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Old 02-18-2018, 01:18 PM
ToughLife ToughLife is offline
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Three sided coin ( basic math question)

In this PDF there is a comparison of different models describing 3-sided coins.

Now,on page 3

Model 1 is pretty clear, to calculate the probability (of the flipped coin landing on its side )
they consider coin side surface area to the total surface area ratio.

Model 2 is a bit confusing

it seems that they divide one side of a perimeter (h) by the sum of three other sides 2(2R)+h
rather than the sum of all four sides
intuitively it should be 2(2R) +2 h

What am I missing ?

Related question: how do they define cross sectional length?
Is it just a half of the diagonal, as per drawing?
And if that's the case,why would the probability be proportional to that length?
I don't find it very intuitive.

Thanks
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Old 02-18-2018, 01:23 PM
ToughLife ToughLife is offline
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Sorry ,wrong forum. Intended for GQ.
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Old 02-19-2018, 02:32 PM
Hypnagogic Jerk Hypnagogic Jerk is offline
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I didn't read the linked paper very closely yet, but this recent video by Matt Parker discusses three-sided coins, or more precisely, how thick a cylinder would need to be to have a probability of exactly one-third to land on each of its ends, and on its side.
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Old 02-19-2018, 03:30 PM
ToughLife ToughLife is offline
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Thanks, I've seen this video, but it's not what I'm asking.
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Old 02-20-2018, 02:22 PM
Bone Bone is offline
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Moved from GD to GQ.

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Old 02-20-2018, 02:41 PM
markn+ markn+ is offline
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What four sides? There are three sides, the head face (length 2R), the tail face (length 2R) and the edge (length h).
  #7  
Old 02-28-2018, 07:35 AM
ToughLife ToughLife is offline
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557 views and no answer ?
Am I asking a wrong question ?
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Old 02-28-2018, 07:48 AM
Jasmine Jasmine is offline
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Quote:
Originally Posted by ToughLife View Post
557 views and no answer ?
Am I asking a wrong question ?
Since you appear to be frustrated and a little peeved, I made the mistake of clicking the link you left and going to the PDF. My God, I've never read anything dryer or more boring!

Is this a course you have to take towards a degree? If so, my "answer" is, DROP THE COURSE!
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  #9  
Old 02-28-2018, 08:08 AM
ToughLife ToughLife is offline
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Jasmine ,
no, my question has nothing to do with courses.
I was reading this
and found a link to that PDF file at the end of the article.

The reason I'm asking is I believe that the authors, serious mathematicians, had decided to consider model 2
on page 3.
It's supposed to be a pretty basic model, but, embarrassingly, I don't follow their logic.
  #10  
Old 02-28-2018, 08:13 AM
Thudlow Boink Thudlow Boink is offline
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Quote:
Originally Posted by Jasmine View Post
Since you appear to be frustrated and a little peeved, I made the mistake of clicking the link you left and going to the PDF. My God, I've never read anything dryer or more boring!
In my limited experience, it's pretty typical (not bad) for the type of article it is (an academic-style journal article). But it's certainly not written for a popular audience, and it would take some time and effort to digest.

I spent a few minutes wrestling with it, and specifically with the formula the OP asks about, and I don't understand it enough to come up with a useful answer. I may, or may not, come back to it later and try again when I have more time.
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Old 02-28-2018, 09:10 AM
Colibri Colibri is offline
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Quote:
Originally Posted by Jasmine View Post
Since you appear to be frustrated and a little peeved, I made the mistake of clicking the link you left and going to the PDF. My God, I've never read anything dryer or more boring!

Is this a course you have to take towards a degree? If so, my "answer" is, DROP THE COURSE!
Moderator Note

Jasmine, if you don't find a question interesting, there is no need to threadshit. This kind of answer is not useful in General Questions. No warning issued, but don't do this again.

Colibri
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  #12  
Old 02-28-2018, 09:14 AM
Jasmine Jasmine is offline
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Quote:
Originally Posted by Colibri View Post
Moderator Note

Jasmine, if you don't find a question interesting, there is no need to threadshit. This kind of answer is not useful in General Questions. No warning issued, but don't do this again.

