Probability - reduced sample space 
In a bridge game, 52 cards are dealt equally to players E, W, N, S.
If N and S have a total of 8 spades among them, what is the probability that E has 3 of the remaining 5 spades?

And the answer is:
$$\frac{(^5C_3)(^{21}C_{10})}{^{26}C_{13}}=0.339$$
Reasoning goes:

*

*Choose 3 spades for E (first term in numerator)

*Choose the other 10 cards out of the 21 cards (second term in numerator)

*Denominator: choose 13 out of 26 cards for E.

And so my question is: what is the reduced sample space here actually? Why is it 26 cards as the sample space for both numerator and the denominator? Why not 52 cards (i.e. all four players)?
 A: You need to find the probability that E has 3 of the remaining 5 spades. Now between them N and S  have a total of 8 spades, out of their combined total of 26 cards. This is why you have a reduced sample space, since you are now looking for the probability of the spades being distributed in a "smaller deck" of 26 cards between E and W, and can ignore the "deck of 26 cards" held by N and S.
Then we have that $^5C_3$ is the number of different combinations of $3$ spades out of the remaining $5$ (which is the number of spades we need for E). Then out of the $26-5=21$ remaining non-spade cards you have $^{21}C_{10}$ combinations to choose from to make up Es other $10$ cards. So now multiply these together to give the number of all possible hands that give E 3 spades, and then divide by the total number of combinations of $13$ cards from $26$, which is $^{26}C_{13}$, to give the probability, which gives the answer you have.
A: Another way to solve the problem using a reduced sample space focuses only on the $5$ spades
Each of E and W have $13$ slots available, and we want $3$ spades for E, $2$ for $W$
$\boxed{.}\boxed{.}\boxed{s}\boxed{.}\boxed{s}\boxed{s}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\quad\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{.}\boxed{s}\boxed{.}\boxed{s}\boxed{.}\boxed{.}\quad$
Thus $Pr = \dfrac{\binom{13}{3}\binom{13}{2}}{\binom{26}{5}}$ 
