Hi Jack, thanks for your answer, but I don't think that it is quite right. I have an answer which I obtained from several million simulations and it does not match what your suggested method yields. In fact, if one follows the advice you gave I believe that the answer you get is about -0.11. I think that your approach is mostly correct though, but I don't think the second expression should have n-3 in the exponent, shouldn't it just be n? I tried with this too and evaluated 300 partial sums, but it was still about 3% off. Thoughts?

I think that for $n=3$ the second sum has a value of $0$ instead of $1$. (this is because for $n=3$ the probability of rolling none of $1$, $2$, $3$, $4$, and $5$ is $1$ instead of $0$) However, this only accounts for a $.5$% error, not $3$%.

Ah, yes I agree. Strange. Oh well, I feel as if I am closer now. Perhaps tomorrow's attempts will yield fruit.

I just ran my own numerical trials. Wolfram Alpha calculates my (corrected) sum to be $13489/21600$, or around $.6245$, and my simulation of $100000$ trials had $62434$ successes.

Is that using my suggestion of changing the terms in the second expression from n-3 to n?

No, the $n-3$ is correct.

Hi Jack.

Thanks for the help by the way.

do you understand why it's n-3?

Well I believe it is because we know that at minimum we will roll the dice 3 times to obtain 3 consecutive sixes, thus offsetting the sum. Is this the correct reasoning?

sort of

Here's the idea: you're trying to break up the probability into chunks depending on how long it took you to roll 3 6's. To make sure you don't count anything twice, you only put something with 3 6's into a chunk if it wasn't in any earlier chunks. This happens if the last of the n rolls is a 6 and two of the rolls before it were also 6's.

Then you evaluate the probability that one of these chunks of length n has all the numbers from 1 to 6. It obviously has 6, so ignoring the 6's, you get a random sequence of length n-3 which needs to have everything from 1 to 5, which is where PIE comes in.

Ahhhh, gotcha. Thanks a lot for the help Jack. I believe I will be able to suss out the problem myself now. Your answer of 0.6245 is correct by the way. I wrote a Monte Carlo simulation to evaluate the answer and was attempting to work my way towards it. Thanks again. I marked your answer as correct.

Glad to be helpful!