20 sided die and 12 sided die. I have a $12$ sided die and you have a $20$ sided die. We each get two rolls and we can each chose to stop rolling on either one of the rolls, taking the number on that roll. Whoever has the higher number wins, with the tie going to the person with the $12$ sided die. What is the probability that the person with the $20$ sided die wins this game?
I am able to easily calculate the Probability incase each player was allowed only $1$ roll by doing:
Let $A$ be the person with the $20$ sided die
$$P(A \text{ win}) = P(A \text{ win} | \text{roll} > 12)\cdot P(\text{roll} > 12) + P(A \text{ win} | \text{roll} \leq 12) \cdot P(\text{roll} \leq12)$$
I found $P(A \text{ win} | \text{roll} \leq 12) = \frac{11}{24}$ since both have an equal chance of winning except in case of a tie where $B$ wins which happens with Probability of $\frac{1}{12}$.
$$P(A \text{ win}) = 0.675$$
I'm however, stuck on how to approach it when both players are allowed an optional reroll.
 A: A rerolls

*

*always on a throw of $6$ or less.


*sometimes on a roll of $[7,12]$.
A $7$ for example wins $\frac5{12}$ times. A reroll increases these odds with probability $\frac{13}{20}$. This appears to have no definitive answer.

*

*never on a throw of $13$ or more.

A: I did the maths, and it turns out there actually is a right answer, if both players are playing to maximise their chance of winning (and both players know that). $A$ should reroll on a 7 or lower, and $B$ should reroll on a 9 or lower.
We can calculate $P(X \ge Y)$ by considering separately the four cases where the 12-sided and 20-sided dice are rolled once or twice each. Assuming $m \ge k$, we have
\begin{align}
P(X\ge Y|k,m)&=
\frac{km\sum_{i=1}^{12} i}{12^2\times 20^2}
+\frac{m\sum_{i=k+1}^{12} i}{12\times 20^2} 
+ \frac{k\sum_{i=m+1}^{12} (i-m)}{12^2 \times 20} 
+ \frac{\sum_{i=m+1}^{12} (i-m)}{12 \times 20} \\
&=\frac{78km}{12^2\times 20^2} 
+ \frac{78m-\frac{k(k+1)}{2}m}{12\times 20^2} 
+ \frac{k\frac{(12-m)(13-m)}{2}}{12^2\times 20}
+ \frac{\frac{(12-m)(13-m)}{2}}{12\times 20}.
\end{align}
Taking $m<k$ is clearly non-optimal for $B$, but you can handle it by replacing the $\sum_{i=m+1}^{12}(i-m)$ in the last term with $\sum_{i=k+1}^{12}(i-m)$, i.e. subtracting $\frac{\frac{(k-m)(k-m+1)}{2}}{12\times 20}$ from the final result.
Here is a table of the probability (multiplied by $12^2\times 20^2$ for clarity) that $A$ will win, for each $k$ (columns) and $m$ (rows).

I've highlighted the minimum probability in each column in yellow, and the maximum in each row in green. If $A$ chooses $k$ as 1-7 or as 11, then $B$ should choose $m$ as 9, to maximise their probability of winning. But if $A$ chooses $k$ as 8-10, then $B$ should choose $m=10$. Similarly, $A$ should choose $k=7$ if they expect $B$ to choose $m$ between 5 and 7 or between 9 and 10 (actually, if they know $B$ will choose $m=10$, it's an equally good idea for $A$ to choose $k=6$ or $k=7$, but if there's even a tiny chance $B$ might choose $m=9$ then $A$ should probably choose $k=7$). Thus, if both players are playing rationally, they will settle on $A$ playing $k=7$ and $B$ playing $m=9$. In that case, $A$ wins with probability $\frac{12594}{57600}=\frac{2099}{9600}$, and $B$ wins with probability $\frac{7501}{9600}$.
If one of the players chooses something else, it will decrease their odds of winning even if the other player keeps the same strategy. But the other player might be able to improve their odds even further by adjusting to their opponent's mistake. For example, if $A$ chooses $k=8$ while $B$ chooses $m=9$, then $A$'s probability of winning goes down to $\frac{12552}{57600}$, but $B$ can capitalise on $A$'s aggression by choosing $m=10$, and decrease $A$'s odds of winning further to $\frac{12480}{57600}$.
