I will interpret "after $n$ rolls" as meaning that we want the probability that our next $n-1$ rolls do not yield a $3$, but the $n$-th roll does give a $3$..

The probability of a non-$3$ on any toss is $\frac{5}{6}$.  So the probability of $n-1$ consecutive non-$3$'s followed by a $3$ is
$$\left(\frac{5}{6}\right)^{n-1}\frac{1}{6}.$$

The reason that we simply multiply is that the results on successive tosses of the die are independent.  The die does not remember what results it has come up with in the past.

We can also get the result by using a counting argument. Imagine tossing a die $n$ times, and recording the results. So a record of $4$ tosses might read $4,5,4,1$.

There are $6^n$ possible records of the results of $n$ tosses, all equally likely.  There are $5^{n-1}$ records in which the first $n-1$ results are a non-$3$ and the last result is a $3$. So our probability is
$$\frac{5^{n-1}}{6^n}.$$

Continuous Dice rolls
LQG formulations have a minimum length/area. Since say, a Planck area can always
be boosted, any minimum area in space can be shrunk. Do LQG proponents worry about local Lorentz invariance violation, and if not, why not?  In LQG, does considering length to be a quantum operator really get rid of the boost problem?
