# Counting in Probability

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 total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?

Discussion: East gets 3 spades out of remaining 5 in $\binom{5}{3}$ ways. The sample space should have all possible ways East can get spades. All possible outcomes are $\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5}$ i.e., East gets no spade, East gets exactly one spade, …and so on till East gets all 5 remaining spades.

Therefore the desired probability is $\frac{\binom{5}{3}}{\binom{5}{0}+\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}} = 0.3125$. But this does not agree with the answer Sheldon Ross gets! So, where is the problem?

The problem is in counting. The number of ways three spades can be selected from five, in this case is not exactly $\binom{5}{3}$. It is in fact, $\binom{5}{3}\binom{21}{10}$ because there could be more ways to rearrange the rest of the cards. I cannot ignore this factor since it is different proportionately when compared with the elements in the denominator. In the denominator, the number of ways East can get no spades is not $\binom{5}{0}$. Instead, it is $\binom{5}{0}\binom{21}{13}$ because now he can choose 13 cards from remaining 21 non-spade cards. Similarly, when he has one spade, he can rearrange rest of the cards in $\binom{21}{12}$ ways. Of course, East has five choices to pick that one particular spade. Therefore, East has $\binom{5}{1}\binom{21}{12}$ ways to have one spade!

So, this works:

$\frac{\binom{5}{3}\binom{21}{10}}{\binom{5}{0}\binom{21}{13}+\binom{5}{1}\binom{21}{12}+\binom{5}{2}\binom{21}{11}+\binom{5}{3}\binom{21}{10}+\binom{5}{4}\binom{21}{9}+\binom{5}{5}\binom{21}{8}} = 0.339$.

Two morals from this story: 1) In most introductory probability questions, we make the mistake of counting incorrectly. 2) There are often multiple ways of solving the same problem. Try the problem yourself in your own way and double check with the answer to ensure your approach is correct. Do not read probability book like a novel.