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My best probability puzzle

Posted By: ah_clem
Date: Friday, 16 August 2024, at 1:39 p.m.

In Response To: My best probability puzzle (Timothy Chow)

It seems intuitively obvious that if there's only one prize the chances are even. And since there's only fifteen possible places for the prize to hide, a simple brute-force proof is easily accomplished if you don't accept the symmetry argument. Six wins for each, with three ties.

How might a two-prize scenario change things? It must, since why would anyone bother to publish this puzzle? But I'm feeling lazy this morning and don't want to do a brute-force "analysis" of the 210 possible prize placements.

Speaking as a topologist, I'm failing to see how the geometry of the box layout could affect anything unless the organizers have some sort of bias in their prize placement. So the short answer (which must be wrong by QF) is that search order doesn't matter.

And if the search order does make a difference, there are 15! possible search orders, so the "real" question would be how to choose the search order to maximize the chances of winning.

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