# Problem 7, Ch1 from Blitzstein and Hwang, Intro to Probability

I am trying to solve this problem.

Two chess players, A and B are going to play 7 games. Each game has three possible outcomes : a win for A (which is a loss for B), a draw(tie) and a loss for A (which is a win for B). A win is worth 1 point, a draw is worth 0.5 points and a loss is worth 0 points.

(a) How many possible outcomes for the individual games are there such that the player ends up with 3 wins, 2 draws and 2 losses?

(b) How many possible outcomes for the individual games are there, such that A ends up with 4 points and B ends up with 3 points?

My attempt:

(a)This is the same as the number of permutations of the WWWDDLL. There are $$7\choose{3}$$ ways to decide where to put the W, then $$4\choose2$$ ways to decide where to put the D and $$2\choose2$$ ways to decide where to put the L.

This gives $${7\choose3}\cdot{4\choose2}\cdot{2\choose2}=\frac{7!}{3!2!2!}=210$$ possibilities.

(b) A must end up with 4 points, and by symmetry B must end up with three points.

We are then interested to find the number of non-negative solutions to the equation

$$1\cdot{x_{1}}+0.5\cdot{x_{2}}+0\cdot{x_{3}}=4$$

This amounts to having 3 distinguishable boxes, into which 7 indistinguishable particles can be put. Each particle can be either a win, a draw or a loss.

For example,

 W  D  L


|***|**|**|

There are 7+3-1=9 slots and we need to select 2 out of the 9 slots, where two |'s can go. But, a draw counts only as 0.5 points. So, I don't think this yields the correct answer.

It would be nice, if someone could verify part (a) and help with part(b).

Edit 1

There is a also a part (c) to the question.

(c) Now assume that they are playing a best-of-7 match, where the match will end when either player has 4 points or when 7 games have been played, which is first. For example, if after the 6 games, the score is 4 to 2 in favor of A, then A wins the match and they do not play a 7th game. How many possible outcomes for the individual games are there, such that the match lasts for 7 games and A wins by a score of 4 to 3?

My attempt:

c) If the match is to last till the 7th game and A to win by a score of 4 to 3, the 7th game should not be a loss for A. Thus, we have

• 4 wins, 3 loss (WWWLLW) - $${7\choose4}\cdot{3\choose3}-{6\choose4}\cdot{2\choose2}$$
• 3 wins, 2 draws, 2 loss (WWWLLDD) - $${7\choose3}\cdot{3\choose2}-{2\choose2}\cdot{2\choose2}$$
• The first is fine. For the second, let's work with B, since $3$ is smaller than $4$. Could be $6$ draws or $4$ or $2$ or $0$. Count all separately. For example, $4$ draws is for reasons you are familiar with $\binom{7}{4}\binom{3}{1}$. Jul 17 '16 at 10:21
• @AndréNicolas , so its the sum of 1) DDDDDDW - (7 Choose 1) x (1 Choose 1) 2) DDDDWWL- (7 Choose 4) x (3 Choose 1) x (1 Choose 1) 3) DDWWWLL - (7 Choose 2) x (5 Choose 3) x (2 Choose 2) 4) WWWWLLL - (7 Choose 4) x (3 Choose 3) Jul 17 '16 at 11:28
• Yes, the case numbers are right, then we compute and add up. The $\binom{1}{1}$, $\binom{2}{2}$, and so on are all $1$, so unnecessary though harmless. There are other more "general" kinds of approaches, for example using generating functions, but for small numbers like these the approach you carried out is efficient. The answer by Shagnik uses essentially the same sort of case analysis. In the answer you wrote out, you forgot to add up at the end, Jul 17 '16 at 12:57
• @AndréNicolas, is the answer to part(c) also correct? Jul 18 '16 at 1:43