Timeline for Generalization of a mind-boggling box-opening puzzle
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| S Oct 21 at 18:43 | vote | accept | Dominic van der Zypen | ||
| Oct 21 at 17:05 | comment | added | Timothy Chow | @SeanEberhard The reason I picked two presents for the version on Kalai's blog was because I thought it was particularly interesting that neither player has an advantage with just one present, but one player does have an advantage when there are two presents. But you're right that there's nothing all that special about two presents. The Monthly version asks the reader to determine all $k$ for which one player has an advantage if there are $k$ presents. | |
| Oct 21 at 14:02 | answer | added | Ilmari Karonen | timeline score: 9 | |
| Oct 21 at 12:07 | vote | accept | Dominic van der Zypen | ||
| S Oct 21 at 18:43 | |||||
| Oct 21 at 12:05 | history | became hot network question | |||
| Oct 21 at 10:09 | comment | added | Sean Eberhard | I am not sure why there is such an emphasis (both in this question and its papertrail) on two presents. The question is simpler and just as "mind-boggling" with one present. In this case the connection with intransitive dice is even closer. | |
| Oct 21 at 10:02 | answer | added | Sean Eberhard | timeline score: 4 | |
| Oct 21 at 8:02 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
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| Oct 21 at 7:54 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
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| Oct 21 at 6:06 | comment | added | Dominic van der Zypen | @DanielAsimov Yes the ${n\choose 2} = n(n-1)/2$ possible placements of the two presents are assumed to be equally likely. | |
| Oct 21 at 2:19 | answer | added | Timothy Chow | timeline score: 9 | |
| Oct 21 at 0:22 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Oct 20 at 18:46 | comment | added | Daniel Asimov | Is there any assumption that all possible placements of the two presents are equally likely? | |
| Oct 20 at 15:20 | answer | added | Iosif Pinelis | timeline score: 8 | |
| Oct 19 at 18:34 | comment | added | LSpice | Re, ah, sorry, I missed that in the post. I think that $\binom{[n]}2$ might be a more common notation. | |
| Oct 19 at 18:01 | comment | added | user14111 | Of course the relation is not transitive. Consider $(1,2,3,4,5,6)$, $(3,4,5,6,1,2)$, and $(5,6,1,2,3,4)$. | |
| Oct 19 at 17:18 | comment | added | Dominic van der Zypen | @LSpice no, $[n]^2 = \{\{x,y\}: x\neq y\}$, so $[n]^2$ has ${n\choose 2} = n(n-1)/2$ elements, and indeed, any $P\in [n]^2$ is a subset of $[n]$, so for $a\in S_n$, the concept of $a^{-1}(P)$ is meaningful. | |
| Oct 19 at 16:55 | comment | added | Peter Taylor | By considering composition of permutations, it's easy to see that for every $n$ there is a $k$ such that each $\pi \in S_n$ is better than exactly $k$ other elements of $S_n$, so both of the questions have trivial answers (respectively: all of them, and none of them). | |
| Oct 19 at 15:45 | comment | added | LSpice | I think of $[n]^2$ as a set of ordered pairs (although do you mean to allow two presents to be put in one box?), but, when you write $a^{-1}(P)$, you're regarding it as a subset of $[n]$ and meaning by this notation its pre-image in $[n]$, right? | |
| Oct 19 at 10:00 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
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| Oct 19 at 9:58 | comment | added | Dominic van der Zypen | OK thanks @SimonRose for this remark, and my being seriously mind-boggled about this problem could have its cause in what you're saying. | |
| Oct 19 at 9:54 | comment | added | Simon Rose | It's worth noting that the methods chosen aren't arbitrary. To make it clear that different methods yield different results, consider that Anna's strategy could be the order [6, 1, 2, 3, 4, 5] and Berts could be [1, 2, 3, 4, 5, 6]. It should be clear in this case that Bert wins 86% of the time. | |
| Oct 19 at 9:42 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |