A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8 , and B wins if he throws a sum of 8 before A throws a sum of 5 . The probability, that A wins if A makes the first throw, is

Solution

Maths sir · 3 steps

A wins on his first throw, or after one full round where both miss: sum that geometric series.

1. Count the outcomes. Sum 5 comes from (1,4),(2,3),(3,2),(4,1), so P(A hits) = 4/36 = 1/9. Sum 8 comes from (2,6),(3,5),(4,4),(5,3),(6,2), so P(B hits) = 5/36.
2. One full round. In one round A throws and then B throws. Both miss with probability 8/9 × 31/36 = 248/324 = 62/81. A can win on throw 1, 3, 5, and so on, and each later chance needs one more round of both missing.

   P(A wins) = 1/9[1 + 62/81 + (62/81)² + …] = (1/9)/(1 - 62/81) = 1/9 × 81/19

3. So. P(A wins) = 9/19, which is option (b).

The idea to keep. In any take-turns game, the first player's chance is p/(1 - (both miss in one round)). Count each player's winning outcomes separately and carefully.

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