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Topic: 3 darts on a sphere (Read 1318 times) |
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aicoped
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3 darts on a sphere
« on: Dec 9th, 2008, 9:49pm » |
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OK this is an easy riddle, what is the probability that 3 randomly chosen points on a sphere could be cut into the same hemisphere(not necessarily north south, but after the points are chosen can you find a hemisphere of your choosing)?
Ok, I will not put the answer, but you guys should not have any troubles figuring it out.
But my real question is what about 4 points, 5 points, etc? is there an easy formula that derives the right answer. thanks in advance.
Btw, i have been coming to this forum and reading the riddles for about 2 years now.
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ThudnBlunder
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Re: 3 darts on a sphere
« Reply #1 on: Dec 9th, 2008, 9:56pm » |
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For n > 1 random points I think it's (1/2)n-2
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aicoped
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Re: 3 darts on a sphere
« Reply #2 on: Dec 9th, 2008, 10:15pm » |
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That yields a value that is not true for 3, so I don't think that is correct.
but could you tell me where you found that so I could research it, as I may definitely be wrong.
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Noke Lieu
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Re: 3 darts on a sphere
« Reply #3 on: Dec 9th, 2008, 10:40pm » |
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on Dec 9th, 2008, 9:49pm, aicoped wrote:
Btw, i have been coming to this forum and reading the riddles for about 2 years now. |
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In which case, revel in your status change from Lurker to Member!
And if you're going to bring questions like this one... you're doubly welcome!
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aicoped
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Re: 3 darts on a sphere
« Reply #4 on: Dec 9th, 2008, 11:49pm » |
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it is a question i have pondered since first hearing the riddle.
I want to wait to share my personal thoughts until more people have chimed in, as to not ruin the 3 point question, but once we have moved past that, I would love to see a four point answer or higher.
and ty for the kind words.
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pex
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Re: 3 darts on a sphere
« Reply #5 on: Dec 10th, 2008, 12:19am » |
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For three points, isn't it just 100 % ? Because the first two points and the center of the sphere define a plane, which cuts the sphere into two hemispheres. The third point will necessarily lie in one of the two.
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towr
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Re: 3 darts on a sphere
« Reply #6 on: Dec 10th, 2008, 12:55am » |
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Given a hemisphere, it'd be 1/2n; but we aren't given a hemisphere.
We need at least three points to define a hemisphere, so the answer should be greater or equal to 1/2n-3 for n > 2 (and 1 otherwise)
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| « Last Edit: Dec 10th, 2008, 12:55am by towr » |
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River Phoenix
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Re: 3 darts on a sphere
« Reply #7 on: Dec 10th, 2008, 1:56am » |
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on Dec 10th, 2008, 12:19am, pex wrote:| For three points, isn't it just 100 % ? Because the first two points and the center of the sphere define a plane, which cuts the sphere into two hemispheres. The third point will necessarily lie in one of the two. |
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Indeed, this exact question was on the Putnam test.
EDIT: actually for putnam the question was 3 points IN a sphere... this is different?
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| « Last Edit: Dec 10th, 2008, 2:07am by River Phoenix » |
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pex
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Re: 3 darts on a sphere
« Reply #8 on: Dec 10th, 2008, 4:00am » |
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on Dec 10th, 2008, 1:56am, River Phoenix wrote:| actually for putnam the question was 3 points IN a sphere... this is different? |
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Not fundamentally, you can work with the projections on the sphere's surface.
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pex
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Re: 3 darts on a sphere
« Reply #9 on: Dec 10th, 2008, 4:14am » |
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Taking that thought a bit further, we do not need any reference to the sphere at all. We can just pick any n points in three-dimensional space, pretend to project them on the unit sphere, and check whether all of the points lie on one side of some plane through the origin. The projection does not change on which side of such a plane we are.
Thus, the problem can be restated as follows: given n points picked randomly in three-dimensional space, form the n*3 matrix X with the coordinates of each point as its rows. Then, what is the probability that Xc > 0 has a nonzero solution for the 3*1 vector c?
Not sure if this is going to be useful, but I thought I'd share it anyway.
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| « Last Edit: Dec 10th, 2008, 4:16am by pex » |
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aicoped
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Re: 3 darts on a sphere
« Reply #10 on: Dec 10th, 2008, 4:42pm » |
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You are right about the 3 dart problem being that value.
Now as to the four dart, when i conceptualize it, it seems as if you take the "triangle" formed by the first three points then that will be the amount of surface area inverted on the opposite side of the sphere that the fourth point can't be in, so all we need to do to solve 4 points is (assuming my conceptualization is correct) is solve the "average surface area" for three random points on a sphere then subtract that from the surface area of a sphere then divide by the surface are of the sphere which would yield our probability. but as I have no formal math training, it is beyond me to actually crunch those numbers or write up a program that could monte carlo the area or some other such method.
