Timeline for A balls-and-colours problem
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2010 at 16:25 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
added 1386 characters in body
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| Oct 13, 2010 at 13:20 | comment | added | JBL | It seems very likely that there exist functions $f(n, k)$ such that $p_{\lambda} = \sum_{i} f(|\lambda|, \lambda_i)$. For example, it appears that $f(n, 1) = \frac{(n - 1)^2}{n}$ and $f(n, 2) = \frac{(2n - 1)(n - 2)}{n}$. Probably a little more data would be enough to guess the general form and proceed as in A. Rex's comment on the question. | |
| Oct 13, 2010 at 6:04 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
took out mistaken answer and put in some computations
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| Oct 12, 2010 at 23:54 | comment | added | David E Speyer | I also confirm 4 for n=3. It's 1+1+2/3 +(2/3)^2+(2/3)^3 + .... I think you might have taken the ratio in the geometric series to be 1/3 be mistake. | |
| Oct 12, 2010 at 23:48 | comment | added | Hedonist | I agree with Ross Millikan's comment below. I have verified the claimed formula upto n=4. The only approach I can imagine for this problem is to draw out the Markov chain explicitly and find the expected time it takes for the chain to hit the one non-transient state. | |
| Oct 12, 2010 at 23:01 | history | answered | Aaron Meyerowitz | CC BY-SA 2.5 |