Timeline for Probability of event occurring before either of two stopping conditions
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2013 at 16:45 | vote | accept | Simd | ||
| Apr 24, 2013 at 10:46 | comment | added | Douglas Zare | You can continue drawing until you have $\sqrt{n}$. This will underestimate the probability because you might reach $1$ and $2$ successfully, then discard this because you encounter a repeat. The underestimate is still more than $1/2n$. | |
| Apr 24, 2013 at 10:12 | comment | added | Simd | Thank you. Embarrassingly (for me) this is the first argument I came up with but decided (mistakenly) it wasn't precise. The process stops early if it finds a repeat or a $1$ and a $2$. So if we talk about $\sqrt{n}$ samples then we know it doesn't contain a $1$ and $2$ already (and also is not a uniform subset). I feel there is one sentence that will fix this confusion of mine so sorry this is so basic. | |
| Apr 24, 2013 at 0:49 | history | answered | Ori Gurel-Gurevich | CC BY-SA 3.0 |