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January 28

How to calculate answers like these?

What's the probability that 3 random points on the Equator leave a gap over 270°?

What's the chance 3 completely random points have no gaps <40°?

How "clustered" is 40°/50°/270°? Where equally spaced is 0 and "all in the same place" is 1? Sagittarian Milky Way (talk) 07:49, 28 January 2017 (UTC)[reply]

An approach to the first two questions would be to consider this as determining the integral of a weight over the volume of the space of a manifold that has a property that can be calculated at each point, where the weight is given by the joint probability density distribution of the coordinates. The manifold for n points is (S¹)ⁿ, and the weighting function for independent uniformly distributed points is just the uniform distribution over the manifold. The property is, in the first question, that "the closest pairwise separation is over 270°".

Your third question is not well-defined. There are an infinite number of measures of clustering that would produce 0 and 1 as you describe. You would have to choose a specific clustering measure. —Quondum 12:56, 28 January 2017 (UTC)[reply]

Are there measures of clustering with articles? Sagittarian Milky Way (talk) 08:11, 30 January 2017 (UTC)[reply]

Quondum, that was a really obscure way to answer a fairly straightforward question.

The first two questions can be solved by integral calculus, by considering the positions of the points as they are added one at a time. For example, if the central angle between the first and second points is alpha, then the arc measure of the places the third point can go so that all points are covered by an arc of 90 degrees is 180 - alpha if alpha is no larger than 90, and 0 otherwise. So the requested probability is

{\frac {1}{180}}\int _{0}^{90}{\frac {180-\alpha }{360}}\,d\alpha

. --JBL (talk) 15:36, 28 January 2017 (UTC)[reply]

Are you sure it's not

{\frac {1}{90}}\int _{0}^{90}{\frac {180-\alpha }{360}}\,d\alpha

? Sagittarian Milky Way (talk) 08:11, 30 January 2017 (UTC)[reply]

Yes: the angle between the first two points is uniformly distributed in [0, 180]. However, if it lies in (90, 180] then you've already lost (regardless of where the third point goes) and so you get zero contribution in these cases. So the size of the domain (what you divide by when computing a probability) is 180, but the contributions to the numerator (the good values) only come from [0, 90]. --JBL (talk) 14:49, 30 January 2017 (UTC)[reply]

It depends on whether you are trying to answer the more general question in the section heading (depending on what "like" means). Giving a concrete solution such as you've given is often a better introduction, but how to generalize it (for example if the probability density is not uniform, or the criterion is more complex) might not be obvious. Simplifying aspects (e.g. noticing that the first point can be fixed due to the symmetry of the problem) are worth exploiting. —Quondum 17:29, 28 January 2017 (UTC)[reply]

Reminds me of this probability question (with an answer that is not obvious. Note, the author is a professor of linear algebra at MIT so uses linear algebra to do the problem.) VIDEO PDF RJFJR (talk) 16:16, 28 January 2017 (UTC)[reply]

I would venture that "questions like these" mean something of the form:

n points of a circle are chosen randomly (uniform law, independent choices). What is the probability that at least one of the arc length between two consecutive points is above/below x?

I see no obvious way to avoid heavy calculus, i.e. a number of integrals in O(n). But I would not be surprised if there was a trick. Tigraan^{Click here to contact me} 17:30, 30 January 2017 (UTC)[reply]

Yes, the general question of x random points and y° of circumference between consecutive points. Sagittarian Milky Way (talk) 09:32, 31 January 2017 (UTC)[reply]

Presumably you mean "at least y°". I would hardly call this "heavy calculus", since what is being integrated in the uniform case is simply finding the volume of a region with flat boundaries (integrating polynomials), no matter how many points are involved. The only complexity is in determining the shape of the n-dimensional region (within a unit n-cube). Getting a closed form expression for an arbitrary number would be a challenge, and finding it by hand for more that a few points would be very tedious. —Quondum 12:04, 31 January 2017 (UTC)[reply]

This should be 13 at the end, not 14, right?

[1] Thanks. 69.22.242.15 (talk) 12:29, 28 January 2017 (UTC)[reply]

It's not clear what you are referring to; I see no "14". The derivation of the equation of the tangent plane seems to me to be correct. —Quondum 12:40, 28 January 2017 (UTC)[reply]

I don't understand why I keep making arithmetic errors like this. 69.22.242.15 (talk) 10:19, 1 February 2017 (UTC)[reply]