Deriving the probability of winning a dice game with two different sets of dice? Purely out of curiosity, I wanted to determine the probability of winning a dice game when each set of dice is not the same. For example, Player 1 has a dice set of $[d_{20}, d_{4}]$, and Player 2 has $[d_6, d_6, d_6, d_6]$. I can get the average outcome of both sets (13 and 14, respectively), but I can't seem to figure out a way to get the exact probability of one player winning.
When I simulate this scenario in a program, the average win rate is approximately 52.5% in favor of Player 2. I was wondering if there is a way to figure out the average win rate for this scenario and if I could generalize this for any two sets of dice.
 A: The math is do-able, but messy.  You can set up a closed form for the computations, and then write a computer program do the heavy lifting.  So, you don't need to do a simulation as such.
I am assuming that die $dn$ refers to an $n$-sided die, whose (equally likely) sides are $\{1,2,\cdots,n\}.$
This means that the sum of the $[d6,d6,d6,d6]$ dice will be in the range of $\{4,5,6,\cdots,24\}.$
Similarly, the sum of the $[d20,d4]$ dice will be in the range of $\{2,3,\cdots,24\}.$

For $~n \in \{4,5,6, \cdots, 24\}, ~$ let $f(n)$ denote the probability that $[d6,d6,d6,d6]$ sum to $n$.
This can be routinely calculated by a computer program.  Simply have each of the dice loop through the values of $1$ through $6$, so that your program is examining $1296$ cases.  If, in the $1296$ cases, the sum $n$ occurs $r$ times, then $~\displaystyle f(n) = \frac{r}{1296}.$
For $~n \in \{2,3,4,5,6, \cdots, 24\}, ~$ let $g(n)$ denote the probability that $[d20,d4]$ sum to $n$.
This can also be routinely calculated by a computer program.  Simply have one die loop through the values $1$ through $20$, while the other die loops through the values $1$ through $4$, so that your program is examining $80$ cases.  If, in the $80$ cases, the sum $n$ occurs $r$ times, then $~\displaystyle g(n) = \frac{r}{80}.$
For $~n \in \{2,3,4,5,6, \cdots, 24\}, ~$ let $h(n)$ denote the probability that $[d20,d4]$ sum to strictly less than $n$.
For $~n \in \{2,3,4,5,6, \cdots, 24\}, ~$ let $j(n)$ denote the probability that $[d20,d4]$ sum to strictly greater than $n$.
This means that

*

*$h(2) = 0.$

*$h(n) = \sum_{k=2}^{n-1} g(k) ~: 3 \leq n \leq 24.$

*$j(24) = 0.$

*$j(n) = \sum_{k=n+1}^{24} g(k) ~: 2 \leq n \leq 23.$
Then, the chance of Player 2 winning is
$$\sum_{i=4}^{24} f(i) \times h(i).$$
Then, the chance of Player 1 winning is
$$\sum_{i=4}^{24} f(i) \times j(i).$$
Then, the chance of a tie is
$$\sum_{i=4}^{24} f(i) \times g(i).$$
