# Probability of Pairings in Chess Game

1. Nov 18, 2007

### e(ho0n3

[SOLVED] Probability of Pairings in Chess Game

1. The problem statement, all variables and given/known data
The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that

(a) Rebecca and Elise will be paired;
(b) Rebecca and Elise will be chosen to represent their schools but will not play each other;
(c) exactly one of Rebecca and Elise will be chosen to represent her school?

2. Relevant equations
Axioms and basic theorems of probabilitiy.

3. The attempt at a solution
(a) Assuming each possible team is equally likely and each possible pairing is equally likely, the the probability sought is the ratio of the number of possible teams and pairing in which Rebacca and Elise are paired to the total possible number of teams and pairings which should be

$$\frac{\binom{7}{3}\binom{8}{3}3!}{\binom{8}{4}\binom{9}{4}4!} = \frac{1}{14}$$

The book says the answer is 1/18 however.

(b) That should be

$$\frac{\binom{7}{3}\binom{8}{3}3 \cdot 3!}{\binom{8}{4}\binom{9}{4}4!} = \frac{3}{14}$$

The book says the answer is 3/18 however.

(c) That should be

$$\frac{\binom{7}{3}\binom{8}{4}4! + \binom{7}{4}\binom{8}{3}4!}{\binom{8}{4}\binom{9}{4}4!} = \frac{1}{2}$$

The book has the same answer.

My gripe is with (a) and (b). Is the book right or am I right?

2. Nov 18, 2007

### Dick

The book is right. You may be overcomplicating things. The probability R is chosen is 4/8, that E is chosen is 4/9. Once chosen, the odds they will play each other is 1/4. What's the probability of all of those things happening? Change it for the case of 'not play'.

3. Nov 18, 2007

### Dick

BTW. Your counting expressions are also correct. Just the final numbers are wrong.

4. Nov 18, 2007

### e(ho0n3

Yes, I do tend to overcomplicate things. You seem to have a very good intuition for this kind of thing.

I realized where I made the stupid arithmetical mistake. Thanks.