Win, Lose or Draw: 7 Games of Chess

1.) Two chess players, X and Y, play a 7-game series of chess. For each game, there are three possible outcomes for a player, win, lose or draw. A win nets the victor one point and the loser zero points. In the case of a draw, each player is awarded a half-point. How many ways are there for one of the players, say, player X, to get an overall result of 3 wins, 2 draws and 2 losses?

Solution: A simple calculation here... We multiply the number of ways a player can win 3 out of the 7 games, the number of ways that player can get a draw or a loss in 4 of the seven games, and the number of ways that player can get a draw or a loss in 2 out of the 7 games. The factors are:
$\binom{7}{3}, \binom{4}{2}, \binom{2}{2}$

The answer is $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$ = 210

2. How many possible outcomes for the games are there with X ending up with 4 points and Y ending up with 3 points?

Solution: One way is to have X win 4 out of the 7 games and lose 3: $\binom{7}{4}*\binom{3}{3}$. Another way is to have X win 3, draw 2, and lose 2: $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$. Another way is to have X win 2, draw 4 and lose 1: $\binom{7}{2}* \binom{5}{4}* \binom{1}{1}$. And, finally, another way is to have X win 1 and draw 6: $\binom{7}{1}* \binom{6}{6}$

Answer = $ \binom{7}{4}*\binom{3}{3} + \binom{7}{3}* \binom{4}{2}* \binom{2}{2} + \binom{7}{2}* \binom{5}{4}* \binom{1}{1} + \binom{7}{1}* \binom{6}{6}$
= 35 + 210 + 105 + 7 = 357.

3. How many possible outcomes for the games are there where one player ends up with 4 points, the other with 3 points, and the full 7 games are played?

Solution: The 7th game must be played, which means that for one player, say, X to end up with 4 points at the end, he or she cannot lose that 7th game. Therefore, the only possibilities for X to end up with 4 points are either a full-point awarded in the seventh game, or only a half point. If the full point is awarded in the 7th game, that means X must get 3 points in the first 6 games. If only a half-point is awarded in the 7th game, then X must get 3.5 points in the first 6 games.

Let scenario 1 be the scenario where X gets a full point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{3}$ = 20 (3 wins and 3 losses)
$\binom{6}{2}* \binom{4}{2}* \binom{2}{2}$  = 90 (2 wins, 2 draws, and 2 losses)
$\binom{6}{1}* \binom{5}{4}* \binom{1}{1}$ = 30 (1 win, 4 draws, and 1 loss)
$\binom{6}{0}* \binom{6}{6}$ = 1 (0 wins and 6 draws)

Let scenario 2 be the one where X gets only a half point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{1}*\binom{2}{2}$ = 60 (3 wins, 1 draw, and 2 losses)
$\binom{6}{2}* \binom{4}{3}*\binom{1}{1}$ = 60 (2 wins, 3 draws, and 1 loss)
$\binom{6}{1}* \binom{5}{5}$ = 6 (1 win and 5 draws)

Answer = 20 + 90 + 30 + 1 + 60 + 60 + 6 = 267