The symmetric situation of Sally and Bill suggests we solve the problem for arbitrary probabilities.
So, let $s$ be Sally's chance of hitting the target and $b$ be Bill's chance. When it's Sally's turn, let $p(s,b)$ represent her chance of eventually winning. We seek a formula for $p(s,b).$ Because Sally and Bill play identical roles in this game, merely differing in their skills, any formula for $p$ will tell us that $p(b,s)$ is Bill's chance of eventually winning when it's his turn. This is the first simplification afforded by the symmetry.
We also know the turns in the game alternate between the players. Thus, when it's a player's turn, their chance of eventually winning equals their chance of hitting the target plus the chance that their opponent (who gets the next turn) eventually loses. That, of course, is given by subtracting the opponent's chance of winning from $1.$
This gives two symmetrical equations, one for each player:
$$\left\{\eqalign{p(s,b) = s + (1-s)(1-p(b,s))\; \\ p(b,s) = b + (1-b)(1-p(s,b)).}\right.$$
That's the second simplification due to symmetry--and it's enough to go directly to the answer, because plugging the second equation into the first (to eliminate $p(b,s)$) gives an equation relating $p(s,b)$ to $s$ and $b.$ Use algebra to solve it:
$$p(s,b) = \frac{s}{1 - (1-s)(1-b)}.$$
With $s=7/10$ and $b=4/10,$ this gives
$$p(7/10, 4/10) = \frac{7/10}{1 - (3/10)(6/10)} = \frac{70}{82}\approx 85.37\%$$
for Sally to win if it's her turn and
$$p(4/10, 7/10) = \frac{4/10}{1 - (3/10)(6/10)} = \frac{40}{82} \approx 48.78\%$$
for Bill to win if it's his turn.
Another solution method notes this is a Markov chain on the four states "Sally's turn," "Bill's turn," "Sally wins," and "Bill wins." You can set up the $4\times 4$ transition matrix
$$\mathbb{P} = \pmatrix{0 & 3/10 & 7/10 & 0 \\ 6/10 & 0 & 0 & 4/10 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1}$$
and turn the Markov chain theory crank. Because the largest size of the non-unit eigenvalues of $\mathbb{P}$ is $\sqrt{(1-s)(1-b)},$ or approximately $0.4242641,$ whose common logarithm is $-0.37,$ and double-precision arithmetic doesn't quite achieve $16$ decimal places, taking a power of $\mathbb P$ greater than about $16 / 0.37 = 43$ will reveal the answers (in its $(1,3)$ and $(2,4)$ entries). That is,
$$p(s,b) = \lim_{n\to \infty} (\mathbb{P}^n)_{\text{Sally's turn},\text{Sally wins}}= \lim_{n\to \infty} (\mathbb{P}^n)_{1,3}$$
and
$$p(b,s) = \lim_{n\to \infty} (\mathbb{P}^n)_{\text{Bill's turn},\text{Bill wins}}= \lim_{n\to \infty} (\mathbb{P}^n)_{2,4}$$
both converge exponentially rapidly.