In a bridge deal, what is the probability that:
a) West has five spades, two hearts, three diamonds, and three clubs?
b) North and South have five spades, West has two spades, and East has one spade?
c) One player has all the aces?
My strategies thus far:
a) The total number of possible deals in bridge is $T = \frac{52!}{(13!)^4}$ I am given this.
My total favorable outcomes would be:
$F = \frac{{13 \choose 5}{13 \choose 2}{13 \choose 3}{13 \choose 3}}{(13!)^3}$
so,
$F/T = .0129$
b) Here I am a bit stuck. I know that for each hand I must factor in the remaining choices after getting their share of spades, but how do I do that?
$F = {13 \choose 5}{52 – 13 (?) \choose 8}{… \choose …}$
c) Lost here as well. I know how to calculate the probability of each player getting a spade:
$F = \frac{4!48!}{(12!)^4}$ so
$F/T = .105$
but how do I calculate one player getting all four aces?
Best Answer
$(a):$
It is better to use the concept of a hand of $13$, with $\binom{52}{13}$ possible hands
$Pr = \dfrac{\binom{13}5\binom{13}{3}\binom{13}{3}\binom{13}{2}}{\binom{52}{13}}$
$(b):$
Here, the sample space is $\binom{52}{13,13,13,13}$.
[If not familiar with the multinomial distribution, it is simply equivalent to $\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$]
$Pr = \dfrac{\binom{13}{5,5,2,1}\binom{39}{8,8,11,12}}{\binom{52}{13,13,13,13}}$
$(c):$
Any of $4$ players can get $4$ aces and $9$ other cards, thus
$Pr = \dfrac{4\times\binom44\binom{48}9}{\binom{52}{13}}$