Best strategy for rolling $20$-sided and $10$-sided dice 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. 
 A: $B$ should reroll anything $5$ or below as he is likely to improve, while with a $6$ he is more likely to get worse.  He then has $\frac 1{20}$ chance of getting each number $1$ to $5$ and $\frac 3{20}$ chance of any number $6$ to $10$.  
$A$ can then compute the chance that each number wins.  Any number $11$ through $20$ is a guaranteed win, so clearly should be accepted.  If he rolls a second die, each number is equally probable, and he will win $\frac 12$ (that he rolls $11-20$) plus $\frac 1{20} \cdot \frac {17}{20}$ (that he rolls a $10$ and wins) plus $\frac 1{20} \cdot \frac {14}{20}$ (that he rolls a $9$ and wins) plus ... $\frac 1{20} \cdot \frac {4}{20}$ (that he rolls a $5$ and wins) plus ...  The total is 
$$\frac 12+\frac 1{20}\left(\frac{17+14+11+8+5+4+3+2+1}{20}\right)=\frac{53}{80}$$
He should therefore accept a $9$ or higher as these have a higher winning chance than a random roll.  His winning chance is then $\frac 12$ (that the first roll is $11-20$) plus $\frac 1{20} \cdot \frac {17}{20}$ (that he rolls a $10$ and wins)  plus $\frac 1{20} \cdot \frac {14}{20}$ (that he rolls a $9$ and wins) plus $\frac8{20}\cdot \frac {53}{80}$ (that he rolls $8$ or less and wins with the second roll).  The total is 
$$\frac 12+\frac {17+14}{20^2}+\frac 8{20}\cdot \frac {53}{80}=\frac {1348}{1600}=\frac {337}{400}=0.8425$$
A: This is not really an answer but a long comment. I did not have chance to read the article you posted so not sure what is the indifference method you refer. Perhaps is the same method I have in mind: Assume player 1 rolls for the second time if and only if his first number was below $x$. Given this strategy compute the expected winning probability of player 2 given his first roll was $y$, that is $W_2(\text{action of 2},y|x)$, under two scenarios: if she rolls again and if she does not roll again. If $x$ and $y$ were continuous variables, you would equate the two winning probabilities (roll and don't roll for 2) and also set $y=x$ and finally solve for $x$. In the discrete case you have to consider several inequalities: $$ W_2(\text{roll},y=x-1|x)>W_2(\text{don't roll},y=x-1|x)$$ and  $$ W_2(\text{roll},y=x|x)<W_2(\text{don't roll},y=x|x).$$
I would use a computer algebra system (Maple or Mathematica) and solve this by brute force.
