Question

$A$ and $B$ toss a coin alternately till one of them gets ahead and wins the game. If $A$ starts the game. Find the probability that $B$ will win the game.

Answer 1

If A starts the game, A can win $1^{st},3^{rd},5^{th}$ so on throws.

Probability of A's winning in first row = $P\left(A\right)$ = $\frac{1}{2}$

Probability of A's winning = $P\left(A\right)+P\left(\overline{A}\right)\cdot P\left(\overline{B}\right).P\left(A\right)+P\left(\overline{A}\right)\cdot P\left(\overline{B}\right).P\left(\overline{A}\right)\cdot P\left(\overline{B}\right)\cdot P\left(A\right)+........\infty$

Probability of getting A first = $\frac{1}{2}+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^5+....\infty$

$=\frac{\frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}=\frac{2}{3}$

Probability of B getting heads = $=1-\frac{2}{3}=\frac{1}{3}$