Final showdown

Sure. Let $x$ be the probability that Bad survives the initial gun battle with Ugly. Since Bad shoots first, there is a $50$ % probability that he wins on the first shot, and a $50$ % probability that he misses and doesn't win. But if he misses, in order for him to survive, Ugly must miss him, which happens with a $40$ % probability. If this happens it returns them back to the original state, where Bad's chance of surviving is again $x$ . Therefore, $x=0.5+0.5(0.4)x\Rightarrow x=\frac{5}{8}$ . This implies that Ugly's chance of surviving the initial gun battle, on the other hand, is only $\frac{3}{8}$ . Having the first shot really gives you a strong advantage. Then do a similar calculation to calculate the probability of Good surviving a one-on-one battle with the survivor. If Bad survives, then Good will have the first shot. He can either hit or miss. If he hits, he survives. But if he misses, then in order to survive, Bad must miss, which brings them back to the original state, in which Good has the same probability of surviving. This forms the equation $x=0.25+0.75(0.5)x\Rightarrow x=\frac{2}{5}$ . The equation for getting into a battle with Ugly gives $x=0.25+0.75(0.4)x\Rightarrow x=\frac{5}{14}$ . So, Good's chance of survival in that case is $\frac{5}{8} \cdot \frac{2}{5}+\frac{3}{8} \cdot \frac{5}{14}=\frac{1}{4}+\frac{15}{112}=\frac{43}{112}$ .

Hello, Arka Dutta. I showed the computations that gave rise to the answer $38.39$ % in response to Laszlo Mihaly already. I'm not sure it represents the optimal strategy of the players, but it is my belief that it does. Note that my assumption is that Bad and Ugly would choose to fire at each other any time both were still alive. This gave Good an incentive to not fire at either while the other two were still both alive because if he hit one of them, then the other would get the first shot against him. But if Good decides to let Bad and Ugly battle it out first, he is guaranteed to get the first shot against his opponent. I just realized that I did not show how I obtained a $32.98$ % probability of Good surviving for the case in which Good decides to fire at Ugly. So, here is that computation. Assume that Good instead decides to fire at Ugly any time Ugly is still alive. Also assume that Bad will decide to fire at Ugly any time Ugly is still alive and vice-versa. Let $w$ be the probability Good survives. Let $x$ be the probability Good survives given that only Bad and Good remain, and Bad gets to fire first. Let $y$ be the probability that Good survives given that only Good and Bad remain, and Good gets to fire first. Let $z$ be the probability that Good survives given that only Good and Ugly remain, and Good gets to fire first. Analyzing the possible transitions between states, one finds $w=0.25x+0.75(0.5y+0.5(0.6z+0.4w))$ . $x$ , $y$ , and $z$ are independent of each other and can each be found separately. Their equations are as follows: $x=0.5(0.25+0.75x), y=0.25+0.75(0.5y), z=0.25+0.75(0.4u)$ . Solving these equations leads to $x=\frac{1}{5},y=\frac{2}{5}, z=\frac{5}{14}$ . Substituting these in to the original equation and solving for $w$ leads to $w=\frac{157}{476}\approx 32.98$ %.

Thank you for your response. I thought that revenge (If he fired at you first) is a good reason to fire at someone (what other reason is there to shoot into the air except the fear of the other guy firing back at you). Thus giving good a chance of 100% of surviving (ugly will fire back at bad because he fired at him first).