# Probabilistic stacking of blocks [closed]

This is a variation on the stacking problem.
A block is a 1D object of length L and uniformly distributed mass. (with some negligible thickness).
A stack of size n is a series of n blocks placed flat one over the other(i.e. their lengths are parallel).
A stable stack is one that doesn't topple under constant gravity.
A monolayer stack is where at most $$1$$ block is immediately above another block.

We are interested in finding the probability that a monolayer stack constructed ground-up from $$n$$ blocks is stable.
The stack is constructed probabilistically as follows:

1. Place the first block.(say with its left edge at $$0$$)
2. The next block is placed randomly such that it touches the block immediately before, at least somewhere, equally probably i.e. its left edge is in $$(-L,L)$$,equally probably. The probability of the block landing outside is taken $$0$$. (say its left edge is at $$-L and nowhere else)
3. Place the third block immediately above the second with the same probability distribution. (i.e its left edge $$x_3$$ can land anywhere in $$(x_2-L,x_2+L)$$ and nowhere else)
4. and so on till the n-th block.

Of course most such random stacks topple. Question is , what is the probability $$p(n)$$ that the stack stays stable during construction? ## So far...

I have figured out the following constriants
1. $$\forall m \ge2$$, $$x_{m-1}-L (placement region) and
2. $$\forall m \ge1$$,$$x_m-L/2

(2) is motivated by the fact that placing a block shouldn't destabilise the stack below. For this we calculate the COM of the $$block_n$$, then $$block_n+block_{n-1}$$ and so on till all such sub-stacks are found to be stable
3. $$p(n)=\frac{1}{2^{n-1}} \frac ab$$ for some positive rational $$\frac ab \lt 1$$ (from numerical analysis)

Calulating phase space volume subject to above constraints as a function of $$n$$ seems difficult.
Am I missing some physics which would make the calculation straightforward?
So far I have found(with $$L=1$$) (mc denotes via monte-carlo), $$\begin{array} {|r|r|}\hline n & p(n) \\ \hline 1 & 1 \\ \hline 2 & 1/2 \\ \hline 3 & 7/32 (mc:0.218) \\ \hline 4 & 107/1152 (mc:0.0928) \\ \hline 5 & 2849/73728 (mc: 0.0385\pm 0.0004) \\ \hline 6 & mc: 0.0156\pm0.003\\ \hline \end{array}$$

• plz move to math se if needed – lineage Dec 14 '19 at 2:56
• Is your high value for $p(4)$ a typo? What does frommc mean? – G. Smith Dec 14 '19 at 6:01
• @G.Smith had a typo before in the mc value(was 0.926)..its correct now – lineage Dec 14 '19 at 6:03
• Where did the exact fractions come from? – G. Smith Dec 14 '19 at 6:06
• till p(3) by hand, rest mathematica – lineage Dec 14 '19 at 6:07

Each block in the stack, except the bottom has a 1/2 chance of spoiling the stable stack. The odds are $$2^{-(n-1)}$$.
• at this point, the $2^{-(n-1)}$ doesn't seem to match the monte carlo or $p(3)$ – lineage Dec 14 '19 at 5:40