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Given $n$ fair coins, we define round $i$ as flipping all of them and obtain one observation $\vec{o_i}$.

We know the domain $O$ has $2^n$ variations.

If the game is designed as terminate as whenever all $2^n$ variations have being observed. What's the expected number of rounds $I$ for the game to terminate?

Even though all the coins are fair, but not necessarily independent. Assume we design the procedure of producing $\vec{o}$ and we know the correlation $\Sigma^{n \times n}$.

peng yu
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  • Are you familiar with the coupon collector's problem? You expect about $2^n \log(2^n)$ tries if everything is nominal-probability of heads is $\frac 12$ and coins are independent. You allude to other conditions but I don't think in a way that one can make use of. – Ross Millikan May 12 '20 at 17:57
  • In the independent case it is a rephrasing of the coupon collector's problem: see https://en.wikipedia.org/wiki/Coupon_collector%27s_problem. According to the article the expected number of rounds scales like $2^n \log(2^n)$. – Rammus May 12 '20 at 18:04
  • thanks for replying, yeah I'm aware of coupon collector's problem. the difficulty here is dependencies. wondering if there is any general solution for correlated observations – peng yu May 12 '20 at 18:04
  • I assume you mean that observations (e.g. vectors $o_1$ and $o_2$) are still independent, and it's just the coins within an observation (i.e. entries in the vector) that are dependent. If my assumption is correct, then you simply have vectors which are not equi-probable. This is akin to coupon collector problem where the coupons are not equi-probable, which is mentioned in the 4th bullet of this wikiepedia section. – antkam May 12 '20 at 19:50
  • yes, your assumption does hold, thanks for the insight! Do you mind elaborate more and maybe make an answer? – peng yu May 12 '20 at 20:43
  • I think i find some clue from this paper, they provided a bound on the expected number which is related to the rarest observation/ commonest observation * n*log(n). And in my case n is actually 2^n, given the covariance does not degenerate. – peng yu May 13 '20 at 03:31

1 Answers1

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The expected number of rounds corresponds to a coupon collector's problem. But as the question's requirement that coins are not necessarily independent from each other. So it's not going to be a standard coupon collector's problem but a non-uniform coupon collector's problem.

The expected number of rounds is going to depend on the probability of rarest observation $P(\vec{o_r})$ and the most common observation $P(\vec{o_c})$.

Based on this paper.

The expected number is going to be bounded between $O(\frac{P(\vec{o_c})}{P(\vec{o_r})} * 2^n * log(2^n))$ and $O(\frac{P(\vec{o_r})}{P(\vec{o_c})} * 2^n * log(2^n))$

peng yu
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  • It’s a pity that I can’t find any resources from $\Sigma$ to $ P(\vec{o_c}) $ or $P(\vec{o_r})$ – peng yu May 15 '20 at 06:21