Colibri
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Sorry!
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  #13  
Old 02-28-2018, 09:19 AM
brossa brossa is offline
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Their derivation is wrong; you can confirm that by plugging in an eta of 2, which corresponds to a square die. You get a probability of 1/3 for a square die, which is counterfactual.

For 'cross sectional length', let's assume they mean 'perimeter along this cross section'. If they meant something else, it's sloppy that they didn't clarify.

The total perimeter is 2h+2(2R)

Because the two h sides are actually the same side, the probability of landing on an h based on its proportion of perimeter is :

p=2h/(4R+2h) rather than h/(4R+2h)

this simplifies to h/(2r+h)

substitute for eta:

p=eta/(2+eta)

Now, if you plug in eta=2 (square cross-section), you get a probability of 1/2, which is correct. An equal probability of landing on any of four sides, but two of those sides are actually the same side: A,B,B,C rather than A,B,C,D, and we're keeping track of the probability of B.
  #14  
Old 02-28-2018, 09:35 AM
kanicbird kanicbird is offline
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There is technically another possibility, landing on a corner (for that matter head side corner or tail side corner). Almost zero chance of it ever happening, but not zero.
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Old 02-28-2018, 09:42 AM
Thudlow Boink Thudlow Boink is offline
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Originally Posted by kanicbird View Post
There is technically another possibility, landing on a corner (for that matter head side corner or tail side corner). Almost zero chance of it ever happening, but not zero.
I think that, theoretically, the probability is zero, since a corner would have zero surface area.
  #16  
Old 02-28-2018, 10:35 AM
ToughLife ToughLife is offline
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Quote:
Originally Posted by brossa View Post
Their derivation is wrong; you can confirm that by plugging in an eta of 2, which corresponds to a square die. You get a probability of 1/3 for a square die, which is counterfactual.

For cross sectional length, let's assume they mean 'perimeter along this cross section'. If they meant something else, it's sloppy that they didn't clarify.

The total perimeter is 2h+2(2R)

Because the two h sides are actually the same side, the probability of landing on an h based on its proportion of perimeter is :

p=2h/(4R+2h) rather than h/(4R+2h)

this simplifies to h/(2r+h)

substitute for eta:

p=eta/(2+eta)

Now, if you plug in eta=2 (square cross-section), you get a probability of 1/2, which is correct. An equal probability of landing on any of four sides, but two of those sides are actually the same side: A,B,B,C rather than A,B,C,D, and we're keeping track of the probability of B.
Agree, I got,eventually,that far myself as well.

Quote:
For 'cross sectional length', let's assume they mean 'perimeter along this cross section'. If they meant something else, it's sloppy that they didn't clarify.
if they define cross sectional length as a perimeter, why do they "presume that the probability is proportional to that length" ?
That doesn't sound very intuitive to me.

Also, just under the Figure1 they define diagonal length l = sqrt{(R^2 +(h/2)^2)} which adds to the confusion.
  #17  
Old 02-28-2018, 10:36 AM
RTFirefly RTFirefly is offline
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Can't get to the link from my work computer, but ISTM that someone's doing an awful lot of work to come up with a fair 3-sided coin when you can just use a regular 6-sided die, treating opposite sides as the same outcome.
  #18  
Old 02-28-2018, 10:52 AM
Mangetout Mangetout is offline
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Originally Posted by RTFirefly View Post
Can't get to the link from my work computer, but ISTM that someone's doing an awful lot of work to come up with a fair 3-sided coin when you can just use a regular 6-sided die, treating opposite sides as the same outcome.
Me too - and I actually think the problem doesn't have a true solution - because the asymmetry/inequality of shape of the sides means that the fairness of the thing is going to depend on how it is thrown, and other factors such as the surface onto which it falls.

