Call player 1 $A$, and has a probability of winning $a = 0.75$
Call player 2 $B$, and has a probability of winning $b = 0.86$
Now, the recurrences for winning in exactly $n$ steps:
\begin{align*}
P_A\left(n\right) &= \sum_{i=1}^{10} \left(1-a\right)a^{i-1}P\left(n-i\right) \\
P_A\left(10\right) &= a^{10} \\
P_A\left(k\right) &= a^{10}\left(1-a\right)\hspace{20pt}\text{for $k=11\ldots 20$}
\end{align*}
and
\begin{align*}
P_B\left(n\right) &= \sum_{i=1}^{12} \left(1-b\right)b^{i-1}P\left(n-i\right) \\
P_B\left(12\right) &= b^{12} \\
P_B\left(k\right) &= b^{12}\left(1-b\right)\hspace{20pt}\text{for $k=13\ldots 24$}
\end{align*}
Hence, the probability of a draw is:
\begin{align*}
\mathbb{P}\left(\text{Draw}\right) &= \sum_{n=10}^{\infty} P_A\left(n\right)\cdot P_B\left(n\right) \\
&\approx 0.0108907854748
\end{align*}
Probability that $A$ wins is:
\begin{align*}
\mathbb{P}\left(A<B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n>k\right) \\
&\approx 0.328106229131
\end{align*}
Probability that $B$ wins is:
\begin{align*}
\mathbb{P}\left(A>B\right) &= 1- 0.328106229131 - 0.0108907854748 \\
&= 0.6610029853942
\end{align*}
The probabilities $P_A(n)$ and $P_B(n)$ have the following probability generating functions, which can be obtained from a markov chain, or from the recurrences.
They were used for calculating the above probabilities:
\begin{align*}
G_A\left(x\right) &= \frac{a^{10} x^{10}}{{\left(a^{10} - a^{9}\right)} x^{10} + {\left(a^{9} - a^{8}\right)} x^{9} + {\left(a^{8} - a^{7}\right)} x^{8} + {\left(a^{7} - a^{6}\right)} x^{7} + {\left(a^{6} - a^{5}\right)} x^{6} + {\left(a^{5} - a^{4}\right)} x^{5} + {\left(a^{4} - a^{3}\right)} x^{4} + {\left(a^{3} - a^{2}\right)} x^{3} + {\left(a^{2} - a\right)} x^{2} + {\left(a - 1\right)} x + 1} \\
G_B\left(x\right) &= \frac{b^{12} x^{12}}{{\left(b^{12} - b^{11}\right)} x^{12} + {\left(b^{11} - b^{10}\right)} x^{11} + {\left(b^{10} - b^{9}\right)} x^{10} + {\left(b^{9} - b^{8}\right)} x^{9} + {\left(b^{8} - b^{7}\right)} x^{8} + {\left(b^{7} - b^{6}\right)} x^{7} + {\left(b^{6} - b^{5}\right)} x^{6} + {\left(b^{5} - b^{4}\right)} x^{5} + {\left(b^{4} - b^{3}\right)} x^{4} + {\left(b^{3} - b^{2}\right)} x^{3} + {\left(b^{2} - b\right)} x^{2} + {\left(b - 1\right)} x + 1}
\end{align*}
If you want an exact solution and have enough patience, proceed as described in the similar problem here.
Update
Alright, here are the exact values!
Building the simultaneous equations as described in that linked answer, we will end up with 120 equations of 120 variables.
This is the required equation:
$$
A_{m,n} = a\, b\, A_{m+1,n+1}+a\, (1-b)\, A_{m+1,0}+(1-a)\, b\, A_{0,n+1}+(1-a)\, (1-b)\, A_{0,0}
$$
and use suitable conditions for the $A$ and $B$ to get the simultaneous equations.
