Conditional Probability, card question

Question

In the card game bridge, the 52 cards are dealt out equally to 4 players – called East, West, North, and South. If North and South have a total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?

The given solution is $\dfrac{\binom{5}{3}\cdot\binom{21}{10}}{\binom{26}{13}}\doteq 0.339$.

Trying to wrap my head around this solution. Why can I not treat the problem as simply: "Since there are 5 spades remaining, just count the number of ways 5 spades can be split among East and West. "

Answer 1

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.)

Answer 2

Why can I not treat the problem as simply: "Since there are 5 spades remaining, just count the number of ways 5 spades can be split among East and West. "

Sure.   You can also do that.   The answers are equal; as manipulation of the binomial coefficients will show.

There are $\binom{26}{5}$ ways to select places for the 5 spades among the two hands, and among these are $\binom {13}3\binom{13}2$ ways to select exactly three places for spades among East's hand (and two places among West's hand, of course.).$$\dfrac{\dbinom{13}{3}\dbinom {13}2}{\dbinom{26}5}=\dfrac{\dfrac{13!}{3!~10!}\dfrac{13!}{2!~11~}}{\dfrac{26!}{5!~21!}}=\dfrac{\dfrac{5!}{3!~2!}\dfrac{21!}{10!~11!}}{\dfrac{26!}{13!~13!}}=\dfrac{\dbinom{5}{3}\dbinom{21}{10}}{\dbinom{26}{13}}$$

So we could also count ways East can draw 3 from 5 spades and 10 from 21 non-spades, when drawing 13 from 5+21 spades-and-non-spades.