height of domino tower
I did not have a rigorous proof here, but the following analysis should provide some reasoning to your quesiton.
If we denote $\delta_i=x_i-x_{i-1}$, we can define $$ \Delta_M^{(N)}\equiv \frac{1}{N-M}\sum_{i=1}^{N-M}i\delta_{N-i+1}=\frac{1}{N-M}\sum_{i=M+1}^N x_i -x_M $$
Because $\delta_i \sim (i.i.d.) (0,\sigma^2)$, $$ \text{Var}(\Delta_M^{(N)})=\sum_{i=1}^{N-M}\left(\frac{i}{N-M}\right)^2\text{Var}(\delta_{N-i+1}) = \frac{(N-M+1)(2N-2M+1)}{6(N-M)}\sigma^2, \quad \text{mean}(\Delta_M^{(N)})=0 $$
The condition that the tower falls at height $N$ is $$ \text{Condition }a_N:\quad \exists M<N, \text{s.t.} \left|\Delta_M^{(N)}\right|>0.5 $$ $$ \text{Condition }b_N:\quad \forall n<N, \forall k<n, \left|\Delta_k^{(n)}\right|<0.5 $$
So the expected maximum height $$ \langle N \rangle = \sum_{N=2}^\infty N \cdot P(a_N \cap b_N) $$
Unfortunately, $a_N$ and $b_N$ do not seem to be independent to me, and I don't know how to evaluate the joint probability. What is worse, I don't even know how to evaluate $P(a_N)$ or $P(b_N)$. For example,
$$ P(a_N)=P\left(\bigcup_{M<N} \left|\Delta_M^{(N)}\right|>0.5 \right) $$ Again this is hard to evaluate because I don't think $\Delta_M^{(N)}$ are independent random variables within the same $N$.
What I can do is give the following, $$ \forall M<N, \quad P\left(\left|\Delta_M^{(N)}\right|>0.5\right) \le P\left(\left|\Delta_1^{(N)}\right|>0.5\right) $$
Suppose $\delta_i$ follows the normal distribution, a ROUGH estimate of the expected height can PROBABLY be given as $$ \tilde{N} \sigma^2 = \sum_{N=2}^\infty N \sigma^2 P\left(\left|\Delta_1^{(N)}\right|>0.5\right) \approx \sum_{N=2}^\infty N \sigma^2 \chi_1^2\left[(0.5)^2/(N\sigma^2/3)\right]\approx \int_0^\infty x \chi_1^2(0.75/x) dx =\text{const.} $$
Here $\chi_1^2(x)$ is the complementary CDF of chi-squared distribution with one degree of freedom. The $\approx$ sign holds here because $\sigma^2\ll 1$
So this rough estimate has the property $\tilde{N} \propto 1/\sigma^2$
Note: the normal assumption for $\delta_i$ is not needed for the conclusion, because
According to central limit theory, even though $\delta_i$ are not normal, $\Delta_1^{(N)}$ is still approximately normal distributed at large $N$
You can replace $\chi^2$ distribution with the corresponding variance distribution of your distribution without changing the conclusion.