TITLE: Best strategy for rolling $20$-sided and $10$-sided dice
QUESTION [5 upvotes]: There is a $20$-sided (face value of $1$-$20$) die and a $10$-sided (face value of $1$-$10$) dice. $A$ and $B$ roll the $20$ and $10$-sided dice, respectively. Both of them can roll their dice twice. They may choose to stop after the first roll or may continue to roll for the second time. They will compare the face value of the last rolls. If $A$ gets a bigger (and not equal) number, $A$ wins. Otherwise, $B$ wins. What's the best strategy for $A$?  What's $A$'s winning probability? 
I know this problem can be solved using the indifference equations, which have been described in details in this paper by Ferguson and Ferguson. However, this approach is complicated and it’s easy to make mistakes for this specific problem. Are there any other more intuitive methods?    
Note: $A$ and $B$ roll simultaneouly. They don't know each other's number until the end when they compare them with one another.

REPLY [3 votes]: Certainly the strategy for each player must be of the form to reroll below a certain threshold. To find the threshold, we can start from Ross’ approximate solution and iteratively adjust the strategies until they're in equilibrium.
So assume that $B$ rerolls below $6$ and $A$ rerolls below $9$. Then every number of $B$ below $6$ has probability $\frac12\cdot\frac1{10}=\frac1{20}$, and every number $6$ or higher has probability $\frac1{10}+\frac12\cdot\frac1{10}=\frac3{20}$. Every number of $A$ below $9$ has probability $\frac8{20}\cdot\frac1{20}=\frac1{50}$ and every number $9$ or higher has probability $\frac1{20}+\frac8{20}\cdot\frac1{20}=\frac7{100}$.
If $B$ were to adjust her strategy by keeping $5$, she would get a winning chance of $5\cdot\frac1{50}=\frac1{10}$ instead of the current winning chance of $\frac1{10}\sum_{k=1}^{10}\frac k{50}+3\cdot\frac1{10}\left(\frac7{100}-\frac1{50}\right)$, where the second term corrects for the $9$ and $10$ that are incorrectly included in the sum in the first term. This is $\frac{10(10+1)}{2\cdot10\cdot50}+\frac3{200}=\frac{11}{100}+\frac3{200}=\frac{25}{200}=\frac18$. Thus, there is no gain in this strategy change.
On the other hand, if she were to adjust her strategy by rerolling $6$, she would give up the current winning probability of $6\cdot\frac1{50}=\frac3{25}$ to get the above winning probability upon rerolling, $\frac18$. Since this is slightly larger than $\frac3{25}$, she should switch strategies and reroll $6$. (The reason this didn't show up in Ross’ argument is that given $A$'s strategy, $9$ and $10$ are far more likely than the other numbers, so it makes sense for $B$ to try harder to beat them even though it lowers $B$'s average roll.)
On the other hand, keeping $7$ yields a winning probability of $7\cdot\frac1{50}=\frac7{50}\gt\frac18$, so $B$ shouldn't reroll $7$. Thus $B$’s best response to $A$’s current strategy is to reroll $6$ or lower. The probability for $B$ to end up with a number below $7$ will then be $\frac6{10}\cdot\frac1{10}=\frac3{50}$, and the probability for a number $7$ or higher will be $\frac1{10}+\frac6{10}\cdot\frac1{10}=\frac4{25}$.
Now we should check whether $A$’s strategy is the best reponse to $B$’s new strategy. If $A$ were to adjust his strategy by keeping $8$, he’d get a winning chance of $1-3\cdot\frac4{25}=\frac{13}{25}$ instead of the current winning chance of $\frac12+\frac1{20}\sum_{k=1}^{10}(k-1)\cdot\frac3{50}+6\cdot\frac1{20}\left(\frac4{25}-\frac3{50}\right)=\frac12+\frac{3\cdot9\cdot10}{2\cdot20\cdot50}+\frac3{100}=\frac12+\frac{27}{200}+\frac3{100}=\frac{133}{200}$. Since $\frac{13}{25}=\frac{104}{200}$ is lower, there is no gain in this strategy change.
On the other hand, keeping $9$ yields a winning probability of $1-2\cdot\frac4{25}=\frac{17}{25}=\frac{136}{200}\gt\frac{133}{200}$, so rerolling $9$ is no gain, either. Thus, $A$’s current strategy of rerolling below $9$ is still the best response to $B$’s new strategy. The overall winning probability of $A$ for this equilibrium strategy pair is
$$
\frac12+\frac1{20}\left(1-\frac4{25}+1-\frac8{25}+8\left(\frac12+\frac1{20}\left(1-\frac4{25}+1-\frac8{25}+\sum_{k=1}^8(k-1)\cdot\frac3{50}\right)\right)\right)
\\
=
\frac12+\frac1{20}\left(\frac{38}{25}+4+\frac25\left(\frac{38}{25}+\frac{7\cdot8\cdot3}{2\cdot50}\right)\right)=\frac{21}{25}=0.84\;,
$$
a slight drop from Ross’ approximation due to the slight improvement in $B$’s strategy.