Dice probability question

First we have to compute the probability of rolling triplets in any given 5 die roll.

To do this we compute the number of favorable cases and divide it by the total number of possible cases. Since each dice has 6 sides and we have 5 die, there are a total of 6^(5)=7,776 possible cases. Keep in mind this is differentiating between rolls like: (1,1,1,1,2) and (1,2,1,1,1), but this is ok as long as we are consistent.

Now for the favorable cases. I'll do it for a triplet of 1s and then multiply by 6 to obtain all triplets. If we want at least three 1s, we need to fix the result of three die, so we only have 6^(2)=36 possibilities for the other 2 die. But this is fixing a particular set of 3 die, so we must multiply for the number of ways we can choose a set of 3 die in a set of 5 die. This number equals 5!/(3!2!)=120/12=10. Hence the final number of possibilities for a triplet of 1s is 10x36=360. Now for any triplet it's just  360×6=2,160.

Finally, the chance of rolling triplets is diving the two numbers 2,160/7,776 = 0.2777...

This is the chance of doing it just once. The chance of doing it 9 times in a row is (0.2777...)^(9) = 0.0000098464. Approximately 1 in 100,000, which is incredibly unlikely! I suspect I'm wrong somewhere just because of how unlikely it would be, but I don't see my mistake. I'm tempted to set up a simulation to see if the math checks out.

If anyone sees a mistake please correct me!

Edit: I should say that this is the probability of rolling at least triplets, so I'm also including rolling 4 or 5 of a kind as a triplet. If you want to exclude those then the probability is even lower, although not much much lower.
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