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Let's assume that the shape of a certain planet is a ball. Three spaceships land independently on random points of the planet's surface. Every two of them can communicate directly by radio only if the angle whose vertex is in the center of the planet and whose rays go through the points in which the spaceships landed is acute.

What is the probability that every two of those ships can communicate directly or through the third ship?

I hope it's understandable.

Here's what I think:

Given two points on a circle, the probability that they form an acute angle (with vertex in the centre of that circle) is $\frac{1}{4}$. Given a sphere and two points on that sphere (with vertex in the centre), the probabilty that they form an acute angle is $\frac{1}{8}$. Correction - it is $\frac{1}{2}$, because the wlog we place one of the ships in $N$, then the whole northern hemisphere is ok for the second ship to park in order for them to communicate directly.

This reasoning isn't at all sophisticated, that's why I doubt it's correct.

When it comes to the second part - communicating via the third ship, I don't know what to do?

Could you help me a bit?

Thank you!

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    $\begingroup$ The probability that two point form an acute angle is $\frac{1}{2}$ (move the 1st one to the North pole, then the second one is equiprobably in the North - acute - or in the South - obtuse - hemisphere) $\endgroup$ – sds Jun 18 '14 at 14:19
  • $\begingroup$ Oh, I see. You're absolutely right. Thank you. I'll correct my question. $\endgroup$ – Bilbo Jun 18 '14 at 14:32
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I came across this question just now while doing a 2003 past exam paper for probability in the first year of my degree. It is a hard question to have to do in an exam. An easier example (one that then appears on a 2007 paper - maybe we're getting less intelligent) is to ask what the probability is of all three being able to communicate directly with each other, i.e. all three angles must be acute. However, to answer your question:

The best way of doing it I came up with is to find the probability density function of the angle between the first two ships to land and then condition on the value of this, before integrating over this angle in two parts to get the total probability. Unfortunately I think it is slightly too complex a problem to be able to just argue geometrically the whole way through.

Since this is an old question and I doubt anyone will ever read this I won't go into full detail unless someone asks, but that's the bare bones method.

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