Timeline for Average number of rows to fit all elements in a multiset of natural numbers
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2016 at 21:01 | vote | accept | marcella | ||
| Dec 3, 2016 at 20:53 | answer | added | Pat Devlin | timeline score: 2 | |
| Dec 3, 2016 at 20:49 | history | edited | marcella | CC BY-SA 3.0 |
added 93 characters in body
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| Dec 3, 2016 at 20:47 | comment | added | marcella | Yes, you are correct. I will include that right now. | |
| Dec 3, 2016 at 20:46 | comment | added | Pat Devlin | So each element $a_i$ is independent identically distributed over a set $R$. I would include that in the original post. | |
| Dec 3, 2016 at 20:45 | comment | added | marcella | Multiset $S$ is being created by choosing numbers from $R$, $2n$ times with replacement. So, $\sum_{\forall{i\in R}}{P_i}=1$. A number $i$ is either present in the multiset or not. Considering binomial distribution probability of having a number $i$, $k$ times in $S$ is $\binom{2n}{k}(P_i^k)(1-P_i)^{2n-k}$. This is my approach. | |
| Dec 3, 2016 at 19:49 | comment | added | Pat Devlin | Your answer can't be determined without knowing a good deal more about the distribution. For example, suppose $S=(a_1, a_2, \ldots , a_{2n})$, and suppose these are selected so that $a_1$ is distributed over $R$ somehow and then the other elements are fixed so that $a_1 = a_2 = a_3 = \cdots = a_{2n}$. Then all the elements are equal with probability 1. [Or we could put some other distribution on $S$] Could you be more specific? | |
| Dec 3, 2016 at 19:05 | history | asked | marcella | CC BY-SA 3.0 |