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?

  • $\begingroup$ 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? $\endgroup$ Oct 14, 2018 at 15:54
  • $\begingroup$ @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. $\endgroup$ Oct 14, 2018 at 16:18
  • $\begingroup$ So players from pairs can see/know each other's hands? $\endgroup$ Oct 14, 2018 at 18:46
  • $\begingroup$ @SextusEmpiricus In Bridge, after the auction and after the first card is played, the 13 cards of one player called Dummy are placed face up so all can see them, and is played by Dummy's partner who can also see their own 13 cards $\endgroup$
    – Henry
    Oct 27, 2021 at 22:51

1 Answer 1


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
[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.

enter image description here

  • $\begingroup$ I think I agree that this makes much more sense as a model to me - thank you for your help. $\endgroup$ Oct 14, 2018 at 16:40
  • $\begingroup$ 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.) $\endgroup$
    – BruceET
    Oct 14, 2018 at 16:46

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