Post #3179036

2025-09-03 14:00 UTC

All pairs are equally likely! Congratulations to the many people who picked this. Why are the probabilities equal? Because you can perform a linear transformation on *any* nondegenerate triangle that turns it into an equilateral triangle, where the probabilities are obviously equal. But such a transformation preserves the ratio between the areas of any two regions, and hence the probabilities.

Replies (4)

@isaackuo@spacey.space 2025-09-03 14:30

@gregeganSF@mathstodon.xyz What's not necessarily so obvious is that this linear transformation preserves probability distributions. I mean ... it does in this case, but it's not a gimme. For example, if you said - "Pick a random line that intersects the triangle" then it no longer works. But why not? You have to get into the nature of linear transformations and what it means to pick a "random line" ...

Open ##3179044
@SvenGeier@mathstodon.xyz 2025-09-03 19:20

@gregeganSF@mathstodon.xyz Whoa - so my pick (purely on intuition, nothing about maps or anything) was right. I'm actually starting to develop a functional geometric intuition in my older age...

Open ##3179046
@buster@mathstodon.xyz 2025-09-03 22:40

@gregeganSF@mathstodon.xyz I'm lazy, so I put a huge amount of effort into montecarloing it, was surprised, and only then thought (just) hard enough to see that it was (probably) ok to use affine transformations to reduce to the case of the equilateral triangle.

Open ##3179047
@oantolin@mathstodon.xyz 2025-09-04 16:25

@gregeganSF@mathstodon.xyz I thought of it slightly differently, instead of mapping the triangle to an equilateral one, I thought of mapping it to itself but permuting the vertices.

Open ##3179048