Timeline for Probabilities in a riddle involving axiom of choice
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2023 at 18:23 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
fix annoying typo (and other minor cleanups) while already bumped.
|
| Dec 14, 2022 at 0:49 | answer | added | Narutaka OZAWA | timeline score: 1 | |
| Dec 13, 2022 at 22:13 | answer | added | Mateusz Kwaśnicki | timeline score: 2 | |
| Jun 21, 2022 at 22:38 | comment | added | Bennett McElwee | @JohanWästlund Even if there are countably infinite mathematicians, a strategy exists such that at most one mathematician fails. See Can an infinite number of mathematicians guess the number in a box with only one error? | |
| Jan 3, 2014 at 17:34 | comment | added | Denis | I didn't come up with it, but unfortunately can't remember the source. I remember someone mentioning that there has even been a research paper involving the function which takes a sequence as argument, and returns the first index for which the sequence match its representant. It even gives a name to such a function (I remember something like "Gavin function", but google does not seem to agree). | |
| Jan 3, 2014 at 5:58 | answer | added | Aaron Meyerowitz | timeline score: 5 | |
| Dec 19, 2013 at 19:43 | vote | accept | Denis | ||
| Dec 17, 2013 at 12:47 | comment | added | Johan Wästlund | This is beyond mind-boggling! What is the source of this riddle? Did you come up with it? I'll probably want to write about it at some point. The version with infinitely many people (where all but finitely many guess correctly) is described on Michael O'Connor's blog, xorshammer.com/2008/08/23/set-theory-and-weather-prediction and in C.S. Hardin and A.D. Taylor, "A Peculiar Connection Between the Axiom of Choice and Predicting the Future", Am. Math. Monthly 115, (February 2008), 91-96, where it is attributed to Yuval Gabay and Michael O'Connor. | |
| Dec 11, 2013 at 21:07 | answer | added | Alexander Pruss | timeline score: 21 | |
| Dec 10, 2013 at 21:15 | comment | added | Denis | @JoelDavidHamkins You're right, thanks for correcting. | |
| Dec 10, 2013 at 9:55 | comment | added | Joel David Hamkins | ...since he or she will be checking from $M=M_j+1=M_i$ onward, not having observed a deviation, but the previous number at $M_i-1$ does in fact differ from the representative. | |
| Dec 10, 2013 at 9:54 | comment | added | Joel David Hamkins | I agree with domotorp that the silent version does not actually work, but not for the reasons domotorp gives. The policy was that a mathematician remains silent if he or she has observed already a deviation of the sequence from the representative. If all $M_j$'s are the same, then they will all be right, since each $i$ will look from $M=M_i+1$ onward, and guess the (correct) value in place $M_i$. But the problematic case occurs when the max $M_i$ is exactly one more than the next highest $M_j$. In this case, $i$ will not be silent, but will definitely guess incorrectly, ... | |
| Dec 10, 2013 at 5:46 | comment | added | domotorp | @Joel and D K: The silent version does not work if all $M_j$'s are equal. E.g., imagine all sequences are 0's - then noone would guess? If someone would guess, that person might be wrong if the unopened box is not 0 but 1. If the silent version was true then in the one mathematician version he would have a strategy to be correct for sure. | |
| Dec 9, 2013 at 19:52 | vote | accept | Denis | ||
| Dec 12, 2013 at 13:54 | |||||
| Dec 9, 2013 at 18:06 | comment | added | Denis | That is why there is +1 in the definition of $M$. If the max occurs twice, no one will be silent and they will all make a correct guess. | |
| Dec 9, 2013 at 17:37 | answer | added | Tony Huynh | timeline score: 2 | |
| Dec 9, 2013 at 17:34 | comment | added | Denis | Indeed, we could modify the question to say "one is allowed to remain silent, and all the others have to guess the content of a closed box". | |
| Dec 9, 2013 at 17:18 | comment | added | Joel David Hamkins | Another interesting thing about this algorithm is that before opening the final box, in the bad case that $M_i$ is at least as large as the approximation $M$ gotten from the $M_j$'s, the mathematician will actually know this already, having observed that the $i$ sequence deviates above $M$ from the representative. | |
| Dec 9, 2013 at 17:02 | comment | added | Denis | @JoelDavidHamkins Yes in all cases "wrong" means "can be wrong", sorry for the shortcut, they can always be right if they are "lucky". | |
| Dec 9, 2013 at 16:59 | comment | added | Joel David Hamkins | @DK, you mean to say that at most finitely many will be wrong, since it could be that all $M_i$ are equal, or that they are bounded but the maximum occurs at least twice, in which case everyone will be right. | |
| Dec 9, 2013 at 16:52 | comment | added | Denis | with infinitely mathematicians numbered by $\mathbb N$, the guy number $i$ can just start looking at the sequence from index $i$. This way, finitely many will be wrong (the ones with $i\leq M$). | |
| Dec 9, 2013 at 16:48 | comment | added | domotorp | Wow, this is indeed amazing! Cannot the same be achieved (with a different strategy) for infinitely many mathematicians? | |
| Dec 9, 2013 at 16:39 | comment | added | Denis | Yes I find it's one of the most surprising use of the axiom of choice :) | |
| Dec 9, 2013 at 16:36 | comment | added | Joel David Hamkins | I really like this version of the riddle! It is already surprising when using only 2, rather than 100. | |
| Dec 9, 2013 at 16:35 | comment | added | Denis | It is because each sequence $\vec u_i$ matches its representant $\vec v_i$ starting in $M_i$. Therefore, if a mathematician starts looking at $\vec u_i$ from position $M> M_i$, he will be able to guess what is $(u_i)_{M-1}$, since it is equal to $(v_i)_{M-1}$, and since he knows $\vec v_i$. | |
| Dec 9, 2013 at 16:32 | history | edited | Denis | CC BY-SA 3.0 |
added 1 characters in body
|
| Dec 9, 2013 at 16:24 | history | edited | Denis | CC BY-SA 3.0 |
deleted 1 characters in body
|
| Dec 9, 2013 at 16:16 | history | asked | Denis | CC BY-SA 3.0 |