Timeline for What is the probability that two random permutations have the same order?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2019 at 12:33 | vote | accept | thibo | ||
| Oct 9, 2018 at 4:39 | history | edited | Martin Sleziak |
removed the deprecated (discrete-mathematics) tag; see the tag-info: https://mathoverflow.net/tags/discrete-mathematics/info
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| Oct 9, 2018 at 0:12 | answer | added | Sean Eberhard | timeline score: 6 | |
| Oct 8, 2018 at 23:41 | comment | added | Sean Eberhard | Another reference to add: arxiv.org/abs/1809.10912 | |
| Feb 6, 2016 at 16:22 | comment | added | usul | One observation (sorry if already obvious). Letting $p_j$ be the probability a random permutation has order $j$, you're considering the "collision probability" $\sum_j p_j^2$. It suffices to essentially ignore all $j$ having $p_j = o(1/n^2)$ in this sum: If we consider $S = \sum_{j: p_j \leq 1/n^{2+\epsilon}} p_j^2$, with the constraint $\sum_j p_j \leq 1$, then by convexity $S$ is maximized by setting each $p_j = 1/n^{2+\epsilon}$ and having $n^{2+\epsilon}$ of them, hence $S \leq \frac{1}{n^{2+\epsilon}} = o(1/n^2)$. So you only need consider orders having probability $\Omega(1/n^2)$. | |
| Feb 6, 2016 at 8:29 | answer | added | Brendan McKay | timeline score: 8 | |
| Feb 5, 2016 at 15:49 | answer | added | Igor Rivin | timeline score: 4 | |
| Feb 5, 2016 at 12:40 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 18 characters in body; edited title
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| Feb 5, 2016 at 12:21 | history | edited | thibo | CC BY-SA 3.0 |
edited body
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| Feb 5, 2016 at 12:18 | review | First posts | |||
| Feb 5, 2016 at 12:30 | |||||
| Feb 5, 2016 at 12:15 | history | asked | thibo | CC BY-SA 3.0 |