3 paper dots fall from a 3-hole punch

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In summary, the probability of three paper dots falling from a 3-hole punch and forming a perfect equilateral triangle on the floor x-y plane is zero. This is because the distribution of the dots, as specified by the OP, does not allow for such a specific and precise arrangement. Additionally, the assumption of a continuous distribution also eliminates any possibility of the dots landing in a specific location on the floor. Therefore, there is no meaningful answer to this problem.
  • #1
pullmanwa
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3 paper dots fall from a 3-hole punch. what is the probability they will land forming a perfect equilateral triangle on the floor x-y plane? Assume the dots are point source singularities, real dots take up space.

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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  • #2
Let two of the dots land. They define one side of an equilateral triangle. The third dot must then land in exactly in one of two positions to form an equilateral triangle. What is the probability that it will do so?
 
  • #3
A rather more interesting question would be if the triangle is acute. For the problem to have meaning, you need to specify a probability distribution for the dots, and a uniform distribution on an infinite line or plane isn't possible.
 
  • #4
Ibix said:
Let two of the dots land. They define one side of an equilateral triangle. The third dot must then land in exactly in one of two positions to form an equilateral triangle. What is the probability that it will do so?

same difference... I don't think there is a solution.
 
  • #5
willem2 said:
A rather more interesting question would be if the triangle is acute. For the problem to have meaning, you need to specify a probability distribution for the dots, and a uniform distribution on an infinite line or plane isn't possible.
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
pullnanwa:
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 exact location 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.]
 
  • #7
economicsnerd said:
pullnanwa:
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 exact location 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
pullmanwa said:
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
pullmanwa said:
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
economicsnerd said:
If you don't make the question explicit, don't expect clear answers.
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.
 
  • #11
D H said:
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
Zafa Pi said:
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?
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 the a priori probability 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 ℝ2.

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
D H said:
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 the a priori probability 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 ℝ2.

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
pullmanwa said:
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
willem2 said:
A rather more interesting question would be if the triangle is acute. For the problem to have meaning, you need to specify a probability distribution for the dots, and a uniform distribution on an infinite line or plane isn't possible.

I think that establishing the probability distribution is part of the solution not part of the question.
 
  • #16
MrAnchovy said:
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
Zafa Pi said:
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
MrAnchovy said:
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
Zafa Pi said:
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.
Oh sorry, I misread your post.

Zafa Pi said:
Your question (bivariate normal) is also nice. It seems much more difficult. Can you do it?
Probably :) www.math.brown.edu/~diana/math/acute.pdf interesting paper illustrates some useful concepts (Method 10 in particular).
 
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  • #20
MrAnchovy - The paper you recommended was fun and gets the same answer as I did, but I didn't do it by any of their methods.
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.
 

1. How can 3 paper dots fall from a 3-hole punch?

The 3-hole punch is a tool used to create evenly spaced holes in paper. When the punch is pressed down, it creates holes and ejects small paper circles, also known as "dots." Therefore, it is possible for 3 paper dots to fall from a 3-hole punch if the tool is used to punch 3 holes in a piece of paper.

2. Why do the paper dots fall out of the 3-hole punch?

The paper dots are designed to fall out of the 3-hole punch to prevent them from clogging the tool. If the dots did not fall out, they could build up and prevent the punch from working properly.

3. How do the paper dots come out of the 3-hole punch in a perfectly round shape?

The 3-hole punch is equipped with sharp blades that are precisely positioned to create clean and circular holes in the paper. As the punch is pressed down, the blades cut through the paper and create the paper dots in a perfectly round shape.

4. Is there a specific size or shape of paper dots that come out of a 3-hole punch?

The size and shape of the paper dots depend on the specific 3-hole punch being used. Different brands and models may have slightly different sizes and shapes of paper dots. Additionally, the type and thickness of paper being punched can also affect the size and shape of the paper dots.

5. Can the paper dots from a 3-hole punch be reused or recycled?

Unfortunately, the paper dots from a 3-hole punch cannot be reused as they are small and may not have adhesive properties. However, they can be recycled along with other paper waste. It is always best to check with your local waste management facilities for proper recycling guidelines.

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