Timeline for Lottery in O(1) per participant
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2021 at 4:24 | answer | added | Faré | timeline score: 0 | |
| Apr 25, 2021 at 3:01 | comment | added | Faré | More or less, though I'm not 100% happy about it. But with very high probability, the first few terms of a Taylor expansion of $1-(1-X)^N$ or its inverse $1-(1-X)^{1/N}$ will converge fast enough for the winner to show his ticket score is lower than any of his competitors'. | |
| Apr 23, 2021 at 13:26 | history | edited | Stefan Kohl♦ |
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| Apr 23, 2021 at 13:25 | comment | added | Stefan Kohl♦ | As to your comments, it seems that you know an answer to your question -- do you? | |
| Apr 23, 2021 at 11:25 | review | First posts | |||
| Apr 23, 2021 at 12:24 | |||||
| Apr 23, 2021 at 8:03 | history | edited | Faré | CC BY-SA 4.0 |
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| Apr 23, 2021 at 5:30 | history | edited | Faré | CC BY-SA 4.0 |
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| Apr 23, 2021 at 0:33 | comment | added | Faré | I mean $(1-X/N)^N \approx exp(-X)$. | |
| Apr 23, 2021 at 0:23 | comment | added | Faré | For question B, first notice that for $N$ very large, $(1-X/N)^N \approx exp(X)$, and the winner will draw a very small number, so we should probably make a change of variable along those lines indeed. | |
| Apr 23, 2021 at 0:01 | history | edited | Faré | CC BY-SA 4.0 |
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| Apr 22, 2021 at 23:38 | comment | added | Faré | OK, so question A is simple enough: | |
| Apr 22, 2021 at 19:51 | review | Close votes | |||
| May 3, 2021 at 3:08 | |||||
| Apr 22, 2021 at 17:41 | history | asked | Faré | CC BY-SA 4.0 |