Two players, $A$ and $B$, play a game with $m$ $n$-sided fair dice. $A$ first picks a sequence of nonnegative integers, $\{a_1, a_2, ..., a_n\}$, such that $\sum a_i=m$, and after rolling the dice, computes value $M=\sum a_ix_i$, where $x_i$ is the result of the roll of the $i$th die ($1 \le x_i \le n, \forall i$). B rolls all the dice again, and computes value $N=\sum y_i$, where $y_i$ is the result of the roll of the $i$th die after the reroll ($1 \le y_i \le n, \forall i$). This process is repeated if and only if $M=N$; otherwise, the person with the greater number wins. What values of $a_i$ should $A$ choose to maximize her chance of winning?
In my honest opinion, I think that since the expected value of $M$ does not change as $a_i$ varies, the probability of $A$ winning does not depend on the values of $a_i$, thus any value works. But I have a baseless feeling that if $A$ picks $a_1=n, a_i=0 \forall i \ge 2$, $A$ might have a higher chance of winning.
I thought of this problem with $m=2$, $n=6$ myself. Then, I generalized it.
Consider the simplest possible case where $n = m = 2$. And we will assume that Alice has chosen to just use the first die. So we can ignore her second die, as it is not used. This allows us to easily see all the possible cases:
$$\begin{array}{cccc}A_1&B_1&B_2&A\text{ Score}&B\text{ Score}&\text{Winner}\\\hline 1&1&1&2&2&-\\ 1&1&2&2&3&B\\ 1&2&1&2&3&B\\ 1&2&2&2&4&B\\ 2&1&1&4&2&A\\ 2&1&2&4&3&A\\ 2&2&1&4&3&A\\ 2&2&2&4&4&-\end{array}$$ Since all these are equally likely, Alice has a 50% chance of winning. To get a better feel, you might want to do the similar calculation $n = m = 3$, which has cases where Alice's outcome is not entirely decided (barring ties) by her own roll.
But what you will find is a symmetry: for every case where Alice wins because she rolled high, there is a balancing case where Bob wins because she rolled low, both equally likely. So no, this does not provide Alice with any advantage, but neither does it supply Bob with one.