# Probability in a game of bridge

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 ?

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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
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 $\binom{52}{13}$ ? – Belgi Nov 11 '12 at 22:34
@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. – André Nicolas Nov 11 '12 at 22:53
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. – Ross Millikan Nov 11 '12 at 23:02
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 – Belgi Nov 11 '12 at 23:16