Wikipedia:Reference desk/Archives/Mathematics/2012 July 17

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July 17

Power in the Zero Channel

If I have an arbitrary signal and I want to estimate its power spectrum density, what is the best way to "detrend" the data so that there is no power in the zero channel. The problem is that the signal I am working with has a very large DC offset so when I get a PSD estimate as it is, there is too much leakage into the neighboring frequencies and I get way too much power than there should be. It just dominates everything and is completely useless. So everyone keeps telling me to "detrend" but what does that mean? Should I just subtract the constant arithmetic average of the signal from the signal? I tried that but it doesn't seem to work. Should I fit a straight line using linear least squares fit to the signal and then subtract that? That reduces the power in the zero channel but doesn't really make it small as I think it should be. Maybe subtracting the line is good enough because the power becomes relatively small compared with other channels but it still seems a lot to me in absolute magnitude.

The signal kind of looks like a decaying exponential with tiny oscillations around the "middle". So next I thought of just fitting an exponential and then subtracting that but would that take away power from other channels? The signal is only about a hundred measurements and I am using the multitaper method with 7 slepian functions from MATLAB's builtin pmtm function with nfft being the length of the signal itself (not padding up to the next power of two). Using more slepian functions seems to decrease the DC (although there seems to be a lower bound and I can't get it below that) but it takes forever. If increasing the signal length makes a difference I can do that. But I can't change the resolution of the signal. Any ideas/suggestions/insights would be helpful. Thanks! 174.56.103.148 (talk) 07:15, 17 July 2012 (UTC)

It sure would be helpful if you could provide a pic of the graph, as it is now, and another of how you want it to look after processing. A picture is worth a thousand words and all. If you have pics and need help displaying them here, just let us know. StuRat (talk) 07:17, 17 July 2012 (UTC)

Take the discrete fourier transform, Y_k=Σ_j X_j1^jk/N/√N, (where X_j is the complex conjugate to X_j, and j=(0,...,N−1), and k=(0,...,N−1), and 1^x is shorthand for e^2π√−1x). Detrend by setting Y₀=0. Transform back: X_k=Σ_j Y_j1^jk/N/√N. Bo Jacoby (talk) 16:14, 17 July 2012 (UTC).

I agree with Bo Jacoby. The multitaper method is not appropriate for this sort of signal. You should use a Fourier transform instead -- it's actually a lot simpler, and the Matlab signal toolbox supplies it to you pretty straightforwardly. With a Fourier transform, if you subtract the DC offset, you are guaranteed not to have any power in the zero band. Given the nature of your signal, it's not likely that any sort of frequency analysis will be very informative, but at least a Fourier transform will allow you to do what you say you want to do. Looie496 (talk) 18:12, 17 July 2012 (UTC)

OP here, Bo your idea is pretty cool. I don't know why I didn't think of it. So I can just take the FFT of the original signal straight up, then set the first element to zero and then take the IFFT? Then on this new signal, can I use the multitaper method to estimate the PSD? The reason I am using the multitaper method is because I thought multitaper was a bit more "modern" and "sophisticated" than good old fashioned basic FFT and/or welch-like windows. I just need to get the PSD and if you guys think something besides MTM is more appropriate, let me know! 65.100.24.211 (talk) 19:59, 17 July 2012 (UTC)

A good principle of data analysis is to always use the least sophisticated tool that does the job. Getting the PSD from the FFT is a very simple operation and does not require an IFFT. I don't have access to Matlab presently but I'm sure it gives you a simple function to do this. Looie496 (talk) 20:19, 17 July 2012 (UTC)

If your data points by nature cannot be negative (such as the signal looking like an exponential decay indicates) then you should take the logarithm first. If you take the complex conjugate and divide by the square root of the number of data points, then the backward FFT is exactly the same as the forward FFT, which is nice. If your data set is symmetric, then the fourier transform is real, which is nice. The n-sized data set (x₀,...,x_n−1) is extended to the symmetric (2n−2)-sized dataset (x₀,...,x_n−1,x_n−2,...,x₁) and the transformed data set (y₀,...,y_n−1,y_n−2,...,y₁) is truncated to (y₀,...,y_n−1). The element y₀ is the DC amplitude and the element y_n−1 is the VHF amplitude. (PS. What is PSD?). Bo Jacoby (talk) 03:26, 18 July 2012 (UTC).

