Probability puzzle

[WMDhamnekar]

TL;DR

Two players start 1 meter away from a target. They simultaneously begin moving towards the target at a same constant speed. Target hitting probability of both the players when they are X meters from the target are given. If both players misses or hits the target simultaneously, neither wins and game will be started over. The player whoever hit the target first will win.What is the probability that left player wins, if both player uses optimal strategies?

Two players start 1 meter away from a target. They simultaneously begin moving towards the target at a same constant speed. If the left player shoots when he is X meters from the target, his shot hit with a probability 1-X. If the right player shoots when he is X meters from the target, his shot hits with a probability [$1- X^2$]. Each player has exactly one bullet and may choose to shoot at any time during the walk. If exactly one player hit the target, that player wins. If both players shoot simultaneously and both hits , then neither player wins. Both are sent back to the starting positions and game starts over. Similarly, if both the players miss the shots, the game starts over from the beginning. The players don't have to shoot at the same time and they can see each other at all times. If they both hit the target, but at different times, whoever hits the target first will win. Assuming both players use optimal strategies, what is the probability that the left player wins? Note: Though, both players uses pistol, they are not criminals,violence supporter.

My answer: I don't under stand how to answer this question. I am still working on this question. If any physics forum's members knows the answer to this question may reply with correct answer.

 

[anuttarasammyak]

$$1-x^2=(1-x)(1+x)>1-x$$

I know that the right player has advantage. I do not think the left has a strategy to win with more than 50% probability.