- #1

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don't grid the floor because that will constrict rotations.

I don't think there is an answer to this problem.

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- Thread starter pullmanwa
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- #1

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don't grid the floor because that will constrict rotations.

I don't think there is an answer to this problem.

- #2

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- #3

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- #4

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same difference... I don't think there is a solution.

- #5

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why would it be more interesting if the triangle was acute?

I'm sure if I posed the problem ..."what is the probability they will land forming an acute triangle...", you would have responded "A rather more interesting question would be if the triangle was a perfect equilateral triangle."

- #6

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Say the locations of the three dots are given by random variables [itex]X,Y,Z[/itex] respectively, each of them an [itex]\mathbb R^2[/itex]-valued random variable. You're asking a question about the joint random variable [itex](X,Y,Z)[/itex]. Specifically, you're asking for the probability of the event [itex]\{(x,y,z): ||x-y||=||y-z||=||z-x||\}[/itex].

What willem2 was explaining is that, to answer this question, you need to specify the distribution of [itex](X,Y,Z)[/itex]

If, for example, [itex]Z[/itex] is independent of [itex](X,Y)[/itex], and the distribution of [itex]Z[/itex] has no mass points (i.e. any given

[*He/she may have been saying more, as independence isn't really important here. But conditional distributions seem like more detail than the current thread needs.]

- #7

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Say the locations of the three dots are given by random variables [itex]X,Y,Z[/itex] respectively, each of them an [itex]\mathbb R^2[/itex]-valued random variable. You're asking a question about the joint random variable [itex](X,Y,Z)[/itex]. Specifically, you're asking for the probability of the event [itex]\{(x,y,z): ||x-y||=||y-z||=||z-x||\}[/itex].

What willem2 was explaining is that, to answer this question, you need to specify the distribution of [itex](X,Y,Z)[/itex]

If, for example, [itex]Z[/itex] is independent of [itex](X,Y)[/itex], and the distribution of [itex]Z[/itex] has no mass points (i.e. any givenexactlocation for [itex]Z[/itex] has zero probability), then the event you've named would have zero probability. This* is the content of Ibix's comment.

[*He/she may have been saying more, as independence isn't really important here. But conditional distributions seem like more detail than the current thread needs.]

Let's use some common sense here. Most 3 hole punches are about 30 inches from the floor. I don't think the dots are going to land 1 mile away.... or even 20 feet away.

- #8

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same difference... I don't think there is a solution.

There is an answer - zero, as I hoped you'd see from my restatement, and as economicsnerd said explicitly. Because the position of the third dot is a continuous variable, the probability of it taking any infinitely precisely specified value is zero.

If you want to specify a tolerance (e.g. equilateral plus or minus 1mm) then the answer is non-zero, but a lot more information is needed.

- #9

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I don't think the dots are going to land 1 mile away.... or even 20 feet away.

So then we would need to specify that, for instance, [itex]P(||x||>20)=0[/itex]. But this is a modeling assumption.

If you don't make the question explicit, don't expect clear answers.

- #10

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The OP essentially did "make the question explicit" by saying "don't grid the floor because that will constrict rotations." In other words, the dots fall on the real plane, with some distribution that keeps the dots from falling a mile away, let alone 20 feet away.If you don't make the question explicit, don't expect clear answers.

This means the subset of points that form an equilateral triangle is a space of measure zero. The answer is zero, as Ibix already noted.

- #11

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The OP essentially did "make the question explicit" by saying "don't grid the floor because that will constrict rotations." In other words, the dots fall on the real plane, with some distribution that keeps the dots from falling a mile away, let alone 20 feet away.

This means the subset of points that form an equilateral triangle is a space of measure zero. The answer is zero, as Ibix already noted.

Just what does, "don't grid the floor because that will constrict rotations." mean? How does that restrict the distribution? Why does that imply a continuous distribution, or not 20 feet away, as you assumed?

I think everything that economicsnerd said is correct (and he was the 1st to actually say "zero"). And this implies that the OP was correct (perhaps inadvertently) when he(?) wrote: "I don't think there is an answer to this problem.", because it is not well posed.

- #12

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To me that "gridding the floor" means some kind of discrete distribution. In the case of a uniform grid, the probability isn't just zero, it can't happen. (There's a big difference between "can't happen" and "space of measure zero" events. Pick a random number from U(0,1). What's theJust what does, "don't grid the floor because that will constrict rotations." mean? How does that restrict the distribution? Why does that imply a continuous distribution, or not 20 feet away, as you assumed?

I'm also assuming i.i.d. random variables and a continuous PDF, both of which are reasonable assumptions given the nebulous nature of the opening post. Getting an isosceles triangle is a zero probability event given those assumptions, let alone an equilateral triangle.

- #13

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To me that "gridding the floor" means some kind of discrete distribution. In the case of a uniform grid, the probability isn't just zero, it can't happen. (There's a big difference between "can't happen" and "space of measure zero" events. Pick a random number from U(0,1). What's thea prioriprobability that a number from U(0,1) will be 2? What about 1/2?) Things are rather different in the case of a hexagonal grid. However, the OP said "don't grid the floor", which to me says some distribution on ℝ^{2}.

I'm also assuming i.i.d. random variables and a continuous PDF, both of which are reasonable assumptions given the nebulous nature of the opening post. Getting an isosceles triangle is a zero probability event given those assumptions, let alone an equilateral triangle.

I'm happy you found meaning in, "don't grid the floor because that will constrict rotations." (the PDF is continuous).

However, to me "nebulous nature" = "not explicit" so economicsnerd didn't need to be corrected, plus he already covered and generalized your 2nd paragraph.

Am I beating a dead horse?

- #14

pbuk

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why would it be more interesting if the triangle was acute?

Because the answer to that is not zero (unlike the probability of an equilateral triangle), but it is rather hard to work out.

- #15

pbuk

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I think that establishing the probability distribution is part of the solution not part of the question.

- #16

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I think that establishing the probability distribution is part of the solution not part of the question.

A natural distribution would be uniform on a circle. It's uniform and there are no unnatural boundaries. For the acute case I get 1/4. What do you get?

- #17

pbuk

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A natural distribution would be uniform on a circle. It's uniform and there are no unnatural boundaries. For the acute case I get 1/4. What do you get?

Why should it be uniform? Isn't the edge of the circle an unnatural boundary? The bivariate normal distribution would seem a more natural choice.

- #18

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Why should it be uniform? Isn't the edge of the circle an unnatural boundary? The bivariate normal distribution would seem a more natural choice.

I'm talking about the circle (has no boundary) not the disk.

Take 3 points at random (uniform) on a circle. What's the probability they make an acute triangle?

This strikes me as simply stated aesthetic question that I can also answer.

Your question (bivariate normal) is also nice. It seems much more difficult. Can you do it?

- #19

pbuk

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Oh sorry, I misread your post.I'm talking about the circle (has no boundary) not the disk.

Take 3 points at random (uniform) on a circle. What's the probability they make an acute triangle?

This strikes me as simply stated aesthetic question that I can also answer.

Probably :) www.math.brown.edu/~diana/math/acute.pdf [Broken] interesting paper illustrates some useful concepts (Method 10 in particular).Your question (bivariate normal) is also nice. It seems much more difficult. Can you do it?

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- #20

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You say that you could probably handle the bivariate normal case. Do you have an estimate for that probability and what distribution are you using?

Even tho it's rotation invariant it seems difficult.

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