Check your intuition — Anton Nikolaev

One of these is truly random. Which one?

How to settle it

Using the method R. D. Clarke published in one page in 1946. Divide each square into a grid, count the dots in every cell, and compare how often cells hold 0, 1, 2, 3 dots to what pure chance predicts (the dashed line).

This is how the London bombing speculations were debunked. In 1944 people were certain the V-1s were being aimed — some districts hit again and again, others untouched, with theories about German precision and about which neighbourhoods were targets. Clarke ran the test above on 576 squares of south London and the impacts came out Poisson. Nobody was aiming. The clustering was what randomness looks like.

Why the two histograms have different shapes

Both squares hold exactly the same number of dots, so both histograms have the same average dots per cell. What differs is the spread. Scattering points independently means nothing stops two of them landing in the same cell, so some cells end up crowded — and because the dots are used up elsewhere, others end up empty. Enforcing a minimum distance removes the crowded cells, and removing them removes the empty ones too. The bars pull in toward the middle from both ends at once.

So the histogram is a picture of the variance. Chance is the wider one.

Show the maths

Why chance gives a ratio of one. Drop N points independently and uniformly into M equal cells. Each point lands in a given cell with probability 1/M, independently of the others, so the count in that cell is Binomial(N, 1/M):

mean = N/M = λ variance = N · (1/M) · (1 − 1/M) = λ(1 − 1/M)

so the ratio is exactly 1 − 1/M — here. As M and N grow with λ fixed, the Binomial converges to Poisson(λ), whose mean and variance are both λ. That limit is why the dashed line is a Poisson curve and why the ratio is called one.

Why a minimum distance pushes it below one. This part is exact too. For a stationary point process, the variance of the count in a region W is

Var[N(W)] = λ|W| + λ² ∫∫ [ g(x−y) − 1 ] dx dy

where g is the pair correlation function — how much more or less likely you are to find a second point at distance r from a given one, compared to chance. For a Poisson process g ≡ 1, the integral vanishes, and variance equals the mean.

A hard-core process forbids pairs closer than d, so g(r) = 0 for r < d. Over that range the bracket is −1, the integral is strictly negative, and the variance drops below the mean. The size of the hole in g is the size of the deficit — which is why raising d pushes the ratio further below one, and why the histogram narrows.

No corresponding closed form exists for the mean nearest-neighbour spacing of this process: points are laid down one at a time and rejected on collision, so the density it reaches has no exact expression and is estimated numerically.

Method: R. D. Clarke, “An Application of the Poisson Distribution”, Journal of the Institute of Actuaries 72 (1946), p. 481. Clarke divided 144 km² of south London into 576 quarter-km squares and counted V-1 impacts per square.

Can AIs solve this correctly?

I generated 20 pairs and asked AIs from top labs to classify, here are the results:

Anything but 100% here I would count as a failure. That is what understanding a concept means rather than guessing at it: once you have it, you apply it consistently. Every human who is taught this rule can then get 20 out of 20.

Reading their answers, the models have the correct reasoning but fail to apply it robustly in practice when looking at images.