This question has been posted here before, but I am not getting the correct answer following the intended solution, so I'd like some insight. Reposting for clarity:
Two people, $A$ and $B$, have a 30-sided and 20-sided die, respectively. Each rolls their die, and the person with the highest roll wins. ($B$ also wins in the event of a tie.) The loser pays the winner the value on the winner's die. How does the expected value of the game for player $A$ change when player $B$ can re-roll?
So, I've confirmed that $B$ should re-roll in the case that (s)he rolls some value $v $ such that $v\leq11$. I'm not sure where I'm going wrong in calculating the EV for $A$, though. Assuming one re-roll, I have
$$EV_A = \frac{1}{30}[21+\cdots+30] - \frac{1}{30}\cdot(\frac{1}{9} + \frac{11}{20}\cdot\frac{1}{9})[12+\cdots+20] - \frac{1}{30}\cdot(\frac{11}{20}\cdot\frac{1}{11})[1+\cdots+11]=7.5633$$
However, the correct answer is $5.4275$.
Secondly, what if $B$ was allowed as many as re-rolls as (s)he wants? Would the $EV$ for $A$ be
$$EV_A = \frac{1}{30}[21+\cdots+30] - \frac{1}{30}\cdot\frac{1}{9}[12+\cdots+20] - \frac{1}{30}[1+\cdots+11]=5.766$$
This is because $B$ is guaranteed to land in the range $12-20$; hence, if $A$ rolls between $1-11$, (s)he is surely to lose. Any help would be appreciated!
$\sum_\limits{x=1}^{20}\sum_\limits{y=x+1}^{30}\ P(X=x)P(Y=y)y-\sum_\limits{x=1}^{20}\sum_\limits{y=1}^{x}\ P(X=x)P(Y=y)x$
Without re-rolls
$\frac 1{600}\sum_\limits{x=1}^{20} \frac {(30+x+1)(30-x)}{2}-x^2\\ \frac 1{600}\sum_\limits{x=1}^{20} \frac {(30)(31)-x^2-x}{2} - x^2\\ \frac 1{600}\sum_\limits{x=1}^{20} 465 -\frac {3}{2}x^2 - \frac 12 x\\ \frac 1{600}(465*20 - \frac {20(21)(41)}{4}-\frac {(20)(21)}{4})\\ \frac 1{600}(465*20 - \frac {20(21)(42)}{4})=8.15 $
Allowing for a reroll.
$P(X=x) = \begin{cases} \frac {1}{40} & x\le 10\\\frac {3}{40} & x>10\end{cases}$
$\frac {1}{1200}(\sum_\limits{x=1}^{10}\sum_\limits{y=x+1}^{30}y-\sum_\limits{x=1}^{10}\sum_\limits{y=1}^{x}x+\frac {3}{1200}(\sum_\limits{x=11}^{20}\sum_\limits{y=x+1}^{30}y-\sum_\limits{x=11}^{20}\sum_\limits{y=1}^{x}x)\\ \frac 1{1200}\sum_\limits{x=1}^{10} 465 -\frac {3}{2}x^2 - \frac 12 x+3(\sum_\limits{x=11}^{20} 465 -\frac {3}{2}x^2 - \frac 12 x)\\ \frac 1{1200}((465*10 - \frac {10(11)(22)}{4}) + 3(465*10 - \frac {20(21)(42))}{4}+\frac {10(11)(22))}{4})\\ \frac 1{1200}((465*40 + \frac {10(11)(22)}{2} - 3\frac {20(21)(42)}{4})\approx 5.48$
For multiple re-rolls, you just need to work out the PDF for $X$.