Timeline for A statistical approach to the prisoners problem
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2013 at 6:16 | comment | added | zneak | I just did the graphics and this looks pretty good, actually. | |
| Oct 21, 2013 at 6:14 | vote | accept | zneak | ||
| Oct 21, 2013 at 5:37 | comment | added | zneak | @ShreevatsaR, yes, as the edit says, this comes out assuming a normal distribution and I am aware this isn't right, but the actual distribution is biased to the left, so I'm surprised that the normal distribution underestimates the result (but then, that might just be me being in the middle of my first stats college class). | |
| Oct 21, 2013 at 2:27 | comment | added | ShreevatsaR | @zneak: alex.jordan's latest answer confirms this $183$ number with an entirely different (and exact) method without the independence assumption, so something seems to be wrong with your simulation. Did you get 157 out of your simulation directly, or did you just get it by assuming normal distribution somewhere? If you did the latter, then we can see that the assumption of normal distribution is worse than the independence assumption Ross used here. | |
| Oct 20, 2013 at 5:24 | comment | added | zneak | I made a program to test (see question update), and the answer should be around 157 days for 99% certainty. 183 days is overestimated by almost exactly one standard deviation. | |
| Oct 20, 2013 at 2:44 | comment | added | Ross Millikan | @leonbloy: The usual solution to the Coupon collector problem gives you the expected time, not the time until you are $99\%$ sure you have them all. I don't see how to modify the usual approach, which is why I took this one. This would be easy to model to see how good the independence assumption is. I believe it is very close, even though I can't justify it. | |
| Oct 20, 2013 at 2:16 | comment | added | leonbloy | +1 This is the sensible approximation for your question. This is esentially a coupon collector problem. en.wikipedia.org/wiki/Coupon_collector's_problem | |
| Oct 20, 2013 at 1:53 | comment | added | zneak | I add $E(X)$ to the summation to account for "wasted days", and they make up for 552 of the 641 days, so I would think I got that part right. For the rest, I'll think about your answer for a bit and come back to accept, probably ;) | |
| Oct 20, 2013 at 1:42 | history | answered | Ross Millikan | CC BY-SA 3.0 |