Two players are flipping a coin. If head, player A wins 1 point. If tail, player B wins 1 point. The person who first wins 2 points wins the game. The loser must pay the winner $\$1$.
However, if player A has an option he can use to increase the stake of the game to $\$2$-game, what is the value of such option?
I received this question during an interview but I am unsure of the answer I provided.
The original game has an EV of $0$ and using the option will yield an EV of $0.25$ according to my calculations. The final answer will then be that the option has a value of $0.25$ USD.
Is this way of reasoning correct? Can anybody provide a thorough explanation of the process and show me the right way of approaching this problem?
Yes, your reasoning looks correct to me. There is a 0.5 chance of a H on the first throw. In that event A exercises the option. If the next throw is H he wins 2. If the next throw is T, then his expected win is 0.
On the other hand, there is a 0.5 chance of a T on the first throw. In that event, A does not exercise the option. If the next throw is T he loses 1. If the next throw is H, then his expected win is 0.
So adding that up, with the option his expected win is $0.25\times2-0,25\times1=0.25$, compared with an expected win of 0 without the option. So the option is worth 0.25.
Note that A the only time A benefits from exercising the option is immediately after a H on the first throw. If there is a head and he delays, then either there is a H on the next throw and it is too late - the game is over - or there is a T on the next throw and he has no advantage. Similarly, if there is a T on the first throw his position is worse, so it hurts him to exercise the option.