Timeline for Urn problem. How to calculate this probability?
Current License: CC BY-SA 4.0
15 events
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| Feb 8 at 17:19 | comment | added | jblood94 | @DavBhaji, see this answer. It deals with the distribution of the number of unique elements drawn ($k$ in the link). From that, it is straightforward to get the distribution of the number of resamples by subtracting $k$ from the number of samples (i.e., $m-k$ in the link). | |
| Feb 2 at 22:34 | comment | added | whuber♦ | "Counting the number of resamples ... and divid[ing] by n" doesn't compute a probability: it computes an expectation. The difference becomes obvious when you consider the case where $n$ exceeds $m+1$: the probability of at least one resample is $1$ (that's implied by Pigeonhole Principle) while the expected number of resamples must exceed $1$ (by the extended Pigeonhole Principle). So: which quantity are you looking for? | |
| Feb 2 at 20:13 | history | reopened | whuber♦ | ||
| Feb 2 at 16:34 | history | edited | Dav Bhaji | CC BY-SA 4.0 |
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| Feb 2 at 16:32 | comment | added | Dav Bhaji | Thanks for your comments. I edited the question. @LevG I am not interested in "at least one resample during n draws", rather counting the number of resamples in n draws and dividing it by n. | |
| Feb 2 at 16:29 | review | Reopen votes | |||
| Feb 2 at 20:16 | |||||
| Feb 2 at 16:28 | history | edited | Dav Bhaji | CC BY-SA 4.0 |
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| Feb 2 at 15:21 | comment | added | whuber♦ | @LevG agreed. Please note your interpretation of this question already has answers in many posts here on CV, because it is easily answered by counting samples with and without replacement. | |
| Feb 2 at 14:54 | comment | added | LevG | I understood the question as the second thing you wrote: at least one resample during n draws. But Dav should better specify what he needs. | |
| Feb 2 at 14:49 | history | closed | whuber♦ | Needs details or clarity | |
| Feb 2 at 14:49 | history | reopened | whuber♦ | ||
| Feb 2 at 14:45 | comment | added | user2974951 | Which draw? This matters sine in the first draw there is 0 chance of resampling. Or are you asking what is the probability of getting "at least one resample during n draws"? | |
| Feb 2 at 14:44 | comment | added | whuber♦ | Could you please explain the distinction you are implying between "probability" and "average probability"? | |
| Feb 2 at 14:43 | history | closed | whuber♦ | Not suitable for this site | |
| Feb 2 at 14:40 | history | asked | Dav Bhaji | CC BY-SA 4.0 |