PSD is Power spectral density. I think all that is too complicated -- computing a power spectrum is everyday stuff, and the Matlab signal processing toolkit makes it easy. Looie496 (talk) 03:44, 18 July 2012 (UTC)

two problems with tangent circles

I was solving this, [note this is no homework], but few questions I am completely puzzled. Can anyone explain me to solve Q67 and Q39, this page. Thanks, extra999 (talk) 11:26, 17 July 2012 (UTC)

Q39. Let the diameter in the big semicircle be AB=4. The radius in the small circle is x. Pytagoras on triangle OPS says 1²+(2−x)²=(1+x)². Now you can finish yourself. Bo Jacoby (talk) 12:22, 17 July 2012 (UTC).

Q67. Let the unknown radius of the little circle be x. So the ratio between the small radius and the big radius is q=x/2. So x=2q. The distance from the centre of the big circle to the corner includes an infinite sum of diameters of decreasingly small circles. 2√2=2+2x+2xq+2xq²... = 2+4q+4q²+4q³+... So (1+√2)/2 = 1+q+q²+q³+... = 1/(1−q) = 2/(2−x). So x = 2−4/(1+√2) = 2−4(√2−1) = 6−4√2. Bo Jacoby (talk) 12:22, 17 July 2012 (UTC).

67: Because the two circles are externally tangent (one does not contain the other), the distance between their centers is obviously the sum of the radii. Now consider the bounding lines as coordinate axes, find the Cartesian coordinates of the centers, and use the coordinates to find the Pythagorean distance between them. With a tiny bit of algebraic massage, you have a quadratic equation. Notice, by the way, the geometric mean of its two solutions. —Tamfang (talk) 20:28, 17 July 2012 (UTC)

I can not see a quadratic equation here. Draw two slope radii of circles and a vertical radius of the small one. Imagine some squares and find out that $R{\sqrt {2}}=R+r+r{\sqrt {2}}$ which implies $r=R({\sqrt {2}}-1)/({\sqrt {2}}+1)$. Multiply both enumerator and denominator by $({\sqrt {2}}-1)$, expand, plug $R=2$, reduce, done. --CiaPan (talk) 10:21, 18 July 2012 (UTC)

My quadratic equation is $(R+r)^{2}=2(R-r)^{2}$. —Tamfang (talk) 04:48, 20 July 2012 (UTC)

I hope it's not a problem that I changed the title of this section to something more meaningful than "Help". —Tamfang (talk) 20:11, 17 July 2012 (UTC)

A weird kind of 20 sided dice...

Suppose I make one of these out of three identical, thin, intersecting, "golden-ratio" rectangles. I want to be able to roll it like one of those icosahedral 20-sided dice that Dungeons & Dragons players are so fond of...except that the 'facets' of the icosahedron are implied rather than physically existing.

The question is: Is it a "fair" dice? Will there be an equal probability of it coming up with any of the 20 "invisible facets" uppermost? (Presuming the three rectangles have non-zero mass of course).

I can see that there are two 'classes' of triangular face:

• Eight faces have their three vertices taken one from each of the three different rectangles - like the face that's foremost in the diagram (lets call these "Class A faces").
• Twelve faces have two vertices from one rectangle and one from another ("Class B faces") which connect between the class A faces.

Since there is symmetry here, all class A faces are equally probable - and so are all class B faces - so I presume that the problem boils down to whether there is some preferential weighting of Class A faces compared to Class B.

Bonus question: If it's not "fair", then how bad is the unfairness? (I presume that depends on the masses of the rectangles...but maybe if I make them out of light materials, the error will be negligible?)

