Background
In this post, the three way dual has a terminator whose has 100% percent of hitting the target, things simplify much because that terminator will let the tree-way duel to two-way duel very quickly.
Now if that guy don't have one hundred percent accuracy, what is the strategy that A will take?
Here is the complete description of the current question, you can ignore the above discussion and start from here directly.
$A,B$ and $C$ arrange a three-way duel, $A$ can hit target with accuracy $1/4$, while $B$ can hit with accuracy $1/2$, $C$ can hit with accuracy $3/4$. They shoot in turn, that is $ABCABC\cdots$ until two of them are down. Since $A$ is first to shoot, what is $A$'s strategy that have the largest chance to win? Suppose $B$ and $C$ are also reasonable man.
If he has to shoot it's better to kill C. Whoever he kills the other will shoot at him, so better B shoot at him than C.
But if he is allowed not to shoot, then we can calculate is it better to wait (cos we know B and C will not shoot at A)
So probability for A to survive when refuse to shoot is:($h$ and $m$ mean hit/miss, $D_{AB}$ means probability for A to survive duel AB when A shoots first) $P=B_hD_{AB}+B_mC_hD_{AC}+B_mC_mP$ $$D_{AB}=1/4(1+\frac 3 4 \frac 1 2+(\frac 3 4 \frac 1 2)^2+(\frac 3 4 \frac 1 2)^3+...)=2/5$$ $$D_{AC}=1/4(1+\frac 3 4 \frac 1 4+(\frac 3 4 \frac 1 4)^2+(\frac 3 4 \frac 1 4)^3+...)=4/13$$ So $$P=\frac 1 2 \frac 2 5+\frac 1 2 \frac 3 4 \frac 4 {13}+\frac 1 2 \frac 1 4P$$ $$\frac 7 8P=1/5+3/26=41/130$$ $$P=164/455$$ And lets calculate probability if he decide to shoot: $$P=A_hD_{BA}+A_m(B_hD_{AB}+B_mC_hD_{AC}+B_mC_mP)$$ $$P=\frac 1 4\frac 1 2\frac 2 5+\frac 3 4(\frac 1 2 \frac 2 5+\frac 1 2 \frac 3 4 \frac 4 {13}+\frac 1 2 \frac 1 4P)$$ $$\frac {29}{32}P=1/5+9/104$$ $$\frac {29}{32}P=149/520$$ $$P=596/1885$$
First is slightly better than second, so he should wait(if is allowed to do so)