So we are tossing $10$ dice. Let $X_i$ be the result of the $i$-th toss. Let $Y=X_1+X_2+\cdots +X_{10}$. It seems that you want the variance of $Y$.
The variance of a sum of independent random variables is the sum of the variances. Now calculate the variance of $X_i$. This as usual is $E(X_i^2)-(E(X_i))^2$.
We know that $E(X_i)=3.5$. For $E(X_i^2)$, note that this is
$$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$
We look only at the simplest case, where there is a single die. But to make things more interesting, let the die be $d$-sided, with equal probabilities for the numbers $1,2,\dots,d$. Suppose there are $m$ players. We find the probability that a tie doesn't occur.
Assume that "highest number goes first." So there is no tie if for some $k\ge 2$, one players rolls a $k$ and all other players roll numbers $\le k-1$.
Let the number of players be $m$. First we find the probability that Alicia rolls a $k$ and everybody else rolls a number $\le k-1$.
The probability Alicia rolls a $k$ is $\frac{1}{d}$. The probability everyone else rolls a number $\le k-1$ is $\left(\frac{k-1}{d}\right)^{m-1}$. So the probability of no tie, with Alicia winning, is
$$\sum_{k=2}^d \frac{1}{d} \left(\frac{k-1}{d}\right)^{m-1}.$$
Sum over all players. The probability of no tie is
$$\frac{m}{d^m}\sum_{k=2}^d (k-1)^{m-1}.$$
The remaining sum is a well-known one, with a long history. There are simple formulas for it in the cases $m-1=1,2,3$. For the general case, please see Faulhaber's Formula.
Remark: We can, with some pain, compute the probability of no tie with $3$ players and $2$ dice. We compute the probability that there is no tie and Alicia is the winner, and multiply by $3$.
Alicia can be the clean winner in any one of $10$ ways. She can throw a $3$ and be the clear winner, or throw a $4$ and be the clear winner, or throw a $5$ and be the clear winner, and so on up to throwing a $12$ and being the clear winner. We start computing these various probabilities.
The probability Alicia throws a $3$ is $\frac{2}{36}$. The probability the other two both throw something $\le 1$ is $\left(\frac{1}{36}\right)^2$. Multiply.
The probability Alicia throws a $4$ is $\frac{3}{36}$. The probability the other two both throw something $\le 3$ is $\left(\frac{3}{36}\right)^2$. Multiply.
The probability Alicia throws a $5$ is $\frac{4}{36}$. The probability the other two both throw something $\le 4$ is $\left(\frac{6}{36}\right)^2$. Multiply.
And so on. Add up the $10$ terms we get. One could even extend to $d$-sided dice and $3$ players, and get a closed form formula.
Best Answer
$\newcommand{\Var}{\operatorname{Var}}$
The formula you give is not for two independent random variables. It's for random variables that are as far from independent as you can get. If $X,Y$ are independent, then you have $\Var(X+Y)=\Var(X)+\Var(Y)$. If, in addition, $X$ and $Y$ both have the same distribution, then this is equal to $2\Var(X)$. It is also the case that, as you say, $\Var(X+X)=4\Var(X)$. But that involves random variables that are nowhere near independent.