Timeline for 5 logicians with 1 or 2 hats (Part II) [8,9,10]
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3, 2016 at 9:43 | comment | added | Etoplay | partition_Z says not the "third person will be the first to answer" but "the third person with two hats will be the first to answer" | |
| May 27, 2015 at 15:05 | vote | accept | ghosts_in_the_code | ||
| May 27, 2015 at 15:05 | |||||
| Apr 2, 2015 at 20:24 | comment | added | dmitch | You're ignoring where I said "regardles of X and Y, the third person will be the first to answer". 11222, 12222, 21222, 22222. In all these cases, the third person is the first to "know", so how are they figuring out the total number of hats? | |
| Apr 2, 2015 at 19:17 | comment | added | partition_Z | Finding out the total number of hats in the game is sufficient for the game to be completed. The number is 8(X,Y=1) if first person with 2 hats "knows", 9 if second 2 hatted person "knows" and 10 if the third one "knows". Everyone on the table can gather this information logically and it is trivial to work out the correct number of hats after a two hatted person declares that he "knows". | |
| Apr 1, 2015 at 22:48 | comment | added | dmitch | The game only completes when each person has given a definitive answer. All you've stated is who will be the first to give an answer. That's a long way from "hence, the game always completes". Imagine the situation XY222 where the person with X is first asked. You've proven that regardles of X and Y, the third person will be the first to answer. Since this says nothing about the values of X and Y, how do the first two determine how many they're each wearing in order for the game to complete? | |
| Apr 1, 2015 at 19:01 | review | First posts | |||
| Apr 1, 2015 at 19:17 | |||||
| Apr 1, 2015 at 18:57 | history | answered | partition_Z | CC BY-SA 3.0 |