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An intriguing treasure hunt is in progress. A valuable item was left on the forest floor somewhere within a 500 mile radius circle shown on the project website. Each day the circle shrinks, so we know the treasure is within the new smaller circle.
At least that's the idea. But this made me wonder: is it actually possible for it to be this simple? That is, can such shrinking discs be generated such that on each day, all we know is that the treasure is somewhere within the latest disc?
Consider the problem. You've placed the treasure, and now you want to generate the discs to reveal to the treasure-hunting public. On the first day, generating a 500 mile radius disc, it's clear how to proceed: pick a point within 500 miles of the treasure uniformly at random, then reveal the disc centred at that point. It's easy to see that if all you know is the disc and that it was generated this way, the treasure has an equal likelihood of lying at any point within the disc.
Now consider day 2. You could apply the same method to generate the new smaller disc, picking the centre uniformly randomly in a disc around the true location. But then it's quite likely that the new disc won't be contained in the first disc -- in the worst case, they could barely overlap at all, and you've accidentally revealed almost the precise location.
So let's say that the new disc must be contained within the first. How can you generate it such that someone who remembers the first disc would still assign an equal likelihood to every point of the new disc? Well, you can't.
To illustrate the problem, let's consider a simpler setup where we can easily calculate. Let's make the problem one-dimensional, so the treasure is placed (uniformly at random on) a circle rather than the globe, and we're generating nested intervals rather than nested discs. Let's also discretise the problem, so the treasure has to lie at one of finitely many equally spaced points on the circle. Let's say the first interval should contain 3 points, and the second 2. So we have four possibilities for the location of the treasure ('*') and the two nested intervals around it:
[[* .] . ] [[. *] . ] [ . [* .]] [ . [. *]]
Now, how do we generate the second interval? If the treasure is on the left, we have to put the interval on the left, similarly for the right. If the treasure is in the centre, the only sensible option is to flip a coin to decide between left and right.
Now, suppose I am a treasure hunter, and I see the two intervals, with the second on the left:
[[. .] . ]
What is the probability that the treasure is on the left? One way to calculate this: of the four possibilities listed above, the first and fourth occur with probability 1/3, while the second and third occur with probability 1/6 because of the coin flip; since we know we're in either the first or second, it's twice as likely that we're in the first, so the probability of it being on the left is 2/3.
It's maybe instructive to see this as an instance of Bayes' rule: denoting I as the event that the intervals are as shown, and L and C as the events that the treasure is at the left or the centre respectively,
P(L|I)*P(I) = P(I|L)*P(L) P(C|I)*P(I) = P(I|C)*P(C)
and since P(L)=P(C) (the treasure was placed uniformly at random), the ratio P(L|I)/P(C|I) is equal to the ratio P(I|L)/P(I|C). But P(I|L) = 1/3, while P(I|C) = 1/3*1/2 because of the coin flip. So again, we get that the treasure is twice as likely to be on the left.
Now let's return to discs on the globe. Here, actual calculations would involve integration, so I'll spare us both the pain. But the same principle applies. Suppose the treasure happens to be on the bounding circle of the first disc, say at the extreme left. Then the second disc has to be placed at the unique position such that it includes that point and is contained in the first disc, pushed up against the left of the first disc. But for any other position of the treasure, that extreme left disc is less likely to be generated (at least if our method of generating discs is invariant under rotation -- anything else would be bizarre, and wouldn't actually help). So by the same Bayesian calculation as above, if a treasure-hunter sees a second disc which is touching the boundary of the first, they should assign a higher likelihood to the treasure being at the point where they touch. More generally, if the second disc is close to the boundary, the treasure is more likely to be close to the boundary.
Anyone who's read this far might like to contemplate the Project Skydrop circles from the last 3 days.
Project Skydrop circles 25-27 Sept
And anyone who lives near Mt. Zion in Massachusetts in the US might want to consider a trip, and be ready to credit probability theory.
