Question

Two players A and B play a series of games of chess. The winning player in any game gets 1 point while the losing player gets 0 points. The player who achieves 4 points first, wins the series. If no game ends in a draw, find the number of ways in which the series can be won by A.

Answer 1

Form four cases in which B wins 0 game, 1 game, 2 games and 3 games respectively. Using the formula of combinations, find the possible ways such that B can at least win the above mentioned win the above mentioned number of games before A wins four games and the series.

Complete step by step answer:
Here, we have to find the number of ways in which the series can be won by A. That means A has to achieve 4 points. To do so A must win 4 games before B. So, B can win a maximum of 3 games. Therefore, let us consider the following cases: -
1. Case (i): - B wins 0 game, A wins 4 games.
So, a total of 4 games will be played in this condition and A will win all of them.

1	2	3	4

Therefore, total number of ways = 1.
2. Case (ii): - B wins 1 game, A wins 4 games.
So, a total of 5 games will be played in this condition.

1	2	3	4	5

Here, B must win a game before A wins 4 games, so B must win a game from the first four games.
Therefore, total number of ways = ${}^{4}{{C}_{1}}$ = 4
3. Case (iii): - B wins 2 games, A wins 4 games.
So, a total of 6 games will be played in this condition.

1	2	3	4	5	6

Here, B must win two games before A wins 4 games. So, B must not win these sets of games: - (1, 6) and (5, 6). Hence, B needs to win two games from the first five games.
Therefore, the total number of ways = ${}^{5}{{C}_{2}}$ = 10.
4. Case (iv): - B wins 3 games, A wins 4 games.
So, a total of 7 games will be played in this condition.

1	2	3	4	5	6	7

Here, B must win 3 games before A wins 4 games. So, B must not win these sets of games: - (1, 6, 7); (1, 2, 7) and (5, 6, 7). Hence, B needs to win 3 games from the first 6 games.
Therefore, total number of ways = ${}^{6}{{C}_{3}}$ = 20
Now, the total number of possible ways in which the series can be won by A will be the sum of possible ways in the above four cases.

$\Rightarrow$ Total number of possible ways = 1 + 4 + 10 + 20 = 35.

Note: One may note that there cannot be more than four cases for the above situation. This is because we can clearly see that if we will consider any case after the ${{4}^{th}}$ case then we have to assume that B wins 4 games, which is not possible. If B will win 4 games then the series will be won by B which is not the given condition in the question. So, B can only win a maximum of 3 games.