- 17 Aug '06 09:26Planet X is a ball with centre O.

Three spaceships A, B, C land at random positions on the surface of Planet X, these positions being independent and uniformly distributed.

Once landed, a pair A, B of spaceships can communicate provided that the angle AOB is less than 90 degrees. Similarly for B, C and C, A.

What is the probability that all three spaceships can stay in touch (with, for example, A communicating with C via B if necessary)? - 17 Aug '06 10:07

In stats a uniform distribution is a distribution where all possibilities are equally possible. So A is just as likely to land at a1 as a2 no matter where a1 and a2 are on Planet X.*Originally posted by smomofo***Zero, if they are uniformly distributed. I think. Wouldn't any two make an angle of 120 from center?** - 17 Aug '06 21:46

Would I be right in thinking that if just *two* ships were involved the probability would be 0.5?*Originally posted by SPMars***Planet X is a ball with centre O.**

Three spaceships A, B, C land at random positions on the surface of Planet X, these positions being independent and uniformly distributed.

Once landed, a pair A, B of spaceships can communicate provided that the angle AOB is less than 90 degrees. Similarly for B, C and C, A.

What is the probability that all three spaceships can stay in touch (with, for example, A communicating with C via B if necessary)?

My reasoning being that one ship landing would define a hemisphere in which communication was possible - with the ship at the pole. The other ship would then have a 50/50 chance of landing in the same hemisphere.

I'm not sure this helps with the final answer. Intriguing problem though. - 17 Aug '06 22:54

Yes that's right for two ships: without loss of generality we can suppose ship A has landed at the north pole. Then there is probability 1/2 that ship B lands in the northern hemisphere, and that's the region where the two can communicate.*Originally posted by howardbradley***Would I be right in thinking that if just *two* ships were involved the probability would be 0.5?**

My reasoning being that one ship landing would define a hemisphere in which communication was possible - with the ship at the pole. The other ship would then have a 50/50 chance of landing in the same hemisphere.

I'm not sure this helps with the final answer. Intriguing problem though. - 17 Aug '06 22:58

Alas, it is not 1/8.*Originally posted by smomofo***Well then I change my answer to a probability of 0.125. How 'bout that?**

XanthosNZ is right. Here 'uniform' means that each of the ships is equally likely to land in one place as any other. Also, 'independent' means the place where a given ship lands is not affected by where the other ships have landed. - 18 Aug '06 02:12Ugh. This going to require integration isn't it? It's been waaay too long for that. Plus, I never dealt with polar coordinates well, and I suspect this might be a problem where they are appropriate. I was on to the whole 0.5 thing, at least!

My new answer is: I don't know. For me, it would be fastest to do it by experiment. - 18 Aug '06 09:30

Certainly that is a good idea, and would get you a feel for the sphere case. And the integration is easier over a circle.*Originally posted by XanthosNZ***In approaching this problem you may find it easier to look at the case of a circle instead of a sphere first. Think of the two cases (B lands connected to A and B lands unconnected to A) as seperate and work from there.**

I hope this helps someone. - 18 Aug '06 09:36 / 2 edits

Yep integration is one way. And spherical polar coordinates are probably helpful.*Originally posted by smomofo***Ugh. This going to require integration isn't it? It's been waaay too long for that. Plus, I never dealt with polar coordinates well, and I suspect this might be a problem where they are appropriate. I was on to the whole 0.5 thing, at least!**

My new answer is: I don't know. For me, it would be fastest to do it by experiment.

On the other hand, we all know the surface area of a unit sphere is 4*pi, right (we choose units where the radius of Planet X is 1).

And we know the area of a 'lune' (intersection of two hemispheres) is 2*alpha where alpha is the angle (in radians) between the two great circles forming the lune.

Perhaps this will help...?