August 2026: Andy’s Afternoon Amble

At first glance, this looked like a cool and somewhat tricky math puzzle, but it ended up being disappointingly easy. I managed to scribble down a solution in fifteen minutes or so not long after the puzzle was released, and though the solution was satisfying, I found myself wishing there was more.

Stats:

Difficulty: 3/10

Enjoyability: 5/10

Leaderboard Placing: 15/?

The approach:

Starting from a given hexagon in the flat grid, it’s possible with some geometric reasoning to discern which of the other Hexagons Andy could enter that would immediately allow him to conclude that he’d fallen off the sphere. With Andy’s starting hexagon painted green, the hexagons which would allow him to conclude this have been painted red in the below image.

Andy's Afternoon Amble

Other hexagons have been painted blue (\(b\)), yellow (\(y\)), orange (\(o\)), and pink (\(p\)). Letting \(E\) denote the event that Andy returns to the green hexagon before visiting any red hexagon, the answer to the puzzle is given by \(1 - \mathbb{P}(E)\). Note that the coloured grid could be extended, but since it’s enclosed by a boundary of red hexagons, it’s impossible for Andy to move beyond the currently coloured region without the game ending, so there’s no need to colour anything else.

Letting \(\mathbb{P}(E \mid c)\) denote the probability that event \(E\) occurs given Andy starts in a hexagon of colour \(c\), we have the following identities by various applications of the law of total probability:

  • \(\mathbb{P}(E \mid b) = \frac{1}{3} + \frac{2}{3}\mathbb{P}(E \mid y)\)
  • \(\mathbb{P}(E \mid y) = \frac{1}{3}\mathbb{P}(E \mid b) + \frac{1}{3}\mathbb{P}(E \mid o)\)
  • \(\mathbb{P}(E \mid o) = \frac{1}{3}\mathbb{P}(E \mid y) + \frac{1}{3}\mathbb{P}(E \mid p)\)
  • \(\mathbb{P}(E \mid p) = \frac{2}{3}\mathbb{P}(E \mid o)\)

One can solve this system of linear equations by hand to find \(\mathbb{P}(E \mid b) = 9/20\). Since Andy’s first move must be from the green hexagon to a blue one, we in fact have \(\mathbb{P}(E) = \mathbb{P}(E \mid b)\), hence the answer is \(1 - \mathbb{P}(E) = \mathbf{11/20}\).