Any ideas before I have to resort to making one and rolling it about a thousand times to check?  :-)

TIA SteveBaker (talk) 19:58, 17 July 2012 (UTC)

Its moment of inertia is spherically symmetric (the inertia tensor of any one of the three rectangles wrt aligned coordinates is diag(x, y, z), and the sum of all three is (x+y+z) I), so I think it is probably fair, but I may be missing something. -- BenRG (talk) 20:17, 17 July 2012 (UTC)

Other than experimentally (and hence imperfectly), how do you even define "fair"? * 86.129.16.198 (talk) 23:05, 17 July 2012 (UTC) (* Other than the case of indistinguishable faces, which must be fair by symmetry)

I would say that "fair" dice are dice where the theoretical probability of each possible result is the same (i.e. equal to 1/n, n being the number of possible results). This is accomplished when all the faces are the exact same size, the density of the thing is consistent throughout (so one face does not weigh more than another, for example), etc. Technically, even some small action like scratching a die with one's car keys is enough to upset this balance, not to mention all of the tiny, tiny, imperceptible errors that probably occur during the manufacture of even those nice, sharp-edged casino dice. Of course, such errors are not likely to have any noticeable influence on the game, so they can be safely discounted.  dalahäst ^{(let's talk!)} 03:01, 18 July 2012 (UTC)

Some cases are clearly fair by symmetry, and I suppose some are clearly unfair by obvious lack of symmetry, but in the general case a procedure for calculating the theoretical probability of each possible result would need to be established, and it is not very obvious how to go about doing that. 86.129.16.198 (talk) 03:15, 18 July 2012 (UTC)

According to the presumption of innocence the dice is fair until proven unfair. The burden of proof is on the prosecution. According to the principle of insufficient reason the dice is fair. Bo Jacoby (talk) 04:01, 18 July 2012 (UTC).

The possibilities are not indistinguishable, as Steve pointed out in the original question. He was already applying the principle of indifference where it does apply. You need Newton's laws to answer this question any further than that. It's really more of a science desk question. -- BenRG (talk) 05:18, 18 July 2012 (UTC)

I tend to believe the die is unfair, and this shouldn't be hard to show by simulation. Take a random orientation of the die; find the lowest vertex; find the location of the center of mass with respect to it, and rotate it in that direction until there is a second vertex touching the table; then do the same for the third vertex. This is a fairly good approximation for what happen when you roll it. By running this enough iterations or analytically deriving it, you can find something which is similar to the actual probabilities. -- Meni Rosenfeld (talk) 04:08, 18 July 2012 (UTC)

Ah, I see a few of us know that a "die" is singular and "dice" are plural, but, alas, I lament that, if put up for a vote, the plurality may be against us. Perhaps we should take everyone who makes this mistake and toss them into the center of a black hole ?  :-) StuRat (talk) 05:00, 18 July 2012 (UTC)

I've always said "die", but I think singular "dice" is so widespread and so old that it's rather ridiculous to treat it as anything but correct. This is how languages change. -- BenRG (talk) 05:18, 18 July 2012 (UTC)

It's probably "correct" by now in the UK. In the States it's still jarring. To me it's a nails-on-chalkboard thing; I will never accept it as correct. --Trovatore (talk) 06:58, 18 July 2012 (UTC)

(ec). I stand corrected. English is not my first language and sometimes I spell latin rooted words correctly, writing 'excentricity' for eccentricity and 'exspect' for expect and so on. In order not to spell more correctly than the English I wrote 'the dice' rather than 'the die'. Bo Jacoby (talk) 05:41, 18 July 2012 (UTC).

The vertices form a regular icosahedron, and the ellipsoid of inertia is a sphere. I can't see what else would bias it given the sort of idealizing assumptions one usually makes in mathematics. I guess air resistance could bias it, but your simulation as described won't catch that. -- BenRG (talk) 05:18, 18 July 2012 (UTC)

From a mechanical point of view the faces are indistinguishable. The center of mass is at the geometrical center. When thrown on a plane table only the shape of the convex hull matters, and it is regular. If the table is not plane the die may be unfair. Bo Jacoby (talk) 05:41, 18 July 2012 (UTC).

