This can be calculated exactly

For example:

[0] = 1 (always starts here)

[1] = P(1) (probability of rolling a 1) = 1/6

[2] = P(2) + P(1)*P(1) = 1/6 + 1/36 (you can get here by rolling a 2 or rolling 2x1s)

...

It's a combinatorial problem and should have a solution as a function of n