Question : Three people named $A, B, C$ are doing a shooting contest. $A$ has an accuracy of $100\%$, $B$ has an accuracy of $70\%$, and $C$ has an accuracy of $30\%$. They take turns shooting, however, they decided to let $C$ go first, since he is the worst shooter, and $B, A, C, B, A, ..$ and so on. They shoot at each other, and if you get hit, you die. The last survivor wins the game, and all players try their best to win. What would the probability be for each player to win (at least, who is most likely to win)?
This should also be considered: $C$ would shoot $A$ because he knows that if $A$ shoots him, he $100\%$ dies. $B$ would also shoot $A$, because $C$ would eliminate him for the same reason. $C$ would shoot $B$ because he is the highest risk for $C$, and so on.
This question can be approached with elementary properties of probability and the enumeration of all configurations of the shooting contest.
Note: I mistakenly changed the probabilities from $70\%$ to $\frac23$ and $30\%$ to $\frac13$. The approach will be same, only values will differ.
Let's start by decomposing the original problem into smaller components. At any time in the game there can be:
When there are two contestants remaining, and one of them is C, then the contest is almost over. C never misses. If A or B is shooting, then either they shoot C with probability $\frac13$ and $\frac23$ respectively, and the game ends, or they miss and in the next round C wins.
Consider the third case, where C has been shot and it's A and B that remain. Whoever goes first, the same thing is going on, either they shoot, hit and win, or they shoot, miss and the turn goes to the next player.
The diagram of a binary tree shows these possible outcomes. Each node has two possible children: hit or miss. The game ends on a hit and it continues to the next player on a miss. This can go on for quite a long time. It's unlikely the game will go on too long, as their miss probabilities are only $\frac23$ and $\frac13$, respectively, but the probability of winning needs to take into account the possibility that it goes on forever.
In the case where A goes first, the sequence of probabilities that A wins is: $\frac13, \frac{2}{27}, \frac{4}{243}, ...$ .
The probability of A winning on the $i^{th}$ iteration will be
$$ \frac{2^i}{3^{2i+1}}$$
This is a geometric series and can be summed from $0$ to $\infty$ with a result of $\boxed{\frac37}$.
Similarly, the sequence of probabilities that B wins looks like $\frac49, \frac{8}{81}, \frac{16}{729}, ...$ The $i^{th}$ iteration is given by $$\frac{2^{i+1}}{3^{2i}}$$ where $i$ runs from $1$ to $\infty$. Again, a geometric series that can be summed and results in the probability of B winning is $\boxed{\frac47}$
And, that's reassuring as the sum of these two probabilities must be 1.
This can also be done for the situation where B goes first. Draw the binary tree, assign the probabilities, compute the infinite sums and find out that for that case $P$[A wins] is $\frac17$ and the $P$[B wins] is $\frac67$.
OK, back to the original problem. Now you just need to enumerate the possible outcomes. A diagram is worth 1000 words - but no diagram, so I'll try and get it done in less than 1000 words!
Draw the tree, starting with A's actions. This is no longer a binary tree. Start with the two actions, hit and miss. Notice that the hit action is made up of two possibilities: hit B or hit C. Without additional knowledge as to the choice of target, let the choice of shooting B or C have equal probabilities.
If A hits B, then on the next round C wins by hitting A. (outcome 1, C wins)
If A hits C, then you are in the situation we just analyzed where it's a two player game with B going first. (outcome 2, A or B wins)
If A misses, then it's B's turn to shoot. Use the same two actions, hit and miss. The hit action is made up of two possibilities: hit A or hit C. Let the choice of target have equal probabilities.
If B hits A, then on the next round C wins by hitting B. (outcome 3, C wins)
If B hits C, then you are in the other situation we just analyzed where it's a two player game with A going first. (outcome 4, A or B wins)
Finally, if B misses, then it's C's turn. C makes the game more tractable. C never misses. There's only one action, hit. The hit action is made up of two possibilities: hit A or hit B. Let the choice of target have equal probabilities.
If C hits A, then on the next round, B either hits C and wins, or misses C and loses. (outcome 5, B or C wins)
If C hits B, then on the next round, A either hits C and wins, or misses C and loses. (outcome 6, A or C wins)
When I draw the trees, assign the probabilities and compute them I get:
P[C wins] = P[C wins outcome 1] + P[C wins outcome 3] + P[C wins outcome 5] + P[C wins outcome 6] = (1/2)(1/3) + (1/2)(4/9) + (1/2)(2/27) + (1/2)(4/27) = 1/2
P[B wins] = P[B wins outcome 2] + P[B wins outcome 4] + P[B wins outcome 5] = (1/2)(1/3)(6/7) + (1/2)(4/9)(4/7) + (1/2)(4/27) = 130/378
P[A wins] = P[A wins outcome 2] + P[C wins outcome 4] + P[C wins outcome 6] = (1/2)(1/3)(1/7) + (1/2)(4/9)(3/7) + (1/2)(2/27) = 59/378
I'm not sure how to deal with the additional caveats: "This should also be considered: C would shoot A because he knows that if A shoots him, he 100% dies. ...."
If C would shoot A because if A shoots him he would be 100% dead, then C is going to be very confused. Because, if B shoots C then C would also be 100% dead. SIt's the same condition. So, flip a coin and use equal likelihoods.
Perhaps the equal weighting used in the case of multiple targets could be manipulated to show a preference to which target to choose. That's something I did not pursue.
Another variation on the problem could be to recognize that there is no 100% shooter in the world. The problem's solution would change if C had a probability of hitting the target of, say 99/100, instead of 1.
So, I think I'll leave it like this and await comment ...