Timeline for Randomly put $k$ balls in $2n$ circular boxes, pick $n$ consecutive boxes such that the number of balls is minimum!
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 30, 2016 at 7:12 | comment | added | N. Gast | Douglas: I agree with your reasoning. The problem with my simulations was that at first, I only simulated up to $n=10^6$ (which is $k=14$), for which the value seems to concentrate around 0.33. For $n=1000$, it is closer to $k/2$. | |
| Jul 30, 2016 at 7:06 | history | edited | N. Gast | CC BY-SA 3.0 |
added 673 characters in body
|
| Jul 29, 2016 at 21:48 | comment | added | Douglas Zare | I don't believe the $k/3$ result. The argument you give for the $k/4$ lower bound can be modified very simply to give $(1/2-\epsilon)k$ by using more sectors, e.g., with $10$ equal $\pi/5$ sectors any semicircle contains at least $4$ so for large $k$, the minimum is at least $0.4k$. | |
| Jul 29, 2016 at 14:06 | comment | added | user95536 | I actually haven't tried to simulate it because for a small number $k$, the answer "may be" converge to $k/f(n,k)$ for some function $f$. And if such $f$ is small relative to $n$ and/or $k$ (take an example $f = \log k$), $f$ can't be distinguished with a constant (I suppose). Before that, could you elaborate more for lowerbound $k/4$? Why can't it be $k/c$ for $c>4$ with the same reason? | |
| Jul 29, 2016 at 7:42 | history | answered | N. Gast | CC BY-SA 3.0 |