Timeline for Is there an elementary proof of a better result for the finite guessing-box puzzle?
Current License: CC BY-SA 4.0
10 events
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| Nov 12 at 21:10 | history | edited | kodlu | CC BY-SA 4.0 |
Fixed a typo in Eric's name. Nice question
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| Nov 12 at 20:47 | answer | added | Elliot Glazer | timeline score: 3 | |
| Aug 4 at 1:04 | comment | added | Timothy Chow | A related question is, in what sense must the graph of a function $f\colon \mathbb{R}^n \to \mathbb{R}$ be a "thin" subset of $\mathbb{R}^{n+1}$? Graphs of functions can be a bit strange; for example, they can be dense, so "almost always" probably can't be defined to be "comeagre." If the function is not measurable then I'm not sure what other notion of "almost always" we might want. | |
| Aug 3 at 9:22 | comment | added | Joel David Hamkins | I'm interested only in deterministic strategies and not necessarily with measurable functions. In the infinitary case, after all, the strategies come from the axiom of choice and the point is to show that there's nothing like that going on in the finite case. | |
| Aug 3 at 1:54 | comment | added | Timothy Chow | Just to clarify what a "strategy" is: Whenever a mathematician needs to make a decision, the strategy specifies a probability distribution over the available options (where the probability distribution depends on the "current state of the world")? | |
| Aug 2 at 14:36 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 2 at 14:18 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 2 at 14:07 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 2 at 13:59 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 2 at 13:38 | history | asked | Joel David Hamkins | CC BY-SA 4.0 |