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 ?