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
 A: EDIT:I have included an explicit answer for the scenario that both players' first shots are fired simultaneously below.
Martigan's answer above works well for the scenario that both player's first shots coincide.  Because it is an interesting problem, here below is the other situation, if their shots are offset.
This can be approached via markov chains if they never fire at the same time (i.e. player A's first shot does not coincide with either player B's first shot or player B's second shot).  I will make the additional assumption that neither player gets a substantial head start.
This can then be modeled with a Markov Chain with the following transition diagram:

Represented by the matrix with order $A_w, B_w, A, B_1, B_2$ as 
$\begin{bmatrix} 
1 & 0 & .8 & 0 & 0\\
0 & 1 & 0 & .5 & .5\\
0 & 0 & 0 & 0 & .5\\
0 & 0 & .2 & 0 & 0\\
0 & 0 & 0 & .5 & 0\end{bmatrix}$
This is in the form $A=\begin{bmatrix} I & S\\ 0 & R\end{bmatrix}$ which $\lim_{n\to\infty} A^n = \begin{bmatrix}I&S(I-R)^{-1}\\0 & 0\end{bmatrix}$
Through matrix arithmetic, we get that $(I-R)^{-1} \approx \begin{bmatrix}1.05 & 0.26 & 0.53\\ 0.21 & 1.05 & 0.11\\ 0.11 & 0.53 & 1.05\end{bmatrix}$
and that $\lim_{n\to\infty} A^n = \begin{bmatrix}1 & 0 & 0.84 & 0.21 & 0.42\\
0 & 1 & 0.16 & 0.79 & 0.58\\
0&0&0&0&0\\
0&0&0&0&0\\
0&0&0&0&0\end{bmatrix}$
Thus, if player A shoots first, player A wins with probability 0.84
If player B has two shots before player A, player B wins with probability 0.79
If player B has only one shot before player A, player B wins with probability 0.58

Because of the confusion among other posters, I decided to explicitly solve the interpretation that both snipers' first shots are at the exact same time and there is no tie allowed (if they both hit, they both shoot again).
The transition diagram then is

The transition matrix with order $A_w, B_w, AB_1, B_2$ is: $\begin{bmatrix}
1 & 0 & .4 & 0\\
0 & 1 & .1 & .5\\
0 & 0 & 0 & .5\\
0 & 0 & .5 & 0\end{bmatrix}$
Fundamental matrix then is $\begin{bmatrix}1 & -.5 \\ -.5 & 1\end{bmatrix}^{-1} = \begin{bmatrix}4/3 & 2/3\\2/3&4/3\end{bmatrix}$
And the limiting matrix becomes $\begin{bmatrix}1 & 0 & .5\overline{3} & .2\overline{6}\\
0 & 1 & .4\overline{6} & .7\overline{3}\\
0& 0 & 0 & 0\\
0&0&0&0\end{bmatrix}$
Thus, the chance that player $A$ wins is $0.5\overline{3}$ assuming they both start at the same time, and therefore player $A$ is more likely to win.
A: I suppose that a tie is accepted and both shooter win in that case (they are both the first). 
The first sniper has a chance of missing each turn of $0.2$.
The second one has a chance of missing of $0.25$
Even with the higher rate of shooting, the first one has a higher probability of being the first (tie accepted) to hit the target. 
The problem for which one hit and the other miss is slightly different.
EDIT: case when we want a "clear winner". Let say that if both shoot at the same time, they have to start again the competition from start.
Sniper 1 wins with probability $0.8*0.5=40$% per turn (he gets the target, sniper 2 misses the first shot).
Sniper 2 wins with probability $0.2*0.5+0.5*0.5=35$% per turn (he gets while sniper 1 misses, or he misses the first time but succeeds the second time).
In any case, bet on the first shooter. 
A: 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.