"From a mechanical point of view the faces are indistinguishable."?? Maybe in a spherical cow sense of an ideal plane, "convex hull" and ideal die or some likeness of these, but I doubt, mechanically, this to be the case here. Unlike the A face which has only contact points at the corners, the B face has an edge that can become flush with the table where friction and drag will be created as the die comes to a rest. --Modocc (talk) 06:45, 18 July 2012 (UTC)

The suggested iterated simulation should not differ from that of an ordinary 20-sided die having the same vertices and mass center. I also agree with Ben with regard to air resistance, as its likely unfair because of a substantial difference in drag (thus stability) with respect to the air (one can sort of get a feel for this by imagining blowing on one while its in different positions while it is at rest), and in the asymmetric contact with the table of its edges with respect to the two face types A (no edge, just three corners) and B (with one edge and a corner). But since this die's vertices are equidistant from the mass center which is located at the center of the intersection of the three rectangles, it might be more or less fair if the edges are recessed slightly to prevent them from contacting the table and these dice are tossed in a vacuum, but even then there might be a slight bias simply because of small differences in how the die's kinetic energy is dissipated. --Modocc (talk) 05:59, 18 July 2012 (UTC)

This is a fascinating example, and it's great to see Steve back, by the way.

One thing no one has touched on (unless I missed it) — are we assuming that the rectangles are perfectly rigid? If the rectangles flex and dissipate energy that way when the die bounces, I would expect that to contribute some sort of asymmetry.

It seems to me that there's a chance the die might be "fair" if everything were perfectly rigid, it's in a vacuum, and there's no friction. But then one has to ask, why would it ever stop bouncing? So maybe in that case it would be "fair" because the probability of it coming to rest on any of the implied faces is the same as for any other face, namely zero. --Trovatore (talk) 07:05, 18 July 2012 (UTC)

Unfairness is not proven merely by pointing out that perfect symmetry does not exist in this imperfect world. The model of fairness implies that each of the 20 outcomes has the same probability. Any other model should show and argue for some other distribution of the probability. It is not sufficient to "expect some sort of asymmetry". Bo Jacoby (talk) 09:23, 18 July 2012 (UTC).

It is also not sufficient to expect fairness, thus when it comes to correctness, assigning equal probabilities can be just as blind as assigning unequal probabilities, hence I take issue with the principle of indifference or insufficient reason as being too prejudicial. --Modocc (talk) 11:01, 18 July 2012 (UTC)

Apparently you missed the fact that the faces are not mutually indistinguishable, having nothing to do with imperfections as they exist in the real world. Sławomir Biały (talk) 12:27, 18 July 2012 (UTC)

What then, Sławomir and Modocc, is the correct probability distribution function ? Bo Jacoby (talk) 13:29, 18 July 2012 (UTC).

I don't know Bo. But I am not quite so arrogant as to think that my lack of knowledge implies that all outcomes are equally likely. I also don't know what the probability that a coin will come up "edge" is, but I don't therefore conclude that the three possible outcomes "heads", "tails" and "edge" are equally likely. Sławomir Biały (talk) 13:51, 18 July 2012 (UTC)

I haven't done the sums, but as Steve says there are A faces and B faces. Because of symmetry, the A faces are going to be equilateral triangles and the distance from the cntre of the triangle to the centre of mass is readily calculable - though I haven't done it. Is the B face equilateral or just isosceles? What is the distance from the centre of a B face to the centre of mass - if it differs from the face A case then it isn't fair. If the triangle isn't equilateral then I guess that it isn't fair. Even if distances are equal and the B triangles equilateal there may still be unfairness relating to face-face angles along edges. -- SGBailey (talk) 13:37, 18 July 2012 (UTC)

The (virtual) faces form an icosahedron, so they are all equilateral, all equidistant from the centre of mass, and all indistinguishable in every sense except with respect to the geometry of the rectangles. 86.160.212.146 (talk) 13:45, 18 July 2012 (UTC)

The faces are not indistinguishable. Certainly some moment of the mass distribution (maybe the third moment, although I'm not convinced that even the second moments are the same) will be different at some of the faces. The question is, will these different moments influence the probability, or do they somehow conspire to create a fair die? Sławomir Biały (talk) 13:51, 18 July 2012 (UTC)

The ways in which they are distinguishable involve the geometry or disposition of the rectangles. I was trying to explain that all the other aspects, such as length of edges, distance to centre, dihedral angles, etc., that SGBailey was concerned about, are not points of difference since the faces form an icosahedron. 86.160.212.146 (talk) 13:57, 18 July 2012 (UTC)

Wow! I started off more of a debate than I expected! I would prefer to ignore air resistance and concerns about the friction of edges rather than corners because the dice will be pretty heavy and those things will likely be negligible. I just love the suggestion to make the short edges of the rectangles slightly concave so that contact with a planar tabletop only happens at the vertices! That's obviously a good idea...providing that making those edges slightly concave doesn't make it less fair for reasons of center-of-gravity.). No real-world dice are perfectly fair - they all have numbers or spots etched into them for example...but I'm only really concerned about whether this is likely to be significantly less "fair" than a conventional 20 sided dice.

