Which prison cells will remain open in the following problem involving a drunken jailor? Suppose we have $100$ prison cells initially all locked. A drunken jailor chooses an integer $i$ uniformly randomly from the interval $\left[1,100\right]$ and then opens every $i$-th cell. On any succeeding round he repeats the same procedure and toggles every $i$-th cell (all integer $i$'s are chosen independently). Find the expression for the probability of the $n$-th cell being open on the $j$-th round, $n$ and $j$ being given.

The probability of toggling door $n$ in any particular round is $\dfrac{d(n)}{100}$ where $d(n)$ is the number of divisors of $n$.

Therefore, the $j$ rounds are a binomial process with $$p=\frac{d(n)}{100}$$ and being open means having been chosen an odd number of times.

This is given by $$\sum_{k}p^{2k+1}(1-p)^{j-2k-1}\binom{j}{2k+1}=\frac{1-(1-2p)^j}2$$ for the $p$ given above.