Probability in game of bridge In a game of bridge, find the probability that the North, East, South, and West hands will get respectively $a,b,c,d$ spades.
I tried like this. First I choose $a$ spades from the $52$ cards; then, from the remaining $39$ I choose $b$ spades; then, from the remaining $26$ cards I choose $c$ spades, and the rest can be done in $1$ way only. But am not sure about my solution. Please help.

Imagine dealing in an unusual way, $13$ cards to South, then $13$ to East, and so on.
There are $\binom{52}{13}$ equally likely ways to choose the cards South gets. There are $\binom{13}{a}\binom{39}{13}$ ways to choose $a$ spades and $13-a$ non-spades. So the probability that South gets the right kind of hand can be computed. For the record (we will not do it again) it is $\frac{\binom{13}{a}\binom{39}{13-a}}{\binom{52}{13}}$
Now there are $39$ cards left, of which $13-a$ are spades, and $26+a$ non-spades.
There are $\binom{39}{13}$ equally likely ways to choose the cards East gets.There are $\binom{13-a}{b}\binom{26+a}{13-b}$ ways to give East the right kind of hand.
Now there are $26$ cards left, $13-a-b$ spades and $13+a+b$ non-spades. There are $\binom{26}{13}$ ways to choose North's cards. There are $\binom{13-a-b}{c}\binom{13+a+b}{13-c}$ ways to give North the right kind of hand.
And now it's over.  When we multiply the probabilities and compute the binomial coefficients, there is a pleasant amount of cancellation. 
Remark: We should really give a symmetrical solution. The numbers $a$, $b$, $c$ appear quite asymmetrically in the argument, and poor $d$ did not get mentioned at all. Symmetry reappears when we simplify, and may suggest a much nicer argument.
Out of tiredness, I just gave a solution that's ugly, but works. Maybe tomorrow.
