Sniper probability question 
There are 2 snipers and they are competing each other in game where winner is the one who hits the target first. First sniper's hit probability is 80%, the second's probability is 50%. In the same time second sniper shoots two times faster than the first. 
The question is which sniper has the highest chance to hit the target first?
Thanks in advance,
HARIS

Answer:
Sniper1 hits the target in a  shot $= 0.8$
Sniper1 does not hit the target in a shot$= 0.2$
Sniper 2 can fire two shots in the same time as Sniper 1.  Let us look at the possibilities.
Case 1: Sniper 2 hits the target in the first shot(H)$ = 0.5$
Case 2: Sniper 2 hits the target in the second shot (NH)$ = 0.25$
Case 3: Sniper 2 fails to hit the target in both the shots in the time Sniper 1 hits the first shot(NN)$ = 0.5*0.5 = 0.25$
Effectively Sniper 2 hits the target at the same time as Sniper 1 $= 0.5+0.25 = 0.75$
Then the probability for these scenarios would be 
Case1 : Sniper 1 Wins, Sniper 2 Wins in the first shot $=0.8*.75= 0.6$
Case2 : Sniper 1 Wins, Sniper 2 Loses in the first shot $= 0.8*.25 = 0.2$
Case3 :Sniper 1 Losses, Sniper 2 wins in the first shot $= 0.75*0.2 = 0.15$
Case4 :Sniper 1 losses, Sniper 2 losses in the first shot $= 0.2*0.25 = 0.05$
Shots by Sniper 1 and Sniper 2 are independent and obeys the laws of probability.
In both cases 1 and 4, the game can be viewed as draw and back to square one.  Thus the probability that there are no clear winner after the fist shot is sum of case 1 and case 4 $= 0.65$
The probability that Sniper 1 will win $= 0.8*0.25 + 0.65*(0.8*0.25)+0.65^2*(0.8*0.25) + \cdots \infty$
$= 0.2\frac{1}{(1-0.65)} = 0.5714$
The probability that Sniper 2 will win $= 0.75*0.2 + 0.65*(0.75*0.2)+0.65^2*(0.75*0.2) + \cdots \infty$
$= 0.15\frac{1}{(1-0.65)} = 0.42857$
P(Sniper1 Wins) $= 0.5714$
P(Sniper 2 Wins) $= 0.4286$
This is one way you can simplify the game yet reaches very close solution to @JMoravitz solution unless you the Markov Chain way of solving.
