Question

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 overall
player A 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?
(C) Now assume that they are playing a best-of-7 match, where the match will end as
soon as either player has 4 points. For example, if after 6 games the score is 4 to 2 in
favor of A, then A wins the match and they don’t 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?

Answer 1

A) 210

B) 357

C) 267

Explanation:

A) Among 7 games, we can first choose 3 wins, and then among remaining 4 games, we can choose 2.

To calculate the possibility, we will use Combination.

`C(7,3)*C(4,2) = 35 *6 = 210`

B) Player A can get 4 points with the following cases:

4 wins and 3 loses

3 wins, 2 draws and 2 loses

2 wins, 4 draws and 1 lose

1 win and 6 draws

Indeed, these cases matches for Player B too to get 3 points.

So again, we will use Combination to calculate the possibility.

`C(7,4) + C(7,3)*C(4,2) + C(7,2)*C(5,4) + C(7,1) = 357`

C) Here, we need to find two possibilities after 6 games and add them, while Player A has 3 points and wins the 7th game, and Player A has 3.5 points and draws the 7th game.

`[C(6,3)+C(6,2)C(4,2)+C(6,1)C(5,4)+C(6,6)] + [C(6,3)C(3,1)+C(6,2)C(4,3)+C(6,1)] =[20+90+30+1]+[60+60+6]=141+126=267`