Timeline for Revealing the least helpful of three numbers
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 16, 2024 at 23:54 | comment | added | FirstName LastName | Agreed, this was the result in mind, and, Bndrew chance of victory (1/n) and the (interpolated) distance linear from critical point is also easy to write. | |
| Jul 16, 2024 at 7:12 | vote | accept | fblundun | ||
| Jul 15, 2024 at 5:51 | comment | added | Nitrodon | @FirstNameLastName This is a two-player zero-sum game (since Andrew wins iff Bndrew doesn't win), and each player has a minimax strategy. And yes, this generalizes to any number of boxes. If Bndrew still needs to choose the largest box out of n, the critical point is n^(-1/(n-1)). | |
| Jul 13, 2024 at 23:38 | comment | added | FirstName LastName | @Nitrodon : it seems to me this solution can be generalized to more numbers ... what do you think? | |
| Jul 12, 2024 at 22:25 | comment | added | FirstName LastName | @Nitrodon : For completeness : non-cooperative games can have multiple pure-strategy Nash equilibrium and one may exist here for which Bndrew chance is higher than 1/3. I personally would argue for such better equilibrium to not possibly exist but what is your exact reasoning about this? | |
| Jul 12, 2024 at 2:48 | comment | added | Tom Sirgedas | Fantastic!! This is the elegant answer I was hoping existed! | |
| Jul 12, 2024 at 1:22 | comment | added | Nitrodon | @FirstNameLastName Bndrew has an obvious strategy to guarantee a 1/3 chance of victory, namely ignoring the revealed number and choosing one of the three boxes uniformly at random. I would imagine that any deviation from this strategy would allow Andrew a counter-strategy that works in his favor. | |
| Jul 11, 2024 at 23:57 | history | answered | Nitrodon | CC BY-SA 4.0 |