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 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.
 A: 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.
A: EDIT: I think I solved it! See below; definitely let me know if this doesn't make any sense.
FWIW, my problem was that I wasn't subtracting out redundant scenarios for Case 1.

Can someone tell me where I went wrong with using the "complement" method for the last question?
Basic rationale: We already calculated (b), so we need to subtract out the cases where the series doesn't go to 7 games. Note, these all take the perspective of player 'A' (but there is symmetry, so this point is somewhat trivial).
We know there are four cases that result in the desired 4 pts vs. 3 pts scoreline:


*

*Case 1: 4 wins, 0 draws, 3 losses; we need to get rid of the ways we can have 'A' win in 4 games, win in 5 games, and win in 6 games.

*Case 2: 3 wins, 2 draws, 2 losses; same logic here, but the soonest 'A' can win is in 5-games, so get rid of 5- and 6-game series.

*Case 3: 2 wins, 4 draws, 1 loss; same logic here, but the soonest 'A' can win is in 6-games, so we only need to get rid of 6-game series.

*Case 4: 1 wins, 6 draws, 0 losses; this always results in a 7-game series, so skip (0 scenarios to subtract).


For Case 1:


*

*Winning in 4 games can only happen $1$ way: WWWWLLL

*Winning in 5 games happens in $\binom{5}{4} - 1$ ways because we need to place 4 wins in 5 games, but we already counted the WWWWLLL case.

*Winning in 6 games happens in $\binom{6}{4} - \binom{5}{4}$ ways because we need to place 4 wins in 6 games, but we already counted the $\binom{5}{4}$ ways when the last two games are losses.


For Case 2:


*

*Winning in 5 games happens in $\binom{5}{3}$ ways because we need to place 3 wins, 2 draws in 5 games.

*Winning in 6 games happens in $\frac{6!}{3!2!} - \binom{5}{3}$ ways because we need to place 3 wins, 2 draws, 1 loss in 6 games. We subtract $\binom{5}{3}$ because we already counted the ways of ordering the games such that the last two games are the losses (e.g., WWWDDLL, WWDWDLL, ...)


For Case 3:


*

*Winning in 6 games happens in $\binom{6}{4}$ ways because we need to place 2 wins, 4 draws in 6 games.


Evaluating everything together (i.e., answer from (b) LESS the sum of the cases enumerated above):
$$
357 - (1 + \binom{5}{4} - 1 + \binom{6}{4} - \binom{5}{4} + \binom{5}{3} + \frac{6!}{3!2!} - \binom{5}{3} + \binom{6}{4}) = 357 - (15+60+15) = 267
$$
Thanks, in advance.
