Explanation:

If A starts the game, P(A win the 1st game) = 12
Suppose A does not win 1st game.

For A to participate in the 3rd game, A must lose 1st game and B must lose 2nd game.

Hence this is conditional probability.
P(A win 3rd game given A lost 1st game and B lost 2nd game)
= P(A win 3rd game) × P(A lost 1st game) × P(B lost 2nd game)  
= 1/2 × 1/2 × 1/2 = (1/2)^3
Similarly, P(A win 5th game given A lost 1st game ,B lost 2nd game, A lost 3rd game, B lost 4th game)
= 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = (1/2)^5
This process continues till A or B wins the game.

Assuming game is happening for a large number of times where A and B tosses alternatively.
Therefore P(A wins the game)
= 1/2 + (1/2)^3 + (1/2)^5 +.... 
This is a geometrical progression with first term,

a = 12 and common ratio, r =(1/2)^2 = 1/4. This is sum of term of infinite terms with r<1. 
therefore, P(A wins the game) =1/2 /1−1/4 = 23
P(B wins the game) = 1 - P(A wins the game)  = 1 − 2/3 = 1/3
Therefore,Probability of the player who starts the game wins = 2/3

Hence, option 1 is the correct answer.