In the card game bridge, the 52 cards are dealt out equally to 4 players

Question

I misunderstand conditional probability as in this problem

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?

My calculation is.

Let E denote:" East has 3 of the remaining 5 spades" and F:" North and South have a total of 8 spades among them".

Since $ P(E|F)= \frac {P(EF)} {P(F)} $

For $P(F)= \frac {{13 \choose 8} * {39 \choose 18}} {{52 \choose 26}}$ =0,161

And $P(EF)= \frac {{13 \choose 8}*{39 \choose 18}*{5 \choose 3}*{18 \choose 10}}{{52\choose 26}*{26 \choose 13}}$ =0,32. But the answer is 0,339. What is wrong with my calculation?

Answer 1

Your calculation of $P(EF)$ is not correct and should be:$$P(EF)=\frac{\binom{13}{3,2,8}\binom{39}{10,11,18}}{\binom{52}{13,13,26}}=\frac{\frac{13!}{3!2!8!}\frac{39!}{10!11!18!}}{\frac{52!}{13!13!26!}}$$

There are e.g. $\frac{13!}{3!2!8!}$ ways to divide $13$ spades in such a way that East gets $3$ and West gets $2$ of them.

Answer 2

We need to condition on the $26$ cards received by East and West. If North and South have eight spades between them, they also have $26 - 8 = 18$ non-spades. That means of the $26$ cards East and West receive, $13 - 8 = 5$ are spades and $39 - 18 = 21$ are non-spades.

We wish to find the probability that East receives three of the five remaining spades. East must receive $13$ of the remaining $26$ cards. If East receives $3$ of the remaining five spades, then East must also receive $13 - 3 = 10$ of the $21$ remaining non-spades. Hence, the probability that East receives three of the five remaining spades $$\frac{\dbinom{5}{3}\dbinom{21}{10}}{\dbinom{26}{13}}$$

Answer 3

The simplest way to solve it is to just focus on dealing out the necessary number of spades from those remaining in the hands of East and West, and forget about the rest. !

$$Pr = \large\frac{\binom{13}{3}\binom{13}{2}}{\binom{26}{5}} =\frac{39}{115}, \approx 0.33913$$