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Noke Lieu
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Re: 3 darts on a sphere
« Reply #11 on: Dec 10th, 2008, 4:52pm » |
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It's gotten me thinking about points... and the hemispheres.
the only way that I could see it being impossible with 2 points was if they are antipodes. Whilst I accept that it IS possible, my reasoning takes a slight hiccup when I try to justify it.
Given the two points, they are both on the equator/great circle.
That line has 0 thickness.
the points have 0 thickness.
that line definately cuts the sphere into two semispheres, but once that happens, where do the points lie? Do they arbitaritly lie on the hemisphere, or arbitarily split between the two, or exist on the edge of both?
Perhaps it's up to you...?
Gah... good question that probably has an axiomic answer...
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Grimbal
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Re: 3 darts on a sphere
« Reply #12 on: Dec 11th, 2008, 12:37am » |
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These cases are possible, but with probability 0. 
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aicoped
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Re: 3 darts on a sphere
« Reply #13 on: Dec 11th, 2008, 3:59am » |
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Well you split the hemisphere, but as i have always interpreted it, a split "on the points" makes the points still in the hemisphere so 2 points will always be 100% even when opposite each other. And then there fore 3 will also be 100% chance since you can always split on the edge where the two are and the three has to be in the same hemisphere.
Now four is where I need help.
question, if this stays in easy for a while with no answer to 4, can we move it into a harder forum, or do all of the "uberpuzzlers" read all of the forums?
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towr
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Re: 3 darts on a sphere
« Reply #14 on: Dec 11th, 2008, 4:50am » |
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on Dec 11th, 2008, 3:59am, aicoped wrote:| question, if this stays in easy for a while with no answer to 4, can we move it into a harder forum, or do all of the "uberpuzzlers" read all of the forums? |
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Probably both.
Besides you have to consider that the title "uberpuzzler" is automatically assigned based on post-count, which grows much more rapidly if you frequent all forums. 
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aicoped
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Re: 3 darts on a sphere
« Reply #16 on: Dec 11th, 2008, 2:39pm » |
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towr, thanks very much. If my math is done right, then 4 darts is also 100%. and 5 darts is surprisingly 30/32nds. Shows how frail our minds are at guestimating.
EDIT, my math is apparently off, as I continue to read the article it shows the answers for 4 and 5 to be 87.5% and 68.75% respectively, but i can not derive that answer from their formula, so i must be mucking up the math somewhere.
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| « Last Edit: Dec 11th, 2008, 2:49pm by aicoped » |
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towr
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Re: 3 darts on a sphere
« Reply #17 on: Dec 11th, 2008, 2:51pm » |
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on Dec 11th, 2008, 2:39pm, aicoped wrote:| towr, thanks very much. If my math is done right, then 4 darts is also 100%. and 5 darts is surprisingly 30/32nds. Shows how frail our minds are at guestimating. |
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I'm not sure what maths you're doing, exactly. From the formula described in that link, and the table, I get 7/8th for 4 points.
Hmm, perhaps it's confusing because S2, i.e. d=2, is the surface of a 3D sphere.
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towr
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Re: 3 darts on a sphere
« Reply #18 on: Dec 11th, 2008, 2:57pm » |
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on Dec 11th, 2008, 2:39pm, aicoped wrote:| EDIT, my math is apparently off, as I continue to read the article it shows the answers for 4 and 5 to be 87.5% and 68.75% respectively, but i can not derive that answer from their formula, so i must be mucking up the math somewhere. |
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Are you using (n2-n+2)/2n or another formula?
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Noke Lieu
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Re: 3 darts on a sphere
« Reply #19 on: Dec 11th, 2008, 3:42pm » |
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on Dec 11th, 2008, 4:50am, towr wrote:
Probably both.
Besides you have to consider that the title "uberpuzzler" is automatically assigned based on post-count. |
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Oh god yeah. I'm still amused/bemused/concerned about me being an Uberpuzzler. Not enough to actually want to do anything to change it, mind you. 
IMHO there's two sorts of uberpuzzlers around here. Those, like me, who have made the requisite through sheer perserveance and mostly thanks to the What Am I?/What happened? sections.
The other type of uberpuzzler doesn't always have the official title.
I remember when Grimbal and I were new around here- roughly the same time.
The difference in the quality of our posts was quite clear.
It was also clear that Grimbal (and for that matter JocK, or FiBsTeR, or the guest "Asterix", or... you know who you are) was deserving of the "accolade" from a pretty early stage. To draw a comparison from another side of nme- "fight the opponent, not their belt".
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| « Last Edit: Dec 11th, 2008, 4:24pm by Noke Lieu » |
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aicoped
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Re: 3 darts on a sphere
« Reply #20 on: Dec 11th, 2008, 9:43pm » |
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I was using the d3 formula. thanks guys.
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