Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. The probability of A's winning if A starts first is

When rolling a die:

Probability of getting a 3: $\dfrac{1}{6}$

Probability of not getting a 3: $\dfrac{5}{6}$

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A can win in the following ways:

A wins on first turn (A rolls 3 immediately):

Probability $= \dfrac{1}{6}$

A wins on second turn (A fails, B fails, then A rolls 3):

Probability $= \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$

$= \left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6}$

A wins on third turn (A fails, B fails, A fails, B fails, then A rolls 3):

Probability $= \left(\dfrac{5}{6}\right)^4 \times \dfrac{1}{6}$

This pattern continues indefinitely.

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The total probability that A wins:

$P(\text{A wins}) = \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^4 \times \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^6 \times \dfrac{1}{6} + ...$

Factoring out $\dfrac{1}{6}$:

$P(\text{A wins}) = \dfrac{1}{6} \left[1 + \left(\dfrac{5}{6}\right)^2 + \left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^6 + ...\right]$

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The expression in brackets is an infinite geometric series with:

First term: $a = 1$

Common ratio: $r = \left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}$

Sum of infinite geometric series: $\dfrac{a}{1-r}$

$= \dfrac{1}{1 - \dfrac{25}{36}}$

$= \dfrac{1}{\dfrac{11}{36}}$

$= \dfrac{36}{11}$

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$P(\text{A wins}) = \dfrac{1}{6} \times \dfrac{36}{11}$

$= \dfrac{36}{66}$

$= \dfrac{6}{11}$

Therefore, the probability of A winning is $\dfrac{6}{11}$.