Timeline for Need help with expected value calculation in coupon collectors in coupon collectors variant
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2018 at 3:07 | vote | accept | jordanlgraves | ||
| Sep 8, 2018 at 3:07 | answer | added | jordanlgraves | timeline score: 1 | |
| Aug 30, 2018 at 22:02 | comment | added | shadowtalker | @terrigenus if you figure it out, it would be great if you could post an answer to your own question with an explanation of the solution. | |
| Aug 30, 2018 at 21:50 | comment | added | jordanlgraves | I see now. The solution calls for an expected value which will take into account all of the possible configurations that would contain $i$ unique units. These configurations could be overlapping patches or isolated patches. So I need to figure out how to compute how many ways their are to select on a sphere $x$ different caps which cover a given area. | |
| Aug 30, 2018 at 21:30 | history | edited | jordanlgraves | CC BY-SA 4.0 |
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| Aug 30, 2018 at 21:27 | comment | added | whuber♦ | I suspect you're unlikely to get useful answers to this one because of the geometric difficulties. Consider asking about the situation you actually have rather than posing an abstract question. | |
| Aug 30, 2018 at 21:24 | comment | added | jordanlgraves | In my case, it is appropriate to model the surface as discrete units as opposed to a continuous manifold. Updated the question to clarify that. Thanks! | |
| Aug 30, 2018 at 21:21 | history | edited | jordanlgraves | CC BY-SA 4.0 |
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| Aug 30, 2018 at 21:21 | comment | added | whuber♦ | Regardless, I hope this make it clear that your question is geometric in nature, so modeling it with discrete patches needs to be undertaken carefully, lest it overlook that essential fact. In that respect it's not really a Coupon Collector problem anymore. | |
| Aug 30, 2018 at 21:20 | history | edited | jordanlgraves | CC BY-SA 4.0 |
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| Aug 30, 2018 at 21:18 | comment | added | jordanlgraves | Good point. The patch is a spherical cap. I’ll add a graphic to make this more clear. | |
| Aug 30, 2018 at 20:50 | comment | added | whuber♦ | Part of the difficulty is the answer depends on the shape of the patch. To see why, consider two cases. In the first, the patch is a spherical cap that has one quarter the area of the sphere. In the second it is a symmetrical equatorial strip of the same area. The covered area after two independent selections in the first case can be as large as $1/2,$ whereas in the second case--because any two such strips must intersect (with the area of intersection never less than $0.0402\ldots$)--it can never be that great. This makes it clear that the distributions of the areas must differ. | |
| Aug 30, 2018 at 20:42 | history | edited | jordanlgraves | CC BY-SA 4.0 |
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| Aug 30, 2018 at 20:29 | history | asked | jordanlgraves | CC BY-SA 4.0 |