A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins

#### Solution

Total of 7 on the dice can be obtained in the following ways:

(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)

Probability of getting a total of 7 = `6/36=1/6`

Probability of not getting a total of 7 = `1-1/6=5/6`

Total of 10 on the dice can be obtained in the following ways:

(4, 6), (6, 4), (5, 5)

Probability of getting a total of 10 = `3/36=1/12`

Probability of not getting a total of 10 = `1-1/12=11/12`

Let E and F be the two events, defined as follows:

E = Getting a total of 7 in a single throw of a dice

F = Getting a total of 10 in a single throw of a dice

`P(E)=1/6, P(barE)=5/6P(F)=1/12, P(barF)=11/12`

A wins if he gets a total of 7 in 1st, 3rd or 5th ... throws

Probability of A getting a total of 7 in the 1st throw = `1/6`

A will get the 3rd throw if he fails in the 1st throw and B fails in the 2nd throw.

Probability of A getting a total of 7 in the 3rd throw = `P(barE)P(barF)P(barE)=5/6xx11/12xx1/6`

Similarly, probability of getting a total of 7 in the 5th throw =** **`P(barE)P(barF)P(barE)P(barF)P(E)=5/6xx11/12xx5/6xx11/12xx1/6 " an so on"`

Probability of winning of A = `1/6+(5/6xx11/12xx1/6)+(5/6xx11/12xx5/6xx11/12xx1/6)+...=(1/6)/(1-5/6xx11/12)=12/17`

∴ Probability of winning of B = 1 − Probability of winning of A =`1-12/17=5/17`