Good point. A 1 in 6 chance of being the same as previous for dice, a 1 in 16777216 of the Pi's being the same.

And also a 5 in 6 chance of it not being the same, a 16777215 in 16777216 chance of not being the same for a pi. So if P = 16777215/16777216 then having two Pi's the probability of not being the same is P, three Pi's P*P, for four P*P*P etc, P^(N-1)

When the probability of not having two the same is 0.5, the probability of having two the same is also 0.5 ... so we need to find N from P^(N-1) = 0.5

When I brute force try that with a lump of Python code the number of P's needed to get that down to 0.5 the number I come up with is 11 million or so.

The numbers are sound though for a dice with P=5/6, that it determines I'd have to roll four times to get another the same as the first.

If the answer is around 4096, I'm not sure where I'm going wrong, in the theory or my coding.