$A$ 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.

$A: \frac{6}{11}, B: \frac{5}{11}$

The correct answer is Option (2) → $A: \frac{6}{11}, B: \frac{5}{11}$ ##

Let $W$ denote the event 'getting a 6' and $F$ denote the event 'not getting a 6'.

Then, $P(W) = \frac{1}{6}$ and $P(F) = \frac{5}{6}$

If $A$ starts the game first, then:

$P(A \text{ wins in the } 1^{\text{st}} \text{ throw}) = P(W) = \frac{1}{6}$

$P(A \text{ wins in the } 3^{\text{rd}} \text{ throw}) = P(F) \cdot P(F) \cdot P(W)$

$ = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$

$P(A \text{ wins in the } 5^{\text{th}} \text{ throw}) = P(F) \cdot P(F) \cdot P(F) \cdot P(F) \cdot P(W) = \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}$

$∴P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} + \dots$

$= \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{1/6}{11/36} = \frac{6}{11}$

$∴P(B \text{ wins}) = 1 - P(A \text{ wins}) = 1 - \frac{6}{11} = \frac{5}{11}$