Probability puzzle - the 3 cannons (Apologies if this is the wrong venue to ask such a question, but I don't understand how to arrive at a solution to this math puzzle).
Three cannons are fighting each other. 
Cannon A hits 1/2 of the time. Cannon B hits 1/3 the time. Cannon C hits 1/6 of the time.
Each cannon fires at the current "best" cannon. So B and C will start shooting at A, while A will shoot at B, the next best. Cannons die when they get hit.
Which cannon has the highest probability of survival? Why? 
Clarification: B and C will start shooting at A.
 A: I assume they all fire balls simultaneously. The probability for the different outcomes of the first volley are as then as follows (I will keep my eighteenths unsimplified for simplicity):
1) $A$ is hit: $\frac{1}{3} + \frac{1}{6} - \frac{1}{18} = \frac{8}{18}$
2) $B$ is hit: $\frac{1}{2}$.
And these two are independent, so we get the following odds for who's dead before round 2:
1) $A$ and $B$: $\frac{4}{18}$
2) Just $B$: $\frac{5}{18}$
3) Just $A$: $\frac{4}{18}$
4) None (back to square one): $\frac{5}{18}$.
This means that after the first volley where someone died, it is just as likely that just $A$ died as it is that both $A$ and $B$ died, so $C$ certainly has better odds than $B$ (since $C$ does have some probability of winning a duel between the two). We therefore need to look at $A$'s odds of survival given that $B$ just died.
In $4$ out of $9$ cases, $A$ will have died in the same volley that killed $B$, so $C$ is crowned winner. In $5$ out of the $9$, a duel breaks out. We can now analyze the probabilities of the outcomes of a single volley the same way as before:
1) $C$ dies with probability $\frac{1}{2}$
2) $A$ dies with probability $\frac{1}{6}$
This gives the following odds for the outcome:
1) They both die: $\frac{1}{12}$
2) They both survive: $\frac{5}{12}$
3) Only $A$ survives: $\frac{5}{12}$
4) Only $C$ survives: $\frac{1}{12}$
So, if $B$ is (among) the first to die (which has probability $\frac{9}{13}$), the probability is $\frac{4}{9}$ that $A$ was also killed, and $\frac{5}{9}$ that we have a duel between $A$ and $C$. In this duel, $A$ has $\frac{5}{7}$ chance to come out on top, and $\frac{1}{7}$ chance to come out on bottom. All in all, this gives survival chances of $\frac{9}{13}\cdot\frac{5}{9}\cdot \frac{5}{7} = \frac{25}{91}$ for $A$. For $C$, there is the $\frac{4}{13} = \frac{28}{91}$ chance of winning without any dueling, plus the $\frac{5}{91}$ for winning by duel against $A$, and similarily, $\frac{1}{13} = \frac{7}{91}$ for winning by duelling against $B$, and we clearly see who is most often the winner here.
A: A has the greatest chance of survival.
Consider the three possible scenarios for the first round:
On the first trial, define $a$ as the probability that A gets knocked out, $b$ is the probability that B gets knocked out, and $c$ is the probability that C gets knocked out.
Since both B and C are firing at A, the probability of a getting knocked out is:
$$a=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$$
Only A is firing at B, so the probability of B getting knocked out is:
$$b=\frac{1}{2}$$
No one is firing at C, so there is no chance of C being knocked out in the first round:
$$c=0$$
The probability of A or B being knocked out first is therefore even.

Now on the second round, there are one of two possibilities: A and C are left to duel, or B and C are left to duel.
Between A and C, the probability of A defeating C is $\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{6}}=0.75$, and the probability of C defeating A is $0.25$.
Between B and C, the probability of B defeating C is $\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}}=\frac{2}{3}$, and the probability of C defeating B is $\frac{1}{3}$.

Finally, assess the total probability of victory for each cannon, using the rule of product:
Probability of A winning:
$0.75*0.5=0.375$
Probability of B winning:
$\frac{2}{3}*0.5=\frac{1}{3}$
Probability of C winning:
$0.25*0.5+\frac{1}{3}*0.5\approx0.2917$
So it's close, but A has the greatest chance of survivial.
