Calculating $Pr(E\cap F)$, let us define our sample space as all ways of distributing the cards where order within each hand doesn't matter and where north and south's hands are considered collectively as one. There are $\frac{52!}{26!13!13!}$ such distributions, each of which are equally likely to occur.
Let us count $|E\cap F|$ in this sample space. To do so, first choose which eight spades north/south got and then which 18 non-spades north/south got.
Then, choose which three spades from those remaining east got and which 10 non-spades from those remaining east got. All remaining cards will be given to west.
There are then $\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}$ such arrangements implying the probability $Pr(E\cap F)$ is:
$$Pr(E\cap F)=\frac{\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}}{52!/(26!13!13!)}=\frac{15152709}{276092852}\approx 0.05488$$
Continuing on to calculate $Pr(F)$, now we temporarily instead consider the sample space to be just the ways in which we divide the deck in half, i.e. giving 26 cards to north/south collectively where order within the hand is unimportant and giving the remaining cards to east/west. There are $\binom{52}{26}$ such ways to do so, each of which are equally likely to occur.
We count $|F|$ in this sample space as being $\binom{13}{8}\binom{39}{18}$, thus making $Pr(F)=\frac{\binom{13}{8}\binom{39}{18}}{\binom{52}{26}}=\frac{44681065}{276092852}\approx 0.161833$
Taking the ratio $\frac{Pr(E\cap F)}{Pr(F)}$ will arrive at the same answer as given before after simplifications.
$$Pr(E\mid F)=\frac{Pr(E\cap F)}{Pr(F)}=\frac{39}{115}\approx 0.3391304$$
As an aside, let us look at $Pr(E)$ and inspect whether $E$ and $F$ truly are independent.
We let the sample space be all ways in which East is given a hand ignoring how the rest of the cards are distributed. There are $\binom{52}{13}$ ways in which this can happen. Choosing which three spades and which 10 nonspades gives us $\binom{13}{3}\binom{39}{10}$ possible hands and a probability of $Pr(E)=\frac{\binom{13}{3}\binom{39}{10}}{\binom{52}{13}}\approx 0.28633$
(notice that no mention of how spades are distributed among north/south appeared in the calculation of $Pr(E)$. You seem to have confused calculating $Pr(E)$ with $Pr(E\mid F)$)
We compare $Pr(E\cap F)\approx 0.05488$ to $Pr(E)Pr(F)\approx 0.161833\cdot 0.28633\approx 0.04634$. Indeed, even if we take the exact values we see that these are different numbers and therefore $E$ and $F$ are not independent as claimed.
The distribution of cards may have been realized as follows. First we give $13+13$ cards to N+S. If we do not have exactly eight $\spadesuit$ cards among them, we ignore this case, it is not contributing to the conditional probability. Else we go on. There are $26=21+5$ cards remaining.
Let us count then the number of ways to split the five $\spadesuit$ cards for the EW axis. We have the possibilities, and the corresponding number of ways to realize them, determined by the E hand:
- $5=5+0$, totally $\binom 55\cdot \binom {21}8$ distributions,
- $5=4+1$, totally $\binom 54\cdot \binom {21}9$ distributions,
- $5=3+2$, totally $\binom 53\cdot \binom {21}{10}$ distributions,
- $5=2+3$, totally $\binom 52\cdot \binom {21}{11}$ distributions,
- $5=1+4$, totally $\binom 51\cdot \binom {21}{12}$ distributions,
- $5=5+0$, totally $\binom 50\cdot \binom {21}{13}$ distributions.
Each "split" has to be weighted with the corresponding number of distributions. So the probabilities are:
sage: for k in [0..5]:
....: p = binomial(5,k)*binomial(21,13-k)/binomial(26,13)
....: print "5=%s+%s :: probability %s ~ %s" % (k, 5-k, p, p.n())
....:
5=0+5 :: probability 9/460 ~ 0.0195652173913043
5=1+4 :: probability 13/92 ~ 0.141304347826087
5=2+3 :: probability 39/115 ~ 0.339130434782609
5=3+2 :: probability 39/115 ~ 0.339130434782609
5=4+1 :: probability 13/92 ~ 0.141304347826087
5=5+0 :: probability 9/460 ~ 0.0195652173913043
(The $5=5+0$ cases will probably feel not so rare, because they remain a long time in the memory.)
Best Answer
The total number of cases where N and S having a total of $8$ spades is not used in the solution, because it is independent from what the question is asking.
$26\choose 13$ is the number of ways to distribute the remaining $26$ cards between E and W. Out of all those distributions, ${5\choose3}$ ways for the E player to choose from the spades and ${21\choose 10}$ ways to choose from the non-spades.
Thus the final answer to the problem is ${5\choose3}{21\choose 10}\over{26\choose 13}$.
The probability does not depend on which of the $8$ spades are chosen for N and S and therefore it is not a conditional probability type problem.
The problem can be restated as "We have a deck of card consisting of $5$ spades and $21$ non-spade cards and we have two players, what's the probability first player gets $3$ spades?"
It becomes much easier to understand.