In the present work we introduce the problem of determining the probability that a rotating and bouncing cylinder (i.e. flipped coin) will land and come to rest on its edge. We present this problem and analysis as a practical, nontrivial example to introduce the reader to Bayesian model comparison. Several models are presented, each of which take into consideration different physical aspects of the problem and the relative effects on the edge landing probability. The Bayesian formulation of model comparison is then used to compare the models and their predictive agreement with data from hand-flipped cylinders of several sizes

Blais, Brian S.

Kuindersma, Scott R.

English

DigitalCommons@Bryant University

Bryant UniversityDigitalCommons@Bryant UniversityScience and Technology Faculty Journal Articles Science and Technology Faculty Publications andResearch2007Teaching Bayesian Model Comparision with theThree-Sided CoinScott R. KuindersmaBrian S. BlaisBryant UniversityFollow this and additional works at: https://digitalcommons.bryant.edu/sci_jouThis Article is brought to you for free and open access by the Science and Technology Faculty Publications and Research at DigitalCommons@BryantUniversity. It has been accepted for inclusion in Science and Technology Faculty Journal Articles by an authorized administrator ofDigitalCommons@Bryant University. For more information, please contact dcommons@bryant.edu.Recommended CitationKuindersma, Scott R. and Blais, Brian S., "Teaching Bayesian Model Comparision with the Three-Sided Coin" (2007). Science and Technology Faculty Journal Articles. Paper 22.https://digitalcommons.bryant.edu/sci_jou/22Determining the Geometry of a Three-sided Fair Coin:Exploring the Probability of a Coin Landing on its EdgeBrian BlaisScience and Technology DepartmentBryant Universitybblais@bryant.eduScott KuindersmaBryant Universitysrk2@bryant.eduMay 9, 2007AbstractA three-sided fair coin is defined to be a cylinder which, when flipped like a coin, has an equal probabilityof landing on heads, tails or the edge. We present an analysis of this problem from several perspectives.1 IntroductionA three-sided fair coin is defined to be a cylinder which, when flipped like a coin, has an equal probability of landingon heads, tails or the edge. A regular coin, like a penny, has almost zero chance of landing on the edge. On theopposite extreme, a long narrow cylinder like an unsharpened new pencil, has almost zero chance of landing on“heads” or “tails”. Somewhere in between these extremes should exist a cylinder with equal probability of landingon its edge or an end. What is the size of this cylinder or, more precisely, how does the radius, R, of the circularcross section compare with the height, h, of the edge (Figure 1)? Although easily stated, this problem has manypossible solutions, with varying degrees of complexity. In addition, there are a number of intuitive hypotheses thatturn out to be incorrect. As a result, this problem is ideal to introduce students to probability and mechanics.Some definitions:• η ≡ h/R• l =√R2 + (h/2)2We will concern ourselves with the more general question ofWhat is the probability, pedge, for the coin to land on the edge, as a function of the radius, R, and theheight, h?Figure 1: Three-sided coin. (A) A cylinder with radius, R, of the circular cross section and height, h, of the edge.(B) The cross section of the cylinder.1The more restricted question of the fair three-sided coin will be What values of h and R yield pedge(h,R) =1/3?. The result should probably be scale invariant, and thus should be expressible in terms of the ratio, η ≡ h/R,although due to the contraints of flipping, this invariance may not always hold.2 Some of the Previous Methods for Solving the ProblemThe following methods have been reported in a number of sources(?; Murray and Teare, 1993), with a varyingdegree of thoroughness and experimental corroboration.2.1 Equal AreasIn analogy with dice, one presumes that the probability is proportional to the surface area of the edge.pedge(h,R) =2piRh2piR2 + 2piRhpedge(η) =η1 + ηFor the fair coin we obtainη1 + η=13η =122.2 Equal LengthAgain, in analogy with dice, one presumes that the probability is proportional to the cross