A puzzle I’ve wondered about but never got around to solving/verifying:
In the game Pokemon Let’s Go, the Pokemon Abra instantly teleports away if the player is detected within its line of sight.
Let’s say you’re in a square of length $s$ and one Abra randomly spawns in the square, with a random orientation, and has a 60-degree line of sight in front of it. What’s the optimal place in the square to be so that Abra doesn’t teleport away?
If Abra has an unlimited length of sight (i.e. can see as far as possible within the square), it seems like any place in the square is equally good? Since given any place you choose to be, no matter where Abra spawns in the square, there is a uniform 5/6 chance that Abra doesn’t detect you due to its orientation?
However, if Abra does have a limited length of sight (i.e. can only spot you from a distance $d$ away), my initial intuition would be to hide in the corner. But is there an elegant argument here without e.g. using calculus? And are there any convex polygons where hiding in a corner wouldn’t be the optimal place (if the intuition is correct)?
Thanks for reading!
Abra must first be within range of you, i.e. in the disc $D$ of radius $d$ around you. If within range, there is always a $\frac16$ chance its sight sector includes you, i.e. it spots you and Teleports – this is because circles and intersections of convex shapes are all convex. The two events are independent.
By minimising the intersection area of $D$ and the park (square) $P$, you therefore mimimise the chance of Abra spotting you. If $d\ge\sqrt2s$ there is no advantage to hiding in a corner, since $P\subseteq D$ no matter where $D$'s centre is within $P$. Otherwise – and this is obvious, not just intuition – placing $D$'s centre at a corner minimises the probability in question.
This argument naturally generalises to $P$ any convex shape, and indeed any shape if we assume there are no walls around the park.