Timeline for Another colored balls puzzle (part II)
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 21, 2013 at 18:19 | vote | accept | navid | ||
| May 21, 2013 at 18:19 | history | bounty ended | navid | ||
| May 16, 2013 at 19:07 | history | bounty started | navid | ||
| May 15, 2013 at 11:49 | answer | added | Karl Fabian | timeline score: 2 | |
| May 15, 2013 at 8:50 | answer | added | Douglas Zare | timeline score: 2 | |
| May 15, 2013 at 6:17 | answer | added | Ori Gurel-Gurevich | timeline score: 6 | |
| May 14, 2013 at 23:21 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| May 14, 2013 at 18:08 | history | edited | navid | CC BY-SA 3.0 |
added 4 characters in body; added 7 characters in body; Post Made Community Wiki
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| May 14, 2013 at 17:40 | answer | added | Vincent Beffara | timeline score: 3 | |
| May 14, 2013 at 17:29 | comment | added | Russ Woodroofe | A very similar problem has already been posted at mathoverflow.net/questions/41939/a-balls-and-colours-problem but the method of choosing balls is a little different. | |
| May 14, 2013 at 13:30 | comment | added | Barry Cipra | Exact calculations for $n=4$ give expected values of $31/3$ and $53/12$, in agreement with the OP's simulated values (10.33 and 4.416). | |
| May 14, 2013 at 12:36 | comment | added | navid | @GregMartin See update to question. | |
| May 14, 2013 at 12:33 | history | edited | navid | CC BY-SA 3.0 |
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| May 14, 2013 at 12:19 | history | edited | navid | CC BY-SA 3.0 |
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| May 14, 2013 at 11:17 | history | edited | navid | CC BY-SA 3.0 |
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| May 14, 2013 at 11:06 | history | edited | navid | CC BY-SA 3.0 |
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| May 14, 2013 at 4:54 | comment | added | Douglas Zare | My calculation earlier wasn't right. The first case may simply be exponential, not asymptotically greater than $c^n$ for every $c$. | |
| May 13, 2013 at 21:52 | comment | added | Greg Martin | Navid, have you run simulations? | |
| May 13, 2013 at 21:24 | history | edited | navid | CC BY-SA 3.0 |
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| May 13, 2013 at 21:17 | comment | added | navid | To confuse matters again, I have now deleted my reply to the deleted comment. | |
| May 13, 2013 at 19:57 | comment | added | Douglas Zare | I don't have a solution yet, but I'd like to point out that there is a huge difference between these procedures. If you always paint the first ball the color of the second, then when there are two colors, there is a strong restoring force toward equal counts. If you are over $3/4$ for one color, the chance to increase it is under $1/4$. This means it will take at least exponentially long for every base, and I think it is something like $n!$. I think the second procedure takes at most a quadratic number of turns on average. | |
| May 13, 2013 at 19:48 | history | edited | navid | CC BY-SA 3.0 |
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| May 13, 2013 at 19:21 | comment | added | Will Jagy | You really need to choose your friends more carefully. | |
| May 13, 2013 at 19:09 | history | asked | navid | CC BY-SA 3.0 |