Two chess players, A and B, are going to play 7 games. There are three possible outcomes for each game, A wins, A loses, or Tie
Addtionally, a win is worth 1 point, draw 0.5 points and loss 0 points. (a) How many possibilities for the games are there such that player A ends up with 3 wins, 2 draws and 2 losses? (b) How many possible outcomes for the games are there such that A ends with four points and B ends with 3 points. (c) Now, assume the first player to 4 points wins and no further matches are played. How many outcomes are there such that the match lasts 7 games and A wins with a score of 4 to 3?
(a) There are $3^7$ total possibilities for a string of 7 games. 3 of the 7 games need to be wins for A ${7 \choose 3}$, then 4 of the remaining games need to be draws ${4 \choose 2}$ and then the other two games are losses for B ${2 \choose 2}$. The total is then ${7 \choose 3}{4 \choose 2}{2 \choose 2}=210$ for part a.
(b) This should be the sum of the previous answer (210) and the total number of outcomes where A wins 4 games, ${7 \choose 4}$= 35. So, 245.
(c) For this one, I guess I just have to subtract from the answer for (b) because this includes the matches that go 7 games. I think there are two terms that need to be subtracted, ${6 \choose 4}{3 \choose 3}$, which denotes all outcomes where A wins four of the first 6 games and B wins the other 3. The other term ${6 \choose 3}{3 \choose 2}$, which is the result where A wins 3 of the first 6 and draws 2 of the remaining 3 games. So, the final answer would be $245-{6 \choose 4}{3 \choose 3}-{6 \choose 3}{3 \choose 2}=170$
Is it okay to subsume ${4 \choose 4}$ and ${5 \choose 4}$ in ${6 \choose 4}$? It seems like if I added these terms, I would be counting different outcomes two or three times.
Best Answer
I feel like the chess players should have real names... Archibald and Bartholomew, perhaps? Anyway...
a) A gets $3$ wins, $2$ draws, $2$ losses
This is a question similar to "how many anagrams of "REFEREE" are there?" - the answer is to divide the total arrangements by the arrangements of each repeated element, so
$$N_a = \frac{7!}{3!2!2!} = \frac{5040}{24} = 210$$
We agree :-) - if you calculate out your binomials as factorials, you'll see they cancel down to the same form.
b) A gets exactly $4$ points
Of course we can calculate this by taking cases as above; let's try that:
$$N_b = \frac{7!}{4!0!3!}+\frac{7!}{3!2!2!}+\frac{7!}{2!4!1!}+\frac{7!}{1!6!0!} = 35+210+105+7 = 357$$
c) A gets exactly $4$ points and doesn't lose game $7$
As you can see I've interpreted the condition; basically our final answer can be obtaining by adding the number of ways, after $6$ games, to get to $3$ points, and to get to $3\frac 12$ points - followed by a win and a tie respectively.
Let's just see what that looks like...
$$\begin{align}N_c &= \left( \frac{6!}{3!0!3!}+\frac{6!}{2!2!2!}+\frac{6!}{1!4!1!}+\frac{6!}{0!6!0!} \right )\\ &\quad +\left( \frac{6!}{3!1!2!}+\frac{6!}{2!3!1!}+\frac{6!}{1!5!0!} \right ) \\ &= 20+90+30+1\:+\:60+60+6 = 141+126 \\ &= 267 \end{align}$$
Feel free to check my numbers, as this was done in my head and without the aid of a safety net.