One day your rival invites you out to an old fashion duel. The rules are as follows:

-You each take turns shooting each other.

-The duel stops when one person hits the other.

-Both of you are honourable enough to not take shots out of your turn.

-The probability of one person hitting the other is 1/2 and is independent of the probability of the other person hitting you.

Clearly, whoever shoots first has a distinct advantage. So your friends suggests flipping a coin for it. Little did you know that your rival uses a coin rigged in such a way that the probability of getting heads is only 1/3.

If you chose heads, what is the probability of you winning the duel?

(In reply to

Answer by K Sengupta)

At the outset, the person who wins the toss will get to shoot first.

Let P(x) be the probability that the person shooting for the overall xth turn will be successful.

Then,

P(1) = 1/2

P(2) = (1/2)*(1/2) = (1/2)^2

..............

..............

P(m) = (1/2)^m

Now, the person winning the toss will will shoot first and will

shoot in every odd numbered turns, that is, 1st turn, 2nd turn,

3rd turn, and so on.

Let R = Probability that the person shooting first will win

Then,

R = 1/2 + (1/2)^2 + (1/2)^3 + .......

= (1/2)/(1 - 1/4) = 2/3

Accordingly, the probability that the person going second will

win = 1-R = 1/3

Now, it is known that the rival rigged the coin in such a

manner, that the probability of the individual getting heads, and

thus shooting first is 1/3, and so the probability that the rival

will shoot first is 1-1/3 = 2/3.

Consequently, the required probability that the individual will

win the duel is:

(2/3)*(1/3) + (1/3)*(2/3) = 4/9