If I use 3 indistinguishable dice to simulate 1 roll of 1 die. I can use this formula "(SUM % 6) + 1". There SUM is the total amount of the three indistinguishable dice added together, 6 is the max number givin by a sixsided die and 1 is to output 1-6 instead of 0-5.
How to use N indistinguishable dice to simulate 1 roll of N-1 dice? Ex. 3 indistinguishable dice to simulate 1 roll of 2 dice.
Source: standupmaths(Matt Parker) "The Three Indistinguishable Dice Puzzle" @ youtube.com
I doubt that you'll find a nice solution like the sum for the single die, but I believe it will always be possible to do it by an enumerative mapping (though I don't have a proof for that). Here's the most succinctly describable mapping I came up with to simulate two indistinguishable dice using three:
If the roll is $456$, use $13$. Else, if it's of the form $aaa$, use $66$. Else, if it's of the form $aab$ or $ab1$, use $ab$. Else, order the dice in ascending order to form a $3$-digit number $x$ and find $y=(x+3)\bmod13$. If $y\lt6$, use $yy$, else use $1$ and $y-6$.
Here's code I used to check that this yields the desired probabilities for two indistinguishable dice.
P.S.: On second thought, it's a bit silly to use the $1$ and leave all the higher digits to make the modulo computation more difficult; so here's a version that's slightly easier to carry out by hand. (It's also slightly easier to remember because most of the special numbers are now $6$.)
If the roll is $456$, use $36$. Else, if it's of the form $aaa$, use $16$. Else, if it's of the form $aab$ or $ab6$, use $ab$. Else, order the dice in ascending order, add $1$ to the middle die and let $y$ be the remainder of the resulting $3$-digit number modulo $13$. If $y\lt6$, use $yy$, else use $6$ and $y-6$.
Here's the corresponding code.