52 cards are equally given to 4 players Find probability that one of them has 3 spades out of remaining 5. 52 cards are equally given to 4 players $A,B,C,D$.
Together, $A$ and $B$ have a total of 8 spades among them,
and we have to find probability that $C$ has 3 spades out if remaining 5.
In this question, the sample space reduces to 44 cards after we eliminate 8 spades from the cards, so now we have 5 spades left from which 3 will be given to $C$, and the remaining 10 cards can be chosen from the remaining 39 cards so I get the answer as
$$ \frac{\binom{5}{3}\binom{39}{10}}{\binom{44}{13}},$$
is this approach correct?
 A: There are only 26 cards left in the deck. 5 are spades, 21 are not. We use the $\frac{Number- of -favorable -outcomes}{Number- of- possible- outcomes}$ formula. Thus we get $\frac{{5 \choose 3}{21 \choose 10}}{{26 \choose 13}}$.
A: Question: In the card game bridge, the 52 cards are dealt out equally to 4 players - called East, West, North and South. If North and South have total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?
My Approach: My answer to the above question was:
East can get 3 spades out of remaining 5 spades in $\binom{5}{3}$ ways. 
All possible outcomes are $\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}$. i.e., East gets no spades, or 1 spade or... and so on. 
Therefore, the desired probability = $\frac{\binom{5}{3}}{\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}}= \frac{10}{32} = .3125$
However, as per Sheldon Ross' book, the answer is 0.339. I understand how he arrived at the solution. But, I do not see what is wrong in the above approach. Why don't the answers match?
A: It is given that A and B have a total of 8 spades among 26 cards.
∴ In the remaining 26 cards, there are exactly 5 spades.
These 26 cards are distributed equally among C and D [13 each]
P(C has 3 of 5 spades) =  $\frac{{5 \choose 3}{21 \choose 10}}{{26 \choose 13}}$
= 0.339
