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 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.
