defenders, not both in the same hand." Is this statement correct/accurate for the situation at hand? If not, explain why, optionally correcting the statement so that it is both accurate and relevant (if possible). If his statement is correct, is there a better way to say it -- i.e. was his justification misleading?

Bridge players know (or should know) that the probability that 2 specific cards of the opponents are in the same hand is 48%.

Here is the calculation: there are 26 unknown card, let's pick the side that has one of the 2 card we are concerned with. What we want to know is how many distribution of 12 cards out of 25 have the 2nd card of interest.

  • C(25,12) is the number of how to pick 12 card out of 25
  • if you put the card on that player hand, there are only 24 cards left and you need to pick 11: so it's C(24,11)

    That come as a probability of
    P=C(24,11)/C(25,12)=48%

    The number is round as C(n,p)=n!/(p!.(n-p)!) : P=24!/(11!.13!) / ( 25!/(12!.13!) )
    P=24!/(11!)/25!.(12!)
    P=12/25

    That's why if you are missing Kx in a suit you should run the A rather than doing a finesse (if you have no other information of course)
    Intrestingly, as the number of card goes down, the probability of the 2 cards being together goes down: with only 4 cards left it's 1/3.

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