I am currently studying probability from Sheldon Ross' book. And the following example was given in the book.
In a card game, 52 cards are dealt out equally to 4 players - called E,W,N,S. If N and S have a total of 8 spades amongst them, what is the probability that E has 3 of the remaining spades. The answer given in the book was the following:
Here is what I did, and I need someone to point out my flaws.
I said that there are $13\choose8$ possible ways in which the 8 spades can be selected and of the remaining 39 cards(I don't want any more spades hence 52-13=29), there are $39\choose18$ ways in which the remaining can be chosen. Now, of the remaining spades, and the remaining cards, the number of ways in which, E can have 3 of the remaining spades is $5\choose3$$26\choose10$. So my probability is:
$\frac{5\choose3}{13\choose8}\frac{26\choose10}{39\choose18}$
Can someone explain to me where my reasoning is off?
Let's use conditional probability: Let event $A = \text{E has } 3 \text{ of the } 5\text{ remaining spades}$, and event $B = \text{N and S have a total of } 8 \text{ spades amongst themselves}$. Then
$P(A|B) = \frac{P(A \cap B)}{P(B)}$. First, let $n$ be the number of ways to give out the cards without any restrictions.
Now, let's compute the number of ways in which $B$ can happen.
If $N$ has $0$ spades and $S$ has $8$, we have $\binom{13}{0}\binom{13}{8}\binom{39}{13}\binom{26}{5}\binom{26}{13}$, where $\binom{13}{0}$ is the number of ways to choose $N$'s spades, $\binom{13}{8}$ is the number of ways to choose $S$'s spades, $\binom{39}{13}$ is the number of ways to choose $N$'s non spades, $\binom{26}{5}$ is the number of ways to choose $S$'s non spades, and $\binom{26}{13}$ is the number of ways to choose $E$'s cards (then $W$ gets the last $13$).
Generally, we have $\binom{13}{k}\binom{13}{8-k}\binom{39}{13-k}\binom{26}{k+5}\binom{26}{13}$, so $$P(B) = \frac{\sum_{k=0}^8\binom{13}{k}\binom{13}{8-k}\binom{39}{13-k}\binom{26}{k+5}\binom{26}{13}}{n}.$$
Similarly, $$P(A \cap B) = \frac{\sum_{k=0}^{8} \binom{13}{k}\binom{13}{8-k}\binom{39}{13-k}\binom{26}{k+5}\binom{5}{3}\binom{21}{10}}{n}.$$
We can thus compute $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\binom{5}{3}\binom{21}{10}}{\binom{26}{13}} = \frac{39}{115} \approx 34\%.$$