I was given the following question:

In the game of bridge there are four players-A, B, C and D. Players A and C are partners and players B and D are partners. Each player gets $13$ cards. If one player and his partner have $9$ spades between them, what is the probability that the $4$ other spades are split three and one between the two other players?

My work:

I treat players A and C as one player - player $1$, and players B and D I treat as Player $2$.

I thought the answer should be $$2\cdot\frac{\binom{13}{12}\binom{39}{14}+\binom{13}{10}\binom{39}{16}}{\binom{52}{26}}$$

My reasoning is:

1) I double by $2$ because I assume that it is player $1$ with the given $9$ spades, but the problem is symmetric

2) If the spades are split in such a way then player $1$ have $10$ spades or $12$ spades. I then choose the spades player $1$ will have and I complete his hand to $26$ cards from the non-spade cards

3) $\binom{52}{26}$ is the number of ways to choose a hand for player 1

However, I was told that my answer is wrong (I used a calculator to compare with another answer which claims the probability is $0.5$).

Can someone please point out my mistake ? did I not account for something ?


You can ignore the partnership with 9 spades. The others have 4 spades and 22 non-spades between them. You can select a hand for one partner in ${26 \choose 13}$ ways. He can get 1 spade in ${4 \choose 1}{22 \choose 12}$ ways, and can get 3 spades in ${4 \choose 3}{22 \choose 10}$ ways. These are exclusive, so we can add them to get $\frac {{4 \choose 1}{22 \choose 12}+{4 \choose 3}{22 \choose 10}}{26 \choose 13}\approx 49.74\%$ as seen in Alpha

  • $\begingroup$ Why do you choose to ignore the partnership with 9 spades ? Isn't this the same calculation basically ? why the number of ways to choose a hand is $\binom{26}{13}$ and not $\binom{52}{13}$ ? $\endgroup$
    – Belgi
    Nov 11 '12 at 22:34
  • $\begingroup$ @Belgi: Because we don't care what their hands look like. We just give them 9 spades and 17 non-spades, leaving the deck I described to be distributed between the pair with 4 spades. $\endgroup$ Nov 11 '12 at 22:36
  • $\begingroup$ @Belgi: If you want to take the hands of $A$ and $B$ into account, you can think of this as a conditional probability argument. Given that $A$ and $B$ have certain hands, where the combined number of spades happens to be $9$, we can compute the probability of a $3$-$1$ or $1$-$3$ split for $C$ and $D$. This should be roughly (but not very roughly) the probability of $1$ head and $3$ tails or $3$ tails and $1$ head in $4$ tosses of a fair coin. $\endgroup$ Nov 11 '12 at 22:53
  • $\begingroup$ In fact it is very close, as the coin flip is exactly $50\%$ The only difference is you "use up spaces" in the hand that gets $3$ spades, which lowers the chance a bit. $\endgroup$ Nov 11 '12 at 23:02
  • $\begingroup$ Ross what do you mean by "use up spaces" ? I got everything down exept why don't we need to muliply by $2$...I'm always confused by that part $\endgroup$
    – Belgi
    Nov 11 '12 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.