sectional length. Asquare cross section would then have equal probability.pedge(h,R) =h2(2R) + hpedge(η) =η4 + ηFor the fair coin we obtainpedge(η) =13=η4 + ηη = 22.3 Equal Solid AngleEdward Pegg Jr in his Master’s thesis(Edward Pegg, 1997) introduces a Geometrical Model, where the probabilityof certain faces of dice being produced (coins are refered to as “cylindrical dice”) is proportional to the solid anglesubtended by the face. For the edge probabilities we obtainβ ≡ atan(h2R)= atan(η/2)edge area subtended =∫ 2pi0dφ∫ pi/2+βpi/2−βsin θdθdφ2= 4pi sinβpedge(η) = sinβ=η√η2 + 42.4 Activation Energy MotivationIn (Levin, 1983) a similar problem is addressed for regular cubical dice loaded with a brass cylinder to offset thecenter of mass. They also consider a similarly altered cylinder as an example. Due to the symmetries of ourproblem, their result takes on a particularly simple form. If we label the states “heads”, “edge”, and “tails” with1,2, and 3, the probability for each state depends on the gravitational energy of the state, Ui, and the activationenergy, AEi, or the energy needed to leave one state for another one. Levin empirically suggests a form of thefollowingpi ∝ e−UiAEip1 = p3 ∝ e−h/2l−h/2p2 ∝ e− Rl−Rsopedge(h,R) =e−Rl−R2e−h/2l−h/2 + e−Rl−R2.5 Center-of-Mass ArgumentFor a solid object with one point of contact with the floor, the location of the center of mass with respect to thevertical line through the contact point determines which way the object will tip. If the center of mass is to the leftof the vertical line then the object will tip to the left. Likewise, if the center of mass is to the right, then the objectwill tip to the right. If we assume there is no bouncing, then the coin lands on a contact point at a particularangle, θ. Comparing the landing angle to the angle at which the center of mass is directly above the contact point(Figure 2), α, will determine the direction the coin will tip over and come to rest:• if 0◦ < θ < α then the coin will land on heads• if α < θ < 90◦ then the coin will land on the edgewhere α = atan(Rh/2). For h/R ∼ 0 then α ∼ 90◦. For larger h/R, then α gets smaller. Symmetry of the problemallows us to then write that the probability for landing on edge ispedge(h,R) = 1− α90◦ ≡ pewhere the definition of pe as the edge probability from the center-of-mass argument will be used later.For a fair coin we obtainpedge(h,R) =13= 1− α90◦α = 60◦h/R =2√3≈ 1.1553αhRlFigure 2: Coin with the center of mass directly above the impact point. The angle α = atan(2R/h) is the angle atwhich this particular configuration is achieved. It is a geometric property of the coin itself.2.6 Murray and Teare, 1993In Murray and Teare, 1993, a dynamical model of a coin is simulated. They use a straightforward application ofNewton’s laws, on a model cylinder allowed to move in one dimension (vertically), rotate and bounce on the surface.In dimensionless units the position of the center of mass, Z(t), the vertical speed, V (t), and the orientation angule,θ(t) are given byZ(t) = Zo + Vot+12t2V (t) = Vo − tθ(t) = θo + ωtThe coin is numerically simulated until it strikes the ground. At which point the vertical and angular speedsare modified by the bounce. Defining V ′ and V ′′ as the vertical speeds just before and after the bounds, and ω′and ω′′ as the angular speeds just before and after the bounce, they presentV ′′ = V ′ − (1 + γ)k2 V′ + xω′(k2 + x2)ω′′ = ω′ − (1 + γ)x V′ + xω′(k2 + x2)where x is the position of the contact point and γ is coefficient of restitution parameter. For γ = 1 there is noenergy loss, and γ = 0 there is no bounce (all energy is lost on the first bounce). One can show that the velocityof the corner of the coin that strikes the floor, before and after the impact, are related by U ′′ = −γU ′.2.7 Summary of MethodsFigure 3 displays the methods described thus far. Table 1 shows the predictions for the probability of real coinslanding on the edge(?).3 Potential Energy of a Coin: Incorporating BounceThe more detailed simulation of the coin seems to have the following two properties:1. much lower probability of edge landing for thin coins4Figure 3: Theories for the Probability of an Edge-falling Coin. Plotted are the edge probabilities as a function