$4$ people $(A,B,C,D)$ receive $13$ cards each from a normal $52$ deck cards. If $A$ and $B$ have combined $9$ hearts, what is the probability of either $C$ or $D$ alone having the other $4$ hearts in their hand?
My way of thinking was:
there are 5 possibilities:
1) C has 0 hearts and D has 4 hearts
2) C has 1 hearts and D has 3 hearts
3) C has 2 hearts and D has 2 hearts
4) C has 3 hearts and D has 1 hearts
5) C has 4 hearts and D has 0 hearts
The only cases that satisfy are case 1 and 5. Therefore, my probability is 2/5.
People have pointed out that my way of thinking is wrong. What is the right answer to this problem?
Best Answer
Consider it this way:
The group formed by A and B have 9 hearts and another $13*2-9=17$ cards.
The rest of cards ($13*2=26$) are received by C and D. This 26 cards contain the 4 hearts remaining.
Now the question is how many combinations can you made of 13 cards taken from 26 that contains the available 4 hearts.
Take those 4 hearts apart. We have 22 cards remaining. 9 of them go to C (or D).
So C (or D) receives 9 among 22.
Calculate the number of combinations $N= \binom {22}{9}$ These are the "favorable" cases, because all of them have 4 hearts.
The total cases are $\binom {26}{13}$
So, the probability is $\frac {\binom {22}{9}}{\binom {26}{13}}$
But this is the probability for C. The question told "C or D". Then the final probability is $P= 2·\frac {\binom {22}{9}}{\binom {26}{13}}$
As @Barry Cipra commented, it can also be seen as "C has 4 hearts" $N_0=\binom {22}{9}$ or "C has no hearts (D has them) $N_1=\binom {22}{13}$.
And because $N_0=\binom {22}{9} = N_1=\binom {22}{13}$ we have $N=N_0+N_1=2N_0$ and hence a factor of 2.