In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :

Solution

Maths sir · 4 steps

A wins on throw 1, 3, 5, … only if everyone before him missed, so add up a geometric series.

1. One-throw chances. A total of 6 comes from (1,5),(2,4),(3,3),(4,2),(5,1), so P(6) = 5/36. A total of 7 has 6 outcomes, so P(7) = 6/36 = 1/6.
2. One full round. A misses with probability 31/36 and B misses with probability 5/6. Both miss in one round with probability 31/36 × 5/6 = 155/216, and then the game starts over.

   P(A) = 5/36[1 + 155/216 + (155/216)² + …] = (5/36)/(1 - 155/216)

3. Simplify. 1 - 155/216 = 61/216, so P(A) = 5/36 × 216/61 = 30/61.
4. So. The probability that A wins is 30/61, option (d).

The idea to keep. In a turn-taking game, P(first player wins) = p divided by (1 − probability that nobody wins in one full round). Find that one-round miss probability first and the rest follows.

Answer from the NTA final answer key. Solution by PrepWiser.