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I am revisiting a question I asked previously with a slight caveat. In my new situation, I am considering the marbles to always be attached to the same neighbors. Hopefully this will be clearer with a rephrasing of the question.

Suppose, I have a ball with a surface area made up of n equally sized units (like a mesh). I choose at random a patch on the surface of area m. The patch resembles a spherical cap. As shown in the image. I am trying to compute the expected value for the total percentage of the surface that will be selected after I select n patches. Selecting a patch is like a selection with replacement, so I select a patch, record which part of the ball was within the patch, and can select the same patch or overlapping patches on the next selection. For simplicity, I consider the surface as a discretized space of $n$ units with patches being of size $m$.

It would also be helpful if I could figure out how to compute the variance for this expected value.

So, what I have thus far is the expected amount of new area that will be selected at selection $i$, with patch size of $m$ for a ball with n patches:

$$ E(i) = \sum_{k=0}^m\frac{???}{n} $$

I arrive at the expected value by summing the probabilities of selecting $k$ amount of new area for $k=0...m$.

Here, the denominator $n$ is the number of possible patches. I am having difficulty figuring out the numerator. I know it is the number of patches that have exactly $k$ amount of unseen area. But how to compute that escapes me.

spherical cap

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  • $\begingroup$ Part of the difficulty is the answer depends on the shape of the patch. To see why, consider two cases. In the first, the patch is a spherical cap that has one quarter the area of the sphere. In the second it is a symmetrical equatorial strip of the same area. The covered area after two independent selections in the first case can be as large as $1/2,$ whereas in the second case--because any two such strips must intersect (with the area of intersection never less than $0.0402\ldots$)--it can never be that great. This makes it clear that the distributions of the areas must differ. $\endgroup$
    – whuber ♦
    Commented Aug 30, 2018 at 20:50
  • $\begingroup$ Good point. The patch is a spherical cap. I’ll add a graphic to make this more clear. $\endgroup$
    – jordanlgraves
    Commented Aug 30, 2018 at 21:18
  • $\begingroup$ Regardless, I hope this make it clear that your question is geometric in nature, so modeling it with discrete patches needs to be undertaken carefully, lest it overlook that essential fact. In that respect it's not really a Coupon Collector problem anymore. $\endgroup$
    – whuber ♦
    Commented Aug 30, 2018 at 21:21
  • $\begingroup$ In my case, it is appropriate to model the surface as discrete units as opposed to a continuous manifold. Updated the question to clarify that. Thanks! $\endgroup$
    – jordanlgraves
    Commented Aug 30, 2018 at 21:24
  • $\begingroup$ I suspect you're unlikely to get useful answers to this one because of the geometric difficulties. Consider asking about the situation you actually have rather than posing an abstract question. $\endgroup$
    – whuber ♦
    Commented Aug 30, 2018 at 21:27

1 Answer 1

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This answer was what I was looking for. With a single cap, I can get the probability that any point chosen randomly from the surface will not be part of the cap as: $ Pr=(1 - \frac{m}{n}) $ and for $k$ caps: $ Pr=(1 - \frac{m}{n})^k $ which can be interpreted as the percentage of surface that is not covered. Given the expected value for the percentage of the surface that is covered: $ E = S - S(1 - \frac{m}{n})^k $

I will be looking to compute the variance and standard deviation.

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