On solving, we end up with these fractions:
Probability that $A$ wins:
128661468699006334270970545935163402035818351279278563034680248131211951306420818408303815664073402597810233410628379856174473510026480769849512780888470516916707988073632462382065397204020546646154950516579486430667685760596413858936657826384695765120008305093793237448375317
/
392133575272137033562321922529480637556002210277004603721797902898819466151384497738806782625797036923472568273600896561549333417714595893979748161963175736246220352217612811269094471136286210626868629349810370977735607823040083223533835780920071901429274067796963981352729761
Probability that $B$ wins:
259201463927349435654857219669092667793013975916096275424383507240622048922732900261901499593234202578969685479739744275800172189637965378580954101270438987858519192089714696009199518112275578000642985768411378237000904554750834140057317845238200853885146949160477867482653211
/
392133575272137033562321922529480637556002210277004603721797902898819466151384497738806782625797036923472568273600896561549333417714595893979748161963175736246220352217612811269094471136286210626868629349810370977735607823040083223533835780920071901429274067796963981352729761
Probability of a draw:
4270642645781263636494156925224567727169883081629765262734147526985465922230779068601467368489431746692649383232772429574687718050149745549281279804266231470993172054265652877829555819990085980070693064819506310067017507692835224539860109297175282424118813542692876421701233
/
392133575272137033562321922529480637556002210277004603721797902898819466151384497738806782625797036923472568273600896561549333417714595893979748161963175736246220352217612811269094471136286210626868629349810370977735607823040083223533835780920071901429274067796963981352729761
Update 2
I read the question again. It's not mentioned that they must be consecutive hits!
Anyway, the equation won't change much:
$$
A_{m,n} = a\, b\, A_{m+1,n+1}+a\, (1-b)\, A_{m+1,n}+(1-a)\, b\, A_{m,n+1}+(1-a)\, (1-b)\, A_{m,n}
$$
and set the boundary conditions.
Therefore, without the constraint of consecutive hits:
Probability that $A$ wins:
534150450266477653874216085760600428121292900457 / 992438238358118763882506280744260200076845920193
$\approx 0.538220344220283$
Probability that $B$ wins:
309351085995527832774459980818626475477199204593 / 992438238358118763882506280744260200076845920193
$\approx 0.311708148717965$
Probability of a draw:
148936702096113277233830214165033296478353815143 / 992438238358118763882506280744260200076845920193
$\approx 0.150071507061752$
Update 3
We may also compute it from the summations, by fixing the last shot as a hit.
\begin{align*}
\mathbb{P}\left(A<B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n>k\right) \\
&= \sum_{k=10}^{\infty} \binom{k-1}{9}a^{10}(1-a)^{k-10} \cdot \left(\sum_{i=k+1}^{\infty} \binom{i-1}{11} b^{12}(1-b)^{i-12}\right) \\
&= \frac{534150450266477653874216085760600428121292900457}{992438238358118763882506280744260200076845920193}
\end{align*}
\begin{align*}
\mathbb{P}\left(A=B\right) &= \sum_{k=10}^{\infty} P_A\left(n=k\right)\cdot P_B\left(n=k\right) \\
&= \sum_{k=10}^{\infty} \binom{k-1}{9}a^{10}(1-a)^{k-10} \cdot \binom{k-1}{11} b^{12}(1-b)^{k-12} \\
&= \frac{148936702096113277233830214165033296478353815143}{992438238358118763882506280744260200076845920193}
\end{align*}
Best Answer
Take an example where the probability of Robin and Tuck hitting a target respectively are $0.9$ and $1.0$. Your method would give the probability of Robin winning being $\dfrac{0.9}{0.9+1.0}$ when the true probability is $0$.
Going back to your original probabilities of hitting of $0.45$ and $0.38$, you would do better saying the probability of Robin winning overall might be $\dfrac{0.45\times(1-0.38)}{ 0.45\times(1-0.38) + 0.38\times(1-0.45)}$ by looking at the decisive and mutually exclusive events of one hitting and the other not. This gives your $0.5717\ldots$