Ref: http://www.yofx.org/?p=194

Truel solution

Before I present my answer, I'd like to clarify the problem a bit
because the "rules" are a little vague. First, let us assume there can
be one or more additional rounds if two or more are standing at the end
of a round. Second, let's assume the shooters all care about their
survival and will aim at the person they feel is the biggest threat.
Third, let us allow shooters to intentionally miss. Finally, instead of
just asking whom you should shoot at, let's also ask what is the
probability that you will survive?

OK. The best strategy for Red (R) is to intentionally miss on his 1st
shot. He'll let the better shooters try to knock each other off, thus
increasing his chances of survival. This will result in the following
situation, with Black (K) firing at Blue (U).

Round 1 (K shooting)

R K U  ->  R K  (K hits U)  2/3  (see Round 2 case 1)
or 
R K U  ->  R K U  (K misses U)  1/3  ->  R U (U hits K)  (see Round 2 case 2)

Round 2 case 1 (R shooting)

time = 0

R K  -> R  (R hits)  1/3
or
R K  ->  R K  (R misses)  2/3  ->  R K  (K misses)  1/3  (time = 1)
or
R K  ->  R K  (R misses)  2/3  ->  K  (K hits)  2/3

Let p = probability R survives in a duel w/ K. Since p is independent of
time, it is the same at time = 0 and at time = 1.

p = 1/3 + (2/3)(1/3)p

Solving, we get p = 3/7.

Round 2 case 2 (R shooting)

R U  -> R U (R misses) 2/3  -> U  1
or
R U  -> R (R hits) 1/3

So, probability R survives a duel w/ U is 1/3. Let p' = 1/3.

Thus, the probability R survives =

(2/3)p + (1/3)p' = (2/3)(3/7) + (1/3)(1/3) = 25/63 = 0.39682539+