strange duel chances and my analysis
Nope, $B$ has a higher chance of winning. I just distributed percents to each person. Round $1$:
$A$ has a $30\%$ chance of winning, so they get $30\%$
$B$ takes $50\%$ of the remaining $70\%$, so he gets $35\%$
Totals so far: $A-30\%$, $B-35\%$
Round $2$:
$A$ takes $30\%$ of the remaining $35\%$, so he gets $10.5\%$
$B$ takes $50\%$ of the remaining $24.5\%$, so he gets $12.25\%$
Totals so far: $A-40.5\%$, $B-47.25\%$
Round $3$:
A takes $30\%$ of the remaining $12.25\%$, so he gets $3.675\%$
B takes $50\%$ of the remaining $8.575\%$, so he gets $4.2875\%$
Totals so far: $A-44.125\%$, $B-51.5375\%$
By this time, $B$ is already over $50\%$, so we conclude that $B$ has a higher chance of winning. Also, $A$ has a $\frac 6{13}$ chance of winning and $B$ has a $\frac 7{13}$ chance of winning.
So I first encounter this kind of questions, here is my analysis: first attempt: I think A has 30 percent to kill B, then if he succeed, then he will not die, so his probability of death is 0, but if he missed, next round B would take the gun, so A can see he might die at next round is 70 percent of miss multiply 50 percent of B's shot, which equals 35 percent, then next round......... Don't laugh I realized this is an endless loop and you don't know when the game will over because it's a probability game, of course, my statistics grade in school is terrible...
An endless loop, or as we prefer to call it, an iteration.
Let $p_{A\mid A}$ be the probability that $A$ ultimately wins given it is her turn to shoot. This happens if $A$ immediately wins, or if that fails then $A$ survives $B$'s shot and ultimately wins given its then $A$'s shoot again. This defines $p_{A\mid A}$ by the iteration: kill, or miss and live to try again.
$$\begin{align}p_{A\mid A} ~=~& 0.30 + 0.70\cdot0.50\cdot p_{A\mid A} \\[1ex] 20 p_{A\mid A}~=~& 6+7p_A \\[2ex]\therefore\qquad p_{A\mid A}~=~&\dfrac{6}{13}\end{align}$$
And yes, this is a purely theoretical result.
I don't have to worry about dying, so I think the person first shoot has some advantage?
Well if $B$ shoots first then, $p_{B\mid B}=0.5+0.5\cdot 0.7 p_{B\mid B} \implies p_{B\mid B}={10}/{13}$, which means that $A$ has a much worse probability of ultimately winning when $B$ shoots first than if $A$ shoots first.
$$p_{A\mid B} = \dfrac 3{13}$$
So $A$ does have an advantage if she shoots first, but she's still a lousy shooter.