# Another Three Way Duel

here is the problem:

Three men - A, B and C - are fighting a duel with pistols. It's A's turn to shoot.

The rules of this duel are rather peculiar: the duelists do not shoot simultaneously, but instead take turns. A fires at B, B fires at C, and C fires at A; the cycle repeats until there is a single survivor. If you hit your target, you will fire at the next person on your next turn.

For example, A might shoot and hit B. With B out of the game, it would be C's turn to shoot - suppose he misses. Now it's A's turn again, ad he fires at C; if he hits, the duel is over, with A the sole survivor. To bring in a little probability, suppose A and C each hit their targets with probability 0.5, but that B is a better shot, and hits with probability 0.75 - all shots are independent.

What is the probability that A wins the duel ?

So I have made a small tree (see below) and according to it the probability of A winning is $$\frac{9}{16}$$. Unfortenately this is not the correct answer. Any help ?

Let's take a closer look at our game here. We first establish notation. $$X_1X_2X_3$$ represents the situation where 3 players are still alive, and it is $$X_1$$'s turn first, then $$X_2$$, etc. $$X_1X_2$$ represents the situation where there are two players alive, and it's $$X_1$$'s turn to shoot.

First, let's think about who dies first. In order for $$A$$ to win, either $$C$$ or $$B$$ has to die first. What is the probability of that?

We know that when all three players are alive, there are three possible states we can be in

• $$ABC$$
• $$BCA$$
• $$CAB$$

So when a player dies, the states we can be in are

• $$AB$$
• $$BC$$
• $$CA$$

We clearly don't care about the $$BC$$ case since $$A$$ is dead, so let's consider what the probability of reaching either the first or third case is.

1. The case of $$AB$$.

We know that this can only be reached from the state of $$BCA$$ when $$B$$ successfully shoots. The probability of reaching state $$BCA$$ is $$\frac12$$ (if $$A$$ misses his first shot) $$+\frac12\cdot\frac14\cdot\frac12\cdot\frac12$$ (if $$A,B,C$$ all miss their first shot, and $$A$$ misses his second) $$+ (\frac12\cdot\frac14\cdot\frac12)^2\cdot\frac12$$ etc etc. We can use the formula for a geometric sequence to find this sum, which equals $$\frac{8}{15}$$. From this state, we know that the probability of $$B$$ hitting $$C$$ is $$0.75$$, so the total probability of getting to $$AB$$ is $$\color{red}{\frac25}$$.

1. The case of $$CA$$.

We can apply a similar set of arguments, except to find the probability of reaching $$ABC$$, to find that probability of reaching $$ABC$$ is $$\frac{16}{15}$$. Before you yell at me for doing some pseudo-math, we should realize that since $$ABC$$ is the starting point for this game, this probability should only be used as an intermediate step. From there, the probability of $$A$$ shooting is $$\frac12$$, so the probability of the game reaching the state $$CA$$ is $$\color{red}{\frac8{15}}$$.

Now, let's take each of these two cases, and find the probability that $$A$$ survives in both.

1. $$AB$$

Since $$A$$ shoots first, the probability of $$A$$ winning is $$\frac12+(\frac12\cdot\frac14)\cdot\frac12+(\frac12\cdot\frac14)^2\cdot\frac12...=\color{red}{\frac47}$$

1. $$CA$$

Since $$C$$ shoots first, the probability of $$A$$ winning is $$\frac12\cdot\frac12+(\frac12\cdot\frac12)\cdot\frac12\cdot\frac12+(\frac12\cdot\frac12)^2\cdot\frac12\cdot\frac12+...=\color{red}{\frac13}$$

So, our final answer is $$\frac25\cdot\frac47+\frac8{15}\cdot\frac13=\color{red}{\frac{128}{315}}$$

Your flowchart isn't dealing with the extra probabilities coming from the loops, particularly the loop when all three people miss their first shot. These add extra possibilities that result in $$A$$ winning.

For a simpler example consider the situation with just two people, $$A$$ and $$B$$. We can calculate the probability $$X$$ that $$A$$ wins if they shoot first like follows:

• They could shoot first, hit, and win immediately. Probability $$P(A)$$.
• They could shoot and miss, but then $$B$$ also shoots and misses. It's $$A$$'s turn again, and they have just as much chance as winning, $$X$$, as when they started. Probability P(A)(1-P(B))X.

We thus get the equation $$X=P(A)+(1-P(A))(1-P(B))X$$. Solve for $$X$$ to get $$X= \frac{P(A)}{1-(1-P(A))(1-P(B))}=$$.

Can you use this result and a similar method to work out the solution for 3 people?

• $$X=P(A) + (1-P(A))(1-P(B))X$$ isn't it like this ? – Vasil Yordanov Sep 26 '18 at 20:13
• Whoops, yes. I'll correct it. – Chessanator Sep 26 '18 at 20:17
• Following your example I came to: $$X = \frac{1}{8} + \frac{3}{16} + \frac{1}{8}X + \frac{3}{64}X + \frac{1}{16}X \Rightarrow X = \frac{20}{49}$$ which is not the correct answer ... – Vasil Yordanov Sep 26 '18 at 20:25
• Remember that $A$ doesn't win immediately after any step with three people. Instead, they end up in a two way game. – Chessanator Sep 26 '18 at 20:28
• I tried several times but I cannot get the correct answer for 3 people. Can you advise on this ? – Vasil Yordanov Sep 26 '18 at 21:06