Timeline for Probabilities in a riddle involving axiom of choice
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2023 at 1:49 | comment | added | Karl | I've asked a question on MSE here about some of the discussion in these comments. | |
| Dec 15, 2022 at 15:09 | comment | added | Alexander Pruss | I am not sure what $P^*$ is in your comment. I was thinking that $P$ is defined on a complete product $\sigma$-algebra, and then $P'$ is an extension of $P$ to a $\sigma$-algebra that also includes the set of representatives. I was implicitly using the fact that if $\mu$ is a measure on the $\sigma$-algebra $\scr F$ and the $y$ is between the inner and upper $\mu$-measures of $A$, then there is an extension of $\mu$ to the $\sigma$-algebra generated by $\scr F$ and $A$ that assigns $y$ to $A$. | |
| Dec 13, 2022 at 22:20 | comment | added | Mateusz Kwaśnicki | One minor thing: I believe $P'$ is not the extension of $P$, it is rather the restriction of $P$ (to the set $A$ of representatives; that is, $P'(E) = P^*(A \cap E)$), right? | |
| Dec 19, 2013 at 23:02 | comment | added | Denis | yes the order would be: 1)describe the probabilistic strategy 2)opponent choses a sequence 3)probabilistic variable i is instanciated | |
| Dec 19, 2013 at 21:25 | comment | added | Alexander Pruss | But the opponent can win by foreseeing what which value of i we're going to choose and which choice of representatives we'll make. I suppose we would ban foresight of $i$? | |
| Dec 19, 2013 at 19:43 | vote | accept | Denis | ||
| Dec 19, 2013 at 19:43 | comment | added | Denis | How about describing the riddle as this game, where we have to first explicit our strategy, then an opponent can choose any sequence. then it is obvious than our strategy cannot depend on the sequence. The riddle is "find how to win this game with proba (n-1)/n, for any n." | |
| Dec 19, 2013 at 15:05 | comment | added | Alexander Pruss | What we have then is this: For each fixed opponent strategy, if $i$ is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least $(n-1)/n$. That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". | |
| Dec 19, 2013 at 11:54 | comment | added | Denis | ah ok I see where the misunderstanding comes from, it's true that "independently" is ambiguous, because only one random variable is involved here. But I think it still has a mathematical meaning in the sense "it does not depend on the opponent's choice", namely we have $\exists x \forall y$ where $x$ is our strategy and $y$ is our opponent's strategy (i.e. the sequence), and we still win this game because we can choose devise a (probabilistic) strategy that works on all sequences. | |
| Dec 18, 2013 at 15:21 | comment | added | Alexander Pruss | I was assuming that "independently" has the meaning it does in probability theory ($P(AB)=P(A)P(B)$ and generalizations for $\sigma$-fields). But that does require a probabilistic description of the opponent's choice. Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf ) Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though. | |
| Dec 17, 2013 at 15:21 | comment | added | Denis | Our choice of index $i$ is made randomly, but for this we only need the uniform distribution on $\{0,\dots,n\}$. It is made independently of the opponent's choice. | |
| Dec 17, 2013 at 14:47 | comment | added | Alexander Pruss | In the probabilistic variant, I don't see that you can win against any strategy of the opponent. If we are making no probabilistic assumptions whatsoever, then in particular we are not assuming that our choice of index $i$ is independent of the opponent's choice of numbers. | |
| Dec 12, 2013 at 16:16 | history | edited | Alexander Pruss | CC BY-SA 3.0 |
added some more fun stuff
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| Dec 12, 2013 at 15:09 | comment | added | Denis | yes but the point is that we can win again any strategy of the opponent, even if he chooses the sequence after we chose our (probabilistic) strategy. This way we avoid talking about probabilities on sequences. | |
| Dec 12, 2013 at 14:33 | comment | added | Alexander Pruss | But isn't this going to depend on how the opponent chooses the sequence? As Joel Hamkins notes in another comment, if the opponent always chooses the same sequence, then there is a strategy that gets the right answer each time. :-) | |
| Dec 12, 2013 at 13:56 | comment | added | Denis | ok this helps, but when you say "suppose $\vec u$ is chosen randomly",this is where the problem arises. Isn't there a way to ask the riddle so that the notion of probability of victory makes sense, without encountering this problem (for instance agains an opponent who chooses the sequence). | |
| Dec 11, 2013 at 21:07 | history | answered | Alexander Pruss | CC BY-SA 3.0 |