Probability: 100 random points on the Earth?

Question

Assume the Earth is a perfect sphere with a radius of exactly 4,000 miles. 100 points are randomly selected on the surface. What is the probability that no two points are within 100 miles of each other?

You all have hit on the twist in this question: It sounds like the birthday problem, but the complication is that areas will overlap whenever two points are less than 200 miles apart. I don't know how to get the answer when you factor that in - wanted to see if someone here did.

A question I could pose that sounds more like the birthday problem but has the same wrinkle: in a classroom with 10 students, what are the odds no two students have a birthday within two days of each other? Again, there is that packing issue: two students born on June 1 and June 4 are three days apart, but two days overlap.

Zanti in Math Land: The Adventure Continues...

So anyway, as I was thinking about this problem, I suddenly got a revelation on how to factor in overlapping areas: Do the "birthday problem" calculation again, except double the radius. If there are no points within 200 miles of each other, there can be no overlap. More importantly, the Poisson distribution I was using can be applied to find the approximate odds of exactly n pairs of points within 200 miles of each other. Then, for example, once I have calculated the odds of there being exactly one set of points within 200 miles of each other, I can get the odds that these same two points are within 100 miles of each other by multiplying the answer by 3/4 -- given a circle with radius = 2n, then the outer ring of the circle from n to 2n makes up 3/4 of the area. So, this is the formula I came up with

P(0) + P(1) * 3/4 + P(2) * (3/4)² + P(3) * (3/4)³ + ...

= ∑ P(n) * (3/4)^n for n = 0 → ∞

(cont.)

... where P(k) is the probability of exactly k points within 200 miles of each other.


As previously said, the P(k) formula I was using came from an application of Poisson distribution for the birthday problem. That formula is:

P(k) = λ^k / e^λ * k!, where k is the number of hits and λ is the ratio calculated for any Poisson distribution. In this case, λ = (100 * 99 / 2 ) * 200² / (4 * 4000²) = 3.09375. (Notes: 100 * 99 / 2 is the combination of 2 items from a group of 100; 200² = radius of the circles; 4 * 4000² = surface area of the Earth.) So, we can rewrite the summation as:

= (1/e^λ) * ∑ ((λ^n) / n! * (3/4)^n) for n = 0 → ∞

Finally, substituting the 3.09375 value calculated for λ and simplifying, we end up with

.0453 * ∑ (2.3203)^n / n! for n = 0 → ∞

So, I think I have the answer and do the calculation -- and, to my astonishment, I get 46%, essentially the same value blahb calculated WITHOUT considering overlaps! So, now I am trying to figure out where I messed up, and then it occurs to me: with only 100 relatively small circles on a relatively large sphere, the amount of overlapping areas will be a small fraction of the total area occupied by the 100 circles. In other words, factoring in the overlapping area does not change the answer significantly, just a fraction of 1%. My correction to Blahb's "primitive" approach doesn't change the answer significantly. It still comes out 46%.

Answer 1

OK, so the surface area of the Earth will be

4*pi*r^2 = 64000000pi

The area of a 100 mile radius circle is

pi*r^2 = 10000pi

The first point can be anywhere on Earth.

The next point has a probability of (64000000-10000)/64000000 = 0.99984375 of not being within 100 miles of being within the first point.

The next has probability (64000000-20000)/64000000 = 0.9996875 of not being within 100 miles of the other two

Next has probability (64000000-30000)/64000000 = 0.99953125 of not being within 100 miles of the others.

And so on.

You would then multiply these numbers together to get your answer because you would need all of these events to occur.

Using Excel, I get 0.459563971 approximately.

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edit: Professional student, I was using the "Birthday paradox" idea while I was computing this. It was immediately what I thought of.

These are conditional probabilities, I just didn't write out the notation. It would get messy and complicated, but I wrote out in words what the probability was, and it takes into account the previous points, thus making them conditional.

I'm not 100% on this answer either. I'm not an expert in geometry, so there may be something I'm missing in terms of calculating the area covered by the 100 mi radius circles on a sphere. I don't even know if it makes a difference. But if you come up with the correct area, you can just put it in place of the 10000pi and the idea would be basically the same.

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edit: IMO, zanti3 always asks excellent questions!

And I agree with you, professional student, that the answer could be close to 50%. Birthday paradox is pretty cool, no?

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edit: You know what, I might be mistaken. I didn't take into account that these 100 mi radius circles can overlap and still have centers that are more than 100 mi apart. I think you have to use circle of 100 mile diameters in this one. That way any overlap means that they are too close.

area of these circles are 50^2 * pi = 2500*pi

Recalculating, you get 0.823978885 approximately.

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edit: I think that this is more complex than this. I think that my probability may be close to the correct answer, but I think it could be better. I need a better understanding of geometry than what I have.

Answer 2

Figuring out the exact odds is too difficult, because it's a "packing problem". To understand what I mean by this, suppose that instead of 100 points, we have 25,600 points. Then the total surface area of 25,600 circles of 100 miles diameter equals the total surface area of the globe of 4,000 mile radius. Obviously it's not even possible to pack the circles with no gaps. With fewer points, calculating the odds that the circles would just pack with no overlap is tremendously difficult to do. Now, even with just 100 points, the odds are not exactly figured by the following approximation method:

Select any point, covering any circle of diameter 100 miles. The odds that the next circle selected is not within this first circle is simply (1-s/S), where s is the area of the circle, and S is the total surface area fo the globe. Then the next one has odds of (1-2s/S). Etc. Multiply all 100 of them together and get 0.823979 odds of this happening. However, as exact sounding this is, it cannot be the exact true odds because of the complications of trying to "fit" 100 such circles on the surface of the globe.

Answer 3

I don't know how to answer this problem exactly, but it sounds like the "Birthday Paradox". In order to answer this problem exactly, it is probably fairly complicated and probably requires extensive use of the 'pigeonhole principle'.

Because of its similarity to the Birthday Paradox, it is very likely that in any random distribution at least 2 points will be within 100 miles of each other.

So the probability that no two points are within 100 miles of each other is very small, probably less than 1%.

--- EDIT ----

I don't think Blahb31 is quite right. The reason is that there are a ton of conditional probabilities that need to be taken into account. Multiplying these probabilities won't give the right answer because there are many dependencies, it also doesn't account for the Earth being a sphere, it assume that the earth is a 2D surface.

-- EDIT --
Maybe not... Blahb31 may be right... I'll have to think about this.

-- EDIT --

Blahb31... I think you are right now that I've thought about it more. It is a very cool problem IMO, and I think it is cool that the solution is nearly 50%. BTW, I added you to my contacts.

Answer 4

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