Question

A and B toss a coin alternately till one of them gets a head and wins the game, if A starts first, find the probability that B will win the game.

Solution

Let P(A), $straight P left parenthesis straight A with bar on top right parenthesis$ be the probabilities of A's getting the head and not getting the head respectively, then
$straight P left parenthesis straight A right parenthesis space equals 1 half space space space space space space space space space space space space space space rightwards double arrow space space space space space space straight P left parenthesis straight A with bar on top right parenthesis space equals space 1 space minus space straight P left parenthesis straight A right parenthesis space equals 1 minus 1 half space equals 1 half$
Similarly, $straight P left parenthesis straight B right parenthesis space equals space 1 half space and space straight P left parenthesis straight B with bar on top right parenthesis space equals space 1 half$
Let A start the game. He can win in the first throw, 3rd throw, 5th throw and so on.
Probability of A's winning in first throw = P(A) = $1 half$
Probability of A's winning in 3rd throw
$equals space straight P left parenthesis straight A with bar on top right parenthesis thin space straight P left parenthesis straight B with bar on top right parenthesis space straight P left parenthesis straight A with bar on top right parenthesis thin space straight P left parenthesis straight B with bar on top right parenthesis thin space straight P left parenthesis straight A right parenthesis$
$equals space 1 half cross times 1 half cross times 1 half cross times 1 half cross times 1 half space equals space open parentheses 1 half close parentheses to the power of 5 space and space so space on.$
Since all these cases are mutually exclusive
$therefore$ probability of A's winning the game first is
$equals space 1 half plus open parentheses 1 half close parentheses cubed plus open parentheses 1 half close parentheses to the power of 5 plus..... infinity equals space fraction numerator begin display style 1 half end style over denominator 1 minus open parentheses begin display style 1 half end style close parentheses squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because straight S space equals space fraction numerator straight a over denominator 1 minus straight r end fraction close square brackets equals space fraction numerator begin display style 1 half end style over denominator begin display style 3 over 4 end style end fraction space equals 2 over 3$
Since either A or B wins
$therefore$ probability of B's winning the game first = $1 minus 2 over 3 space equals space 1 third$

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Probability

A and B toss a coin alternately till one of them gets a head and wins the game, if A starts first, find the probability that B will win the game.

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