Timeline for An infinite hat puzzle variation—if we don't know our place, can we still be almost all correct?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 1 at 1:19 | comment | added | Joel David Hamkins | Possible objection: there was no requirement that they all follow the same strategy. Reply: all but finitely many must, since otherwise we can swap them around and make them wrong that way. | |
| Jul 31 at 14:01 | vote | accept | Joel David Hamkins | ||
| Jul 31 at 11:46 | comment | added | bof | @Lucenaposition But the solution for $\omega^2$ also works for any ordinal $\gamma\ge\omega^2$, right? Assign arbitrary colors to all hats after the first $\omega^2$, color them all black, whatever, there are sure to be infinitely many errors among the first $\omega^2$ prisoners. | |
| Jul 31 at 8:56 | comment | added | Lucenaposition | It is specified that there are an uncountabel set of prisoners. However, your construction still works for $\gamma=\omega_1$. | |
| Jul 31 at 4:45 | history | answered | paste bee | CC BY-SA 4.0 |