Incidentally...I'm also thinking about making a more normal 6-sided dice from two interlocking rectangles (each having sides of length one and root-two)...similar problem - some faces are formed from two parallel edges and others are an X-shape intersection to two edges. Are these fair?

Thanks again! This is great stuff. SteveBaker (talk) 14:40, 18 July 2012 (UTC)

The alternative construction of using six rigid wires to connect pairs of opposite vertices to the centre would indubitably give fairness - and the inevitable slight springiness would give a pleasing degree of bounce when the arifact was thrown onto a table. It would be harder to make, though. I'd do it by marking the spherical coordinates of each vertex on a foam ball, pushing the wires through, temporarily tieing the vertices together, burning the ball away, securing the wires at the centre then cutting the temporary ties ←86.139.64.77 (talk) 15:43, 18 July 2012 (UTC)

Hard to mark the numbers though... 86.160.212.146 (talk) 17:17, 18 July 2012 (UTC)

The device in the OP didn't have faces to mark, either ←86.139.64.77 (talk) 23:55, 18 July 2012 (UTC)

You could write the numbers on the rectangles according to some suitable scheme. 86.160.212.146 (talk) 00:50, 19 July 2012 (UTC)

Or you could have the numbers removed, for luck. It is OK, as long as you remember where they used to be. --Trovatore (talk) 02:37, 19 July 2012 (UTC)

SteveBaker, your drawing of the icosahedral dice (or die?) is very nice (or nie?). Show us a drawing of the hexahedral thing you have in mind! You ask if it is fair. It has less symmetry than the first one because the moments of inertia are not obviously equal. But that does not imply that it is unfair, so the answer is that it is fair until proven otherwise. Bo Jacoby (talk) 04:44, 19 July 2012 (UTC).

I'm sorry, but I really do think that "fair until proven otherwise" argument is nonsense. 86.146.110.153 (talk) 10:35, 19 July 2012 (UTC)

Don't be sorry, you are entitled to be mistaken. The fair probability distribution has maximum entropy reflecting complete ignorance. An unfair probability distribution reflects some knowledge about why some outcome is less probable than another outcome. That's why. Bo Jacoby (talk) 10:53, 19 July 2012 (UTC).

I'm afraid it's you who are mistaken. Assuming that an arbitrary dice is fair just because no one has proved it unfair is quite clearly incorrect. 86.146.110.153 (talk) 11:19, 19 July 2012 (UTC)

See indifference principle. If you have n alternatives, and know nothing about them, the best assumption is to treat them as equally likely. While dice may not be fair as a general rule, usually your best estimate of the behavior of an unknown die is to treat it as fair. This is likely wrong, but less wrong than an arbitrary other assumption. --Stephan Schulz (talk) 13:19, 19 July 2012 (UTC)

Yes, and it has already been explained why the indifference principle does not apply. FTA (emphasis mine): "The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n." But the sides are not indistinguishable. Indeed, the mass has distinct moments about the two different sets of faces. If anything, that should constitute a proof that the die is not fair, unless somehow these moments conspire to create a fair die. That should require proof. As 86 says, the "'fair until proven otherwise' argument is nonsense". I might as well claim that the three outcomes of a coin toss (heads, tails, and edge) are equally likely. Sławomir Biały (talk) 13:52, 19 July 2012 (UTC)

The logical fallacy that you and Bo are committing is argument from ignorance. Sławomir Biały (talk) 15:05, 19 July 2012 (UTC)

