# Bridge Hand Probabilities

Each of the 4 players in the game of bridge get dealt 13 cards. One player and his partner can see they hold 8 of the heart cards so they know that the 2 remaining hands they can't see hold the remaining 5 heart cards. What is the probability that the hearts are split 0:5 among the other 2 players?

If these other 2 players are called A and B, is it correct to treat each heart card as a Bernoulli trial and the probability for each one that player A has it is 0.5. Then a 0:5 split would be $$2$$ x $$0.5^5$$ since either player A or player B could be the one with all 5 hearts. Am I right in thinking this or is there more going on?

• Is the fact that "One player and his partner can see they hold 8 of the heart cards" independent from the distribution of the heart cards with the opponents? – Sextus Empiricus Oct 14 '18 at 15:54
• @MartijnWeterings The way I interpret the Q, the first pair can see all their 26 cards once they have been dealt out and know they have 8 of the 13 hearts - thus, the other pair must have 5 hearts but don't know how they were split between them. So I think there is independence with the hearts of the opponents i.e. it boils down to how many ways the 5 hearts could be split among the opposing sides 2 hands of 13 cards. – Ben Collister Oct 14 '18 at 16:18
• So players from pairs can see/know each other's hands? – Sextus Empiricus Oct 14 '18 at 18:46

A little more, I think. I'm not a bridge player, but as I understand it, among the remaining 26 cards there are 5 Hearts and 21 non-Hearts. Then the probability that player A was dealt 0 or 5 Hearts among 13 cards is hypergeometric:

$$\frac{{13\choose 0}{13 \choose 5} + {13\choose 5}{13\choose 0}}{{26\choose 5}} = 0.03913 < 2(.5)^5 = 0.0625.$$

sum(dhyper(c(0,5), 5, 21, 13))
[1] 0.03913043
2*choose(13,5)/choose(26,5)
[1] 0.03913043


You are using a binomial model, which does not take account of the non-Hearts. (For every Heart Player A receives, s/he must receive one less non-Heart.) In the figure below, the correct hypergeometric model is shown by vertical bars and the binomial model by small circles.

• I think I agree that this makes much more sense as a model to me - thank you for your help. – Ben Collister Oct 14 '18 at 16:40
• In practice, you would probably need to play hundreds of games before seeing a difference between the two models. For a small sample from a large finite population, the binomial model can be used as an approximation to the hypergeometric. (Here, 5 out of 26 does not quite qualify as a small sample from a large population; some authors use a rule of thumb that sample must be less than 10% of population size.) – BruceET Oct 14 '18 at 16:46