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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 ?

Duel Tree

Thanks in advance!

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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 $$\frac25\cdot\frac47+\frac8{15}\cdot\frac13=\color{red}{\frac{128}{315}}$$

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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?

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

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