Some pseudo-mathematical musings: To a first order approximation, the die will be fair, since the convex hull is a regular icosahedron whose center of mass in its geometric center. Hence all of the equilibria of the die have the same first moment. The higher moments will be different, so the die (most likely) will not be fair when it is actually rolled, since this involves the mechanics of rotation about various axes, etc. It would probably be difficult to detect this lack of fairness in practice, since the deviation from a fair die will be small, although one should in principle be able to estimate it by a calculation. I lack the particular expertise to do this, but many similar calculations have been performed in classical mechanics by people like Richard Montgomery and Jerrold Marsden. Sławomir Biały (talk) 15:21, 19 July 2012 (UTC)

The stated condition for fairness that "the n possibilities are indistinguishable except for their names" is sufficient but not necessary. Probabilities express our state of knowledge, and we do not know any reason why type A outcomes should be less or more frequent than type B outcomes. Maybe people like Richard Montgomery and Jerrold Marsden can contribute to our knowledge, but so far they haven't.

The three outcomes of a coin toss (heads, tails, and edge) are not equally likely because edge is an excited state while heads and tails are ground states. The Boltzmann factor estimates the ratio between the probabilities. Bo Jacoby (talk) 16:04, 19 July 2012 (UTC).

If a coin's edge allow is allowed to be as wide or wider than its faces your argument fails because you do not know what kind of coin you have. In general, we use empirical knowledge to assert fairness. --Modocc (talk) 16:56, 19 July 2012 (UTC)

we do not know any reason why type A outcomes should be less or more frequent than type B: Actually, we do. There is asymmetry present. This implies that, unless a proof is given to the contrary, one outcome will be more likely than another. Our computational inability to determine otherwise does not mean that suddenly both outcomes are equally likely. I hope you can see that this is a ridiculous argument! If someone were to pick a random mass distribution supported in the icosahedron, with center of mass at the geometrical center, the resulting die would almost surely not be fair. So why is this particular mass distribution special? Sławomir Biały (talk) 16:41, 19 July 2012 (UTC)

Bayesian/frequentist smackdown

Are you guys having a Bayesian/frequentist smackdown? It sounds like Bo and Stephan might be operating with the Bayesian definition of probability, which I'll crudely summarize here as "probability is a statement about our current state of knowledge. Absent any additional information, you can assign equal priors to all of the equivalent states. Then, if and when you get new information, you can just calculate updated posteriors". It also sounds like Sławomir is operating from a more "frequentist" perspective, which I'll crudely summarize as "events have fixed, constant and objective probabilities. Just because we're not aware of what they should be, doesn't mean that they change as we learn more." - I think that explains your perspectives, where Bo and Stephan are saying "okay, there are different faces, but we can't think up a way that would make a difference - we can just call them equivalent until someone points why they're not", and Sławomir is saying "The faces are different! We can't say anything about the probabilities until we can determine how that affects the (true, underlying and unchanging) probabilities." -- 71.217.5.199 (talk) 16:56, 19 July 2012 (UTC)

My position is that there is an absolute, for allegorically its the elephant in blind men and an elephant, and I prefer Bayesian probability approach since it takes into account the knowledge of the not so blind men. The best assumptions are usually going to be those that happen to be correct, but these are not always necessary. --Modocc (talk) 18:16, 19 July 2012 (UTC)

The thing is though that I'm not interested in our present state of knowledge about a particular dice - it's useless to me. The problem here is to predict the future: What statistical distribution of numbers will this dice produce? We don't have any lack of knowledge here. The system is fully described. So saying "we don't know the answer so we're going to just guess that it's fair" is a viable philosophical position - but absolutely useless for producing the desired result. In answering this WP:RD question, "The math is too complicated, so I don't know" is a better answer than "I assume it must be fair". SteveBaker (talk) 16:30, 20 July 2012 (UTC)

Bayesian statistics is superior to frequentist statistics. But Bo is misusing it by applying it to the wrong question. If the question was "I've just rolled the die, what is the probability that it landed on face 13?" the answer is 1/20. But the question is "for a given idealized probabilistic process of rolling the die, what is the probability of the event that the die lands on 13?", the answer isn't 1/20, it's a prior distribution over the collection of possible probabilities, with a mean of 1/20. If I spend more time gaining knowledge by studying the probabilistic process in question, I can obtain a more precise answer, and if I solve it completely I will be able to give a single number as an answer (and at that point my answer to the first question will also be different). -- Meni Rosenfeld (talk) 05:16, 22 July 2012 (UTC)

Assume you know nothing about the die except that the outcomes are 1,2,3,...,20.