I think it could quite easily be possible to design a cylindrical 3 sided coin that is 'fair' when dropped from 1 metre, with a 100 rpm spin, perpendicular to the axis, onto smooth glass, and find that it's desperately unfair when dropped from a different height, at a different spin, etc.
  #19  
Old 02-28-2018, 10:53 AM
Thudlow Boink Thudlow Boink is offline
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Quote:
Originally Posted by RTFirefly View Post
ISTM that someone's doing an awful lot of work to come up with a fair 3-sided coin when you can just use a regular 6-sided die, treating opposite sides as the same outcome.
As the link in Post #9 points out, there do exist 3-sided flippable objects.
Quote:
Some real life examples of this are the delicious Toblerone chocolate bars, and a ruler (which is not delicious).
The OP's link claims to be interested in the subject as an example of "Bayesian model comparison."

ETA: Due to the shape and weight of the actual chocolate inside the triangular package, I suspect the three sides of a Toblerone might not all land with equal frequency.

Last edited by Thudlow Boink; 02-28-2018 at 10:54 AM.
  #20  
Old 02-28-2018, 10:57 AM
Leo Bloom Leo Bloom is online now
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I believe our esteemed member Chronos owns/has built his very own 3-sided coin [cite: some previous thread on this]. Maybe he'll show up to add his 2/3 cents.
  #21  
Old 02-28-2018, 11:00 AM
ToughLife ToughLife is offline
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Originally Posted by Mangetout View Post
Me too - and I actually think the problem doesn't have a true solution - because the asymmetry/inequality of shape of the sides means that the fairness of the thing is going to depend on how it is thrown, and other factors such as the surface onto which it falls.

I think it could quite easily be possible to design a cylindrical 3 sided coin that is 'fair' when dropped from 1 metre, with a 100 rpm spin, perpendicular to the axis, onto smooth glass, and find that it's desperately unfair when dropped from a different height, at a different spin, etc.
You are right, they do take that in consideration in models 3-6.

Alternative link
  #22  
Old 02-28-2018, 11:18 AM
octopus octopus is offline
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I think that, theoretically, the probability is zero, since a corner would have zero surface area.
Probability 0 isnít an impossible event though.
  #23  
Old 02-28-2018, 11:19 AM
Leo Bloom Leo Bloom is online now
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Originally Posted by Thudlow Boink View Post
As the link in Post #9 points out, there do exist 3-sided flippable objects.The OP's link claims to be interested in the subject as an example of "Bayesian model comparison."

ETA: Due to the shape and weight of the actual chocolate inside the triangular package, I suspect the three sides of a Toblerone might not all land with equal frequency.

Did some mention Toblerone lattices?

Some exact results for self-avoiding random surfaces
Amos Maritan and Attilio Stella
The formal connection between self-avoiding surfaces (SAS) and suitable lattice gauge theories is discussed in the limit that the number of field components goes to zero. Different gauge models correspond to different rules for weighting the SAS topologies or to different constraints imposed on the boundaries. The fractal dimension of a SAS model on a toblerone lattice in d = 2.58Ö dimensions is calculated exactly. Finally a general qualitative discussion of the behaviour of the SAS in the scaling limit is given in the light of the above and other recent results.
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Old 02-28-2018, 12:39 PM
brossa brossa is offline
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Originally Posted by ToughLife View Post
if they define cross sectional length as a perimeter, why do they "presume that the probability is proportional to that length" ?
That doesn't sound very intuitive to me.
It's a fairly unphysical way of looking at the problem, but it's saying 'let's say that a coin flip is picking a point along the perimeter at random and make a note of which side contains that point.' Under that definition, longer sides get picked more often than short sides in proportion to their lengths.

With that definition, we see that an eta of 1 (ie radius=height) gives a center cross-section perimeter of 6R, with 2R for side A, 2R for side B, and 2R for edge. So a point drawn at random is equally likely to fall on A, B, or E
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Old 02-28-2018, 01:34 PM
brossa brossa is offline
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I should add that the first model treats the coin as a collection of surfaces, and picks a point on the surface, while the second model reduces those surfaces to line segments and picks a point along the line.
  #26  
Old 02-28-2018, 03:08 PM
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Nope, I do have a number of dice of my own design with various numbers of sides, but I've never done a d3.
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Last edited by Chronos; 02-28-2018 at 03:09 PM.
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