Question

A and B throw a die alternately till one of them gets a '6' and wins the game. Find their respective probabilities of winning, if A starts the game first.

Answer 1

Let S denote the success (getting a '6') and F denote the failure (not getting a '6').

Thus, P(S) = `1/6` = p, P(F) = `5/6` = q

P(A wins in first throw) = P(S) = p

P(A wins in third throw) = P(FFS) = qqp

P(A wins in fifth throw) = P(FFFFS) = qqqqp

So, P(A wins) = p + q²p + q⁴p + ..... = p(1 + q² + q⁴ + ....)

= `p/(1 - q^2)`

= `(1/6)/(1 - 25/36)`

= `6/11`

P(B wins) = 1 – P(A wins)

= `1 - 6/11`

= `5/11`

So, P(A wins) = `6/11` and P(B wins) = `5/11`.