# Bowling percolation puzzle

FiveThirtyEight has an interesting column, Riddler column, which I follow with great interest. In this post we will take a look at bowling percolation puzzle formulated as Riddler Classic puzzle for August 5th post.

## Puzzle

In short (see the original post for a more detailed formulation), we have $$N^2$$ bowling pins arranged in a rhombus pattern. The first (topmost) pin is always hit by the bowler, but further pins must be knocked down by the pins in front. Actually, let us assume that $$p$$ is the probability (which depends on the bowlers skill) for a falling pin to knock down either of the pins directly behind it (independently of each other).

If there is just one pin behind a fallen pin, it will be knocked down with probability $$p$$. If there two pins, then there is $$p^2$$ probability that both pins will go down and $$2 p (1-p)$$ chance that a single pin (either of the two) will go down.

Given this mechanism, what is the probability to hit the last (bottommost) ping? How does it depend on $$p$$?

## Solution

General analytical solution for arbitrary $$N$$ is quite cumbersome to obtain. Actually, it is an unsolved problem in the percolation theory. Yet this problem is quite approachable to be solve numerically.

Below you should see an interactive app, which allows you to experiment with this puzzle numerically. You can change $$N$$ and $$p$$ values to observe what happens. Smaller $$N$$ are best for visual viewing purposes, otherwise larger $$N$$ should produce the same results plus reveal additional information.

On the left hand side of the app pins are shown. Some of them are highlighted (with red outline). State of the highlighted pins is observed over iterations of the simulation.

On the right hand side of the app there is a plot which shows how empirical probability (frequency) changes as the state of the middle pins of deeper rows is observed. As this dependency appears to be close to exponential, on the y axis of the plot we have base 10 logarithm of the empirical probability itself. Note that for larger $$p$$ the dependence flattens out: if $$p$$ is larger than some critical value, then the system will percolate (bottommost pin will be knocked down) with non-zero probability.