TITLE: Conditional Probability: Bridge Hand given north and south has 8 spades
QUESTION [9 upvotes]: i apologise as i'm pretty certain i could find the actual answer myself on these boards (as in the solution to the problem) but if possible i'd like it if someone can confirm whether my thinking is correct.
"in the card game bridge, the entire 52 cards are dealt out equally to 4 players, E/W/N/S
if N and S have a total of 8 spades among them, what is the probability E has 3 of the remaining 5 spades?"
this is example 2c on page 60 in "A First Course in probability 8th edition"
and gives the solution as $$\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}\approx .339$$
later in the book it asks to recalculate this using conditional probability (question 3.3 page 102)
Compute the conditional probability that E has 3 spades given North and South has a combined total of 8 spades.
my answer is then as follows:
let E be the event that E has 3 spades. and F the event that N and S has 8 spades.
then $P(E)=\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}$
as it is given that 26 cards have already been delt to N and S, of which 8 of the 13 spades have also been delt. this means for E to occur we have to choose 3 of the remaining 5 spades and any combination of 10 cards remaining from the 21 left over (that is 26 remaining cards minus the 5 spades). all divided by the reduced sample space.
(this is one of the parts in which i want to be certain)
as we only care about E we can consider N and S to be one person which will be delt 26 cards, 8 of which must be spades, giving
$P(F)=\frac{{{13}\choose{8}}{{39}\choose{19}}}{{52}\choose{26}}$
this reasoning seems right to me as the two of them should recieve any combination of 26 cards so long as 8 of them are spades. (we dont care if N has all 8 or if S has all 8)
then our conditional Probability is
$$P(E|F) = \frac{P(E\cap F)}{P(F)}$$
my major issue is i'm not actually sure how i would go about calculating $P(E\cap F)$ in this instance, logically i argue that the two events P and F are independent of each other, meaning
$$P(E\cap F)=P(E)P(F)$$ and so making our final conditional probability $P(E|F)=P(E)$ which yes does give me the right answer, but unless the question specifically states that these are independant of each other i would then have to prove that $P(E\cap F)=P(E)P(F)$ holds surely?
any clarification would be greatly appreciated. thank you.

REPLY [8 votes]: For calculating $P(E\cap F)$ think of the number of ways that $13$ spades can be divided in this context such that exactly $3$ come in hands of $\bf E$ and exactly $2$ in hands of $\bf W$. 
There are $\binom{13}3\binom{13}2\binom{26}8$ ways, and this on a total of $\binom{52}{13}$ ways. 
This results in: $$P(E\cap F)=\frac{\binom{13}3\binom{13}2\binom{26}8}{\binom{52}{13}}$$
Sortlike you can find $P(F)$ as:$$P(F)=\frac{\binom{26}5\binom{26}8}{\binom{52}{13}}$$
Working this out with application of $P(E\mid F)P(F)=P(E\cap F)$ leads to:$$P(E\mid F)=\frac{13!13!5!21!}{3!10!2!11!26!}=\frac{\binom53\binom{21}{10}}{\binom{26}{13}}$$