height of domino tower Suppose you are building a domino tower using identical pieces of unit length. You place a new domino piece, one at a time, on the top of the tower. However there is a random error in the placement of dominoes, so a new piece is placed some small random distance off the center of the previous piece, $x_{n+1}=x_n + \delta$, where $x_n$ is the horizontal position of nth piece. The characteristic size of the random error is $\sigma\ll$ 1, e.g., $\delta$ is drawn from a Gaussian distribution, $exp(-(\delta/\sigma)^2)$, or from a uniform distribution $-\sigma/2 < \delta < \sigma/2$, or from some other. Obviously some part of the tower will be eventually off balance and it will fall. Question: How does the expected maximum height of the tower, $\langle N \rangle$, depend on $\sigma$? Experimentally (on a computer) I see that the scaling law is very simple: $\langle N \rangle \propto 1/\sigma^2$ (for any distribution function used for $\delta$). How can this be derived analytically?

 A: 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.