• If the question was "I've just rolled the die, what is the probability that it landed on face 13?" the answer is "1.000000 if it landed on 13 and 0.000000 otherwise".
• If the question was "what is the probability that it will land on face 13 the first time I toss it?" the answer is "0.050000".
• If the question was "what is the probability that it will land on face 13 the second time I toss it, provided it landed on face 13 the first time?" the answer is "0.095238".
• If the question was "what is the probability that it will land on face 13 the second time I toss it, provided it didn't land on face 13 the first time?" the answer is "0.047619".

Bo Jacoby (talk) 12:46, 22 July 2012 (UTC).

It might be easier to guess the reasoning behind these numbers if they were presented as 1/20, 2/21, 1/21 rather than 0.050000, 0.095238, 0.047619. —Tamfang (talk) 18:30, 22 July 2012 (UTC)

Bayesian probabilities represent a subjective state of knowledge. If you rolled the die and I know it landed on 13 then the probability is 1. If you rolled the die and I don't know the result, the probability (for me) is 1/20. -- Meni Rosenfeld (talk) 03:49, 23 July 2012 (UTC)

Two persons sharing knowledge should agree on the Bayesian probabilities. In that sense it is objective. Frequentists only consider probabilities for future events such that nobody knows the result. They consider the probability that the die will land on 13, not the probability that the die did land on 13. Bo Jacoby (talk) 12:36, 23 July 2012 (UTC).

Can moments of the mass distribution over order 2 be ignored?

Suppose we assume simple Newtonian physics, and, in particular, a constant gravitational acceleration in the region of interest. Can we then say that the dynamics of a perfectly rigid object such as an idealized version of this die, in that environment, in interactions that never penetrate its convex hull, are completely determined by its mass, center of mass, and inertia tensor? If not, can anyone give an example of a phenomenon where third- or higher-order moments have some sort of physical effect, given the restrictions of the model above? -- The Anome (talk) 18:48, 19 July 2012 (UTC)

1. 71.217.5.199 is right. I am a Bayesian.
2. Anome is right. The hamiltonian of a rigid body does not depend on third- or higher-order moments.
3. Modocc is right. If a cylindrical 'coin' is thicker than its diameter then the edge outcome is probable.
4. Sławomir, is face type A or face type B the more probable ?
5. Perhaps the probability that a convex polyhedron will land on face number i may be approximated by (the solid angle of face number i as seen from the center of mass)/(4π). (I retract the suggestion to use Boltzmanns law because die tossing is far from thermodynamic equilibrium).

Bo Jacoby (talk) 22:11, 19 July 2012 (UTC).

I have computed the inertia tensor. It's proportional to the identity, so the Hamiltonian is spherically symmetric and the die is fair. Sławomir Biały (talk) 23:11, 19 July 2012 (UTC)

Luke 15:7. Bo Jacoby (talk) 06:57, 20 July 2012 (UTC).

Well, even as a Bayesian you can still work on your prior... The Hamiltonian is not spherically symmetric because the di(c)e has to land on a flat surface and come to rest in a stable or metastable position; the surface (and the gravitational force) breaks the symmetry. Imagine taking a cube and chipping off a bit at the corners, so you end up with a polyhedron with large octagonal and small triangular (trigonal?) sides. The moment of inertia is still spherically symmetric, but the thing will more often come to rest on one of the octagonal sides and only rarely on a triangular side (Bo's fifth point). For the icosahedron in question, the situation is less obvious and from a statistical point of view the assumption of fairness (or complete ignorance) is justified if you don't want to spend a lot of effort on the physics, even more so when you're dealing with a material realization of the thing with faults and imbalances and everything. --Wrongfilter (talk) 07:51, 20 July 2012 (UTC)

Wrongfilter, what is your suggestion for an improved prior? Bo Jacoby (talk) 08:33, 20 July 2012 (UTC).

