how can I calculate the probability to get triples or better when throwing n 6-sided dice? I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.
I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.
Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...
I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).
Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.

Hint
You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = \binom{6}{1}\\
f_2 = \binom{6}{2}2!+\binom{6}{1}\\
f_3=\binom{6}{3}3!+\binom{6}{1}\binom{5}{1}\frac{3!}{2!}\\
f_4=\binom{6}{4}4!+\binom{6}{1}\binom{5}{2}\frac{4!}{2!}+\binom{6}{2}\frac{4!}{2!2!}\\
f_5=\binom{6}{5}5!+\binom{6}{1}\binom{5}{3}\frac{5!}{2!}+\binom{6}{2}\binom{4}{1}\frac{5!}{2!2!}\\
f_6=\binom{6}{6}6!+\binom{6}{1}\binom{5}{4}\frac{6!}{2!}+\binom{6}{2}\binom{4}{2}\frac{6!}{2!2!}+\binom{6}{3}\frac{6!}{2!2!2!}\\
f_7=\binom{6}{1}\frac{7!}{2!}+\binom{6}{2}\binom{4}{3}\frac{7!}{2!2!}+\binom{6}{3}\binom{3}{1}\frac{7!}{2!2!2!}\\
f_8=\binom{6}{2}\frac{8!}{2!2!}+\binom{6}{3}\binom{3}{2}\frac{8!}{2!2!2!}+\binom{6}{4}\frac{8!}{2!2!2!2!}\\
f_9=\binom{6}{3}\frac{9!}{2!2!2!}+\binom{6}{4}\binom{2}{1}\frac{9!}{2!2!2!2!}\\
f_{10} = \binom{6}{4}\frac{10!}{2!2!2!2!}+\binom{6}{5}\frac{10!}{2!2!2!2!2!}\\
f_{11}=\binom{6}{5}\frac{11!}{2!2!2!2!2!}\\
f_{12}=\binom{6}{6}\frac{12!}{2!2!2!2!2!2!}$$
For $k>13$ we have $f_k=0$
The number of all possible throws $\omega_n$ is:
$$\omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=\frac{f_n}{\omega_n}$$
and probability of getting triple:
$$q_n=1-p_n$$
