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in a game of contract bridge, partner and me together have 11 cards in a suite. The remaining 2 cards in the same suite can be distributed amongst our opponents as either 1-1 or 2-0.

What is the probability that it will be distributed as 1-1 and what is the probability it will be distributed as 2-0?

Once the above is solved, how can you extend it if more cards are missing. That is, let us assume that partner and me have 8 cards between us. How do you calculate the probability for the distributions (5-0, 4-1, 3-2)?

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    A rough rule of thumb for uneven breaks is the chance that cards will break n-m is somewhat less than 100/(n-m)% for n>m+1. This overstates the chances somewhat and gets worse for bad breaks. So 4-1 is less than 33% (actually 28.26%) and 4-2 is less than 50% (actually 48.45)2010-11-16

2 Answers 2