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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.
Solution
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`.
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