I don't know. But as Steve said in the original question, we do know that there are two classes of triangular faces. The question was, and still is, how to use that knowledge to predict more accurate probabilities for whether the icosahedron will fall on one or the other type of face. --Wrongfilter (talk) 11:10, 20 July 2012 (UTC)

Imagine that the convex hull of Steve's die is filled with some mysterious opaque and massless substance. Then you have a perfect regular icosahedron with center of mass located in the geometrical center and inertia tensor proportional to the identity tensor. There is no way that you can distinguish A-faces from B-faces any more. This thing has the same convex hull and the same Hamiltonian as Steve's die, so it moves in exactly the same way, both when tossed and when hitting the table. The regular icosahedron is fair, and so is Steve's die. Q.E.D.. Bo Jacoby (talk) 14:30, 20 July 2012 (UTC).

Aha! That's the answer I needed! Many thanks! SteveBaker (talk) 16:30, 20 July 2012 (UTC)

Perfected your prior, I see. Now if we could only assume that such approximations are sufficiently valid. Given that golf balls are dimpled to improve flight, and the seams of heavy baseballs affects their flight, see knuckle ball, I'd would want Steve's dice tested for systemic bias due to drag effects, and perhaps modify them to eliminate any, before using them in a lottery. The die has to settled into a stable non-rotating state where it cannot role anymore, and if one configuration has less drag as it rolls it will tend to not lose momentum as quickly and settle down. Small random imperfections may not affect the die, but this systematic one might be significant, and dice are generally fairly lightweight (for practical reasons) thus the problem might be significant. Modocc (talk) 15:40, 20 July 2012 (UTC)

We could roll the die in a vacuum. Rckrone (talk) 16:02, 20 July 2012 (UTC)

I'll grant that air resistance and different friction (and 'sticktion') effects might come about because of the different amount of contact with the ground between class A and class B surfaces. However, for my purposes where the die is small and fairly heavy - I don't think they'll be significant enough to concern me. After all, no dice is ever perfectly constructed - they have dimples or numbers on them that alter the center of gravity, the air drag on the facets and the friction for each face is different...the real world is complicated. However, I don't need that kind of perfection - I just need people who play Dungeons & Dragons not to notice that class A faces come up twice as often as class B faces! SteveBaker (talk) 16:30, 20 July 2012 (UTC)

Its been decades since I've played, but I have enjoyed the game, and thus hope these do work out well. With a single, standard die, most minor mass and surface imperfections will tend to cancel each other of course, and people normally assume that their dice are not loaded in any obvious special way, and certainly with any game that is not as high-stakes as the lottery, and with a small sample size with limited use, no one will notice or care about anything as small as a one percent difference. That said, should there ever be a larger audience though, and a larger tested sample size of these unique dice with a proven bias, even if the difference is negligible, there will be folks that will tend to favor the luckier sides or favor the standard dice if these are perceived as being fairer. I do think this die looks really cool, and it should be relatively easy for a manufacture to test and modify. :-) Modocc (talk) 17:47, 20 July 2012 (UTC)

Experimental results

• I have made one of these out of card, and I observe a bias towards "Class A" faces (currently 181 out of 300 throws). When you throw it, it also "feels" as if that result will be more likely. I encourage others to try this and see if they get the same results. 86.179.1.131 (talk) 19:23, 20 July 2012 (UTC)

The die has 8 A-faces and 12 B-faces, so 300 throws should give 120±8 A-faces and 180±8 B-faces. Are you sure you haven't switched labels A and B? Bo Jacoby (talk) 00:06, 21 July 2012 (UTC).

The case I am calling "A" is the case when the dice is resting on three points of three different rectangles, per the OP's original description. 86.179.1.131 (talk) 00:35, 21 July 2012 (UTC)

Your result 181:119 indicate that the A-probability is 0.603±0.028. (beta distribution). The A-probability for a fair icosahedron is 0.400 (=8/20), which is 7.25 standard deviations below mean. (0.603-7.25*0.028=0.40). So you have - beyond reasonable doubt - proven unfairness of your die! What are the dimensions of your cards? Bo Jacoby (talk) 07:04, 21 July 2012 (UTC).