Lets say I am spinning a bottle while sitting in a group of 10 people. I want to know the minimum number of times I would need to spin the bottle to have a 50% chance of hitting everyone in the circle at least once. How would I go about determining this minimum number of spins? Would the method to calculate this change with a very small (say 1%) or large (say 99%) chance, or with a much larger number of people (say 1000)?

I have attempted to look up some formulas but I am really not sure how to model this problem mathematically/in terms of probabilities. I know that it is impossible to determine if the percentage is set to 100% (no guarantee you'll ever hit each unique person). Hopefully this same math question hasn't been duplicated elsewhere with a different framing, the closest I could find was here.

EDIT: A little more searching and apparently this is a well defined problem... It seems searching for it in terms of dice rolls yielded much better results and based on my understanding should be equally applicable (replacing hitting a person with the bottle with getting a certain number on the die).

See here: https://en.wikipedia.org/wiki/Coupon_collector%27s_problem Will update when I think I have a real answer.

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    $\begingroup$ This is related to the coupon collector's problem. With $10$ equally probable values sampled with repetition, you need a sample size of at least $27$ to have a probability of at least $50\%$ of seeing them all (in a sense this is the median compared with a mean of about $29.28968$ and a mode of $23$); at least $13$ to have a probability of at least $1\%$ and at least $66$ to have a probability of at least $99\%$ $\endgroup$ – Henry Dec 19 '16 at 8:06
  • $\begingroup$ Thank you very much for this answer. I have read through the Wikipedia page and I understand how the expected value (what you refer to as the mean) is calculated. I am still, however, a bit fuzzy on how you came up with the number associated with the exact probabilities of 1%, 50% and 99%. I see a reference to Markov and Chebychev probability bounding equations but It is not clear to me how to come up with the answers you did of 27, 13, and 66. $\endgroup$ – scuba Dec 19 '16 at 9:03

In this answer, it is shown that the probability of not hitting everyone by the $m^\text{th}$ spin would be $$ 1-p=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\left(1-\frac kn\right)^m\tag{1} $$ Thus, the probability of hitting everyone by the $m^\text{th}$ spin would be $$ p=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(1-\frac kn\right)^m\tag{2} $$ According to the answer cited above, the expected duration is $nH_n\sim n\log(n)+\gamma n$. If we set $m=n\log(n)+nx$, then $$ \begin{align} p &=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(1-\frac kn\right)^{n\log(n)+nx}\\ &\approx\sum_{k=0}^n(-1)^k\binom{n}{k}\frac1{n^k}e^{-kx}\\ &\approx\sum_{k=0}^n(-1)^k\frac1{k!}e^{-kx}\\[9pt] &=e^{-e^{-x}}\tag{3} \end{align} $$ Therefore, we get the approximation $$ m=n\log(n)-n\log\left(\log\left(\frac1p\right)\right)\tag{4} $$

For $p=1\%$, we need $m\approx n\log(n)-1.52718\,n$.

For $p=50\%$, we need $m\approx n\log(n)+0.366513\,n$.

For $p=99\%$, we need $m\approx n\log(n)+4.60015\,n$.

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