TITLE: Possibility that all lights $\mathbf{X}=(X_1,X_2,\cdots)$ turn off again with every time turn a light with its number $n\sim\text{geom}(\frac{1}{2})$.
QUESTION [14 upvotes]: Problem: Let $\mathbf{X} = (\mathbb{Z}_2)^\mathbb N$, i.e., $\mathbf{X} = (X_1,X_2,\cdots,X_N,\cdots)$, $X_i\in \{0,1\} $. It can be considered as countable lightbulbs. $0$ means off, $1$ means on. We start with $\mathbf{X}_0 = 0$. Keep generating independent geometric random variables, whose distribution are $geom(1/2)$. Denote them as $K_1, K_2,\cdots$. Now let $\mathbf{X}_m$ (for $m \ge 1$) be as follows
$$(\mathbf{X}_m-\mathbf{X}_{m-1})_k = \mathbf{1}(k = K_m), $$
i.e, in the $m$-  th turn, we only change the status of the $K_m$-th light bulb. Then what is the probability of all lights being off again, i.e.,
$$\mathbb P(\exists m>1, \mathbf{X}_m =0)$$

My first intuition below was wrong I’m afraid.

 My intuition is that since $1/4+1/16+1/64+\cdots=1/3$, the final answer might be $1/2$. Since all lights being off means all lights are encountered even times, I tried to use generating function but it doesn't work since it's hard to derive a correspondence between all lights are encountered even times and generating function coefficients, and brute force calculation seem also hard due to the same reason that there're too many cases to consider, and I'm stuck here. Are there any other thoughts to deal with this question? Thanks!

New intuition: We may define a function $f(x_1,x_2,\cdots)$ as the probability that status $\mathbf{X}=(x_1,x_2,\cdots)$ will eventually become all $0$, then our goal is to calculate $\sum_{n=1}^{\infty}\frac{1}{2^n}f(0,\cdots,0,1,0,\cdots)$ where $1$ only appears on $n$-th term, and we can achieve all the transporting equations of $f$, which is an uncountable dimensional equation system. We can see immediately that the equation system has a solution $f(x_1,x_2,\cdots)\equiv 1$. Can we conclude that this solution is unique (Then the answer to this problem is $1$)?
Continue the intuition above, we define $$g(x_1,x_2,\cdots)=f(x_1,x_2,\cdots)-1,$$ moreover we define $g(0,0,\cdots,0,\cdots)=0$.
Obviously $f\le1$, hence $g$ can achieve a maximum at $(0,0,0,\cdots,0,\cdots)$. Note that after changing all the equations involving $f$ into $g$, the constant term will be cancelled and the coefficients at the righthand side is strictly larger than $0$ and the sum of all coefficients are always $1$. For example, the equation
$$f(1,0,0,\cdots,0,\cdots)=\frac{1}{2}+\frac{1}{4}f(1,1,0,\cdots,0,\cdots)+\frac{1}{8}f(1,0,1,\cdots,0,\cdots)+\cdots$$
is changed into
$$g(1,0,0,\cdots,0,\cdots)=\frac{1}{2}g(0,0,0,\cdots,0,\cdots)+\frac{1}{4}g(1,1,0,\cdots,0,\cdots)+\frac{1}{8}g(1,0,1,\cdots,0,\cdots)+\cdots$$
We want to use the maximum principle, but we lack an equation with $\text{LHS}=g(0,0,0,\cdots,0,\cdots)$, are there any ways to supple this equation?
Edit: Another possible intuition from here: ignore $g(0,0,0,\cdots,0,\cdots)$, we only focus on the remaining term. We wish to discard the term that has infinity $1$ since the probability of getting this term is $0$. If the maximum happens on one remaining element we're done, what we don't want is if the maximum happens on the limiting term, and we wish to prove that this won't happen because the limit is (somewhat?) decaying. (I don't know how to proceed here, but I'll post my intuitions these days anyway)

REPLY [4 votes]: Let $u_k$ be the probability that all lights are off after $2k$ steps
We decompose that event according to the number of flips done for the first bulb, using the law of total probability and the property that a truncated geometric distribution is again geometric. Letting $2j$ be the number of flips done for the first bulb, we get the following recursion:
$$  u_k =\frac{1}{2^{2k}} \sum_{j=0}^k \binom{2k}{2j} u_{j} \iff u_k=\frac{1}{4^k-1} \sum_{j=0}^{k-1} \binom{2k}{2j} u_{j}\tag{1}$$
with the initial condition $u_0=1$.
The solution [*] can be obtained explicitly, and is surprisingly simple:
$$u_{k}=\frac{1}{2k+1} \tag 2$$
Then, the sum $\sum_{k=0}^\infty u_k$ diverges; and by the properties of Markov chains, the state is recurrent, and the probability of returning to zero is $1$.
If we wish to compute the probability of returning with finite steps: Letting $f_k$ be the probability of getting the first "return to zero" (first time lights are off) after $2k$ steps (with $f_0=0$), we have the know relationship $$ u_k = f_1 u_{k-1} + f_2 u_{k-2} +\cdots +f_k u_0 \tag 3$$
I've not found an explicit formula for $f_k$. The first values are
$$
\begin{array}{[c|c|c]}
k & u_k & f_k\\
1 & 0.333333 & 0.333333 \\
2 & 0.200000 & 0.088888 \\
3 & 0.142857 & 0.046561 \\
4 & 0.111111 & 0.030194 \\
5 & 0.090909 & 0.021797 \\
\end{array}
$$
The convergence is very slow, though,  $\sum_{j=1}^k f_k \to 1 - \frac{a}{\log(k)}$ and the probability of having returned to zero after $1400$ steps is still below $0.8$.



Added: the solution of $(1)$:

Let $S(x)= \left( 1+x \right)^{2k}=\sum_{i=0}^{2k} \binom{k}{i}  x^i $ and
$$G(x)= \frac12 \left( S(x) + S(-x) \right)  = \sum_{j=0}^{k} \binom{2k}{2j}  x^{2j}$$
Then evaluating $\int_0^1 G(x) dx $ we get $$  \sum_{j=0}^{k} \binom{2k}{2j} \frac{1}{2j+1}={2}^{2k} \frac{1}{2 k+1} $$
Hence indeed $(2)$ is solution of $(1)$, with the right initial condition $u_0=1$.