Let A : Event of A getting a head 

⇒ \(\bar{A}\) : Event of A not getting a head 

∴ P(A) = \(\frac{1}{2}\) and P(\(\bar{A}\)) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)

Similarly, B : Event of B getting a head 

\(\bar{B}\) : Event of B not getting a head.

So P(B) = \(\frac{1}{2}\) and P(\(\bar{B}\)) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\).

Let A start the game. A's wins if he throws a head in the 1st throw or 3rd throw or 5th throw or throw and so on.

Probability of A's winning in 1st throw = P(A) = \(\frac{1}{2}\)

Probability of A's winning in 3rd throw = P(not A). P(not B). P(A) = P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)

= \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\) = \(\big(\frac{1}{2}\big)^3\)

Similarly, probability of A's winning in 5th throw =  P(\(\bar{A}\)) x P(\(\bar{B}\)) x  P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)

= \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\) = \(\big(\frac{1}{2}\big)^5\) and so on. Since all these events are mutually exclusive,

p(A winning the game first) = \(\frac{1}{2}\) + \(\big(\frac{1}{2}\big)^3\) + \(\big(\frac{1}{2}\big)^5\) + \(\big(\frac{1}{2}\big)^7\) ... 

= \(\frac{1}{2}\)\(\bigg[\) 1 + \(\big(\frac{1}{2}\big)^2\) + \(\big(\frac{1}{2}\big)^4\) + \(\big(\frac{1}{2}\big)^6\) + ...\(\bigg]\)

= \(\frac{1}{2}\)\(\bigg[\frac{1}{1-\big(\frac{1}{2}\big)^2}\bigg]\) = \(\frac{1}{2}\) x \(\frac{1}{\frac{3}{4}}\) = \(\frac{1}{2}\) x \(\frac{4}{3}\) = \(\frac{2}{3}\).

(∵ For an infinite GP, S_∞ = \(\frac{\text{First term}}{\text{1-common ratio}}\))

Since A and B are mutually exclusive events, as either of them will win, P(B winning the game first) = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\).