Timeline for Probabilities in a riddle involving axiom of choice
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2013 at 19:32 | comment | added | Alexander Pruss | By the way, can we prove that $M_j$ and $M$ are nonmeasurable, in the case where the adversary's sequence is iid? I think I can prove that $M_j$ is nonmeasurable under the special condition that the representatives are chosen in a shift-invariant way (i.e., if $\tau$ is a shift to the left, with left-hand-most element deleted, and $r(A)$ is the representative of equivalence class $A$, then $r(\tau A)=\tau r(A)$)--by Zorn such a choice is possible. | |
| Dec 10, 2013 at 21:17 | comment | added | Denis | @JoelDavidHamkins That is why I prefer the adversary formulation: you fix a strategy, and then the opponent chooses a sequence. | |
| Dec 10, 2013 at 8:11 | comment | added | domotorp | Now I understand, imho the answer is given in Alexander's excellent first comment. | |
| Dec 9, 2013 at 20:21 | comment | added | Joel David Hamkins | Perhaps it helps to clarify things to point out that if the whole sequence really is fixed, then there is an even better strategy: just announce the values of all the boxes without looking at any of them. | |
| Dec 9, 2013 at 20:18 | comment | added | domotorp | Why was this answer accepted? I disagree with the arguments. 1. We need the strategy to work for every f with 99% chance, so no need to put a probability space on the functions. 2. You pick i, so it is not a problem that they are independent. | |
| Dec 9, 2013 at 19:54 | comment | added | Denis | @AlexanderPruss Thank you for explaining the Brown-Freiling argument, I understand your point better. Would this change if we change the riddle to "an opponent selects a sequence, and then you play with a (possibly probablistic) strategy"? Or in the other order (opponent plays after) ? | |
| Dec 9, 2013 at 19:52 | vote | accept | Denis | ||
| Dec 12, 2013 at 13:54 | |||||
| Dec 9, 2013 at 19:35 | comment | added | Alexander Pruss | @DK: Brown-Freiling argument: Assume CH. Let $\prec$ be a well-order of $[0,1]$. You and I are assigned i.i.d. uniform $X$ and $Y$ in $[0,1]$. One loses if one has the lower number. I say: No matter what you have, say $y_0$, $\{ x : x \prec y_0 \}$ is a countable set, so it has probability zero, and so almost surely I'll lose. But you conclude that almost surely you'll lose. This seems an analogue to your argument. If you think it disproves CH, work instead in ZFC with a translation-invariant extension of Lebesgue measure that assigns zero probability to all subsets of cardinality $<c$. | |
| Dec 9, 2013 at 19:24 | comment | added | Alexander Pruss | I now think the i.i.d. normally distributed counterexample doesn't work. The problem is that you're looking at a "randomly" indexed variable (I guess something like $X_{100M + i}$, where $i$ is randomly distributed over $\{0,...,99\}$ and $M$ is chosen by the algorithm), but the "random" index may not be a measurable function ($i$ is measurable, but $M$ presumably won't be). | |
| Dec 9, 2013 at 19:21 | comment | added | Denis | I don't get why we need a probability measure on the sequences. Why can't we say that "winning with probability at least $\frac{N-1}{N}$" means that no matter the sequence chosen by an adversary opponent, we will win with probability at least $\frac{N-1}{N}$? | |
| Dec 9, 2013 at 17:53 | comment | added | Alexander Pruss | That "for [each] fixed sequence, the probability of failing is at most $1/N$" basically says something like that $P(F|S)=1/N$ for each sequence $S$. But you can't infer that $P(F)=1/N$ unless you've got a probability measure on the whole space conglomerable with respect to the partition induced by the $S$s. (I bet the probabilities are going to be at best finitely additive, and if we have merely finitely additive probabilities, we can have failures of conglomerability.) I am reminded of the Brown-Freiling argument against CH (mdpi.com/2073-8994/3/3/636). | |
| Dec 9, 2013 at 17:41 | comment | added | Denis | I'm not sure I agree. I think we can make sense of "fails with probability at most $1/N$", by saying that for all fixed sequence, the probability (which comes from the strategy) of failing is at most $1/N$. Moreover I don't understand your counter-example, because no matter how you choose the sequence, the strategy still has $\frac{N-1}{N}$ chance of guessing correctly. | |
| Dec 9, 2013 at 17:37 | history | answered | Tony Huynh | CC BY-SA 3.0 |