EDIT: I originally misread the problem. The below is written thinking that each gun has one bullet in it. Sorry! I'll probably make another post dealing with the actual problem.
But the first three steps aren't too bad on paper.
The probability of a bullet on the first shot is clearly 1/6.
So far we have P(guy gets shot) = 1/6 + something.
The complement is 5/6, which we'll use in just a bit.
After this we have 2/3 probability that the probability will be 1/6, and a 1/3 probability that the probability will be 1/5.
So now we have P(guy gets shot) = (1/6) + (5/6)[ (2/3)(1/6) + (1/3)(1/5) ] + something.
Let x = (2/3)(1/6) + (1/3)(1/5) for ease.
It's getting a bit hairy now. Given that no bullet has been fired yet, there is a 2/3 probability that we have 1/6, 1/5, 1/5 as our next probabilities, and a 1/3 probability that we have 1/4, 1/6, 1/6.
Anyway what I come up with is
P(guy gets shot) = (1/6) + (5/6)(x) + (5/6)(1-x)[ (2/3) ((1/3)(1/6) + (2/3)(1/5)) + (1/3) ((1/3)(1/4) + (2/3)(1/6)) ] + something
As you can see it just gets more and more cumbersome, but it's doable.
Edit 2: While writing one of the below posts, I realized that the methodology for getting the third shot above is flawed, although the underlying reasoning is sound. So I grayed it out.