ofthe h/R ratio of the coin. The Murray-Teare results are from simulation, with a γ = 0.4.A)Coin Radius (mm) Height (mm)Quarter 12.1 1.75Dime 8.95 1.35Nickel 10.6 1.95Penny 9.5 1.5B)Number of Edge Landings in 10,000 FlipsCoin Equal Area Equal Length Center of Mass Murray and Teare, 1993∗Quarter 1264 349 460 3Dime 1311 363 479 5Nickel 1554 440 584 17Penny 1364 380 502 7∗ extrapolated from data. The original Murray and Teare, 1993 paper quoted 1/6000 tosses for a nickel.Table 1: (A) Size Data for US Coins. (B) Number of Edge Landings in 10,000 Flips Predicted by Different Methods5Figure 4: Energy (or Height) of the Center of Mass of a Coin versus Orientation Angle.2. higher relative probability than other models for thick coinsPoint 1 seems to be more closely related to ones intuition: flipping a quarter has much less than 1/10 chance oflanding on edge (Equal Areas), and even less than 1/30 chance (Equal Length and Center of Mass). Is there asimpler way to account for this? The physics, given the assumptions, for the Center of Mass argument is correct.The error must lie in the assumptions. There is one obvious assumption: no bouncing is included. What is theeffect of bouncing?In order to address this problem in a simple way, we introduce an energy function for the coin. This is theheight of the center of mass, as one rolls the coin along the surface, starting in a “heads” position. This height, orgravitational energy (in dimensionless units) is shown in Figure 4. It has local minima at the positions of “heads”,“tails” or the “edge”, and maxima when the coin is in the configuration shown earlier in Figure 2.We can incorporate bouncing in the following way. Let ECM(θ) be the center of mass energy as a functionof the orientation of the coin as it strikes the floor, E be the incoming energy of the coin, and γ be the fractionalenergy loss on each bounce, i.e. on each bounce E → γE. Note that this is slightly different than the γ used byMurray and Teare, 1993. In their case the velocities behaved like U → γU at a bounce. Since the kinetic energy ofthe coin goes as the square of the velocity, our γ should be compared to the square of the γ used by Murray andTeare, 1993.As the coin bounces, it loses energy. As long as it has enough energy to escape the energy wells for the threecases, it will keep randomizing its angle. Eventually it will not have enough energy to escape one of the energywells. In general (for h/R < 2) the energy well for “edge” is smaller than that for “heads” or “tails”. There willcome a point when the energy will be large enough to escape the “edge” well, but not a “heads” well. After thatpoint, if the coin has a heads-inducive fall, then it will stay on heads forever. Thus, to land on edge, the coin musthas a series of edge-inducive falls in a row until the energy falls to the point where it cannot escape the “edge”energy well.6We have the following relationshipsα = atan(Rh/2)(probability for an edge-inducive fall) pe = 1− α90(coin energy after n bounces) En = Eoγn(number of bounces, given initial and final energy) n = log(En/Eo)/ log γ + 1(half-length of the diagonal of the coin) l =√R2 + (h/2)2(depth of the “heads” energy well) Eh = l − h/2(depth of the “edge” energy well) Ee = l −RIf we start with energy just able to escape the largest well (Eh), and bounce until we cannot escape thesmallest well (Ee), then we will bounce n times where n isn = log(Eh/Er)/ log γ + 1The probability of bouncing n times, each time with an edge-inducing fall, is simplypedge = pneFigure 5 compares this simple bounce model with the Murray and Teare, 1993 results. The similarity isquite striking, given the simplicity of the bounce model.3.1 Correlated BouncingIn this model, we ignore the energy of the bounce, and simply posit that sequential edge-conducive bounces arecorrelated, and sequential heads-conducive bounces are correlated. This model is comparable to the urn problemdiscussed in (?). Using his notation, we assume that an edge-conducive bounce increases the probability of anotheredge-conducive bounce by an amount ² > 0, while a heads-conducive bounce decreases the probability for anedge-conducive bounce by an amount δ > 0. Thus we have:P (Ek|Ek−1I) = pe + ²P (Ek|Hk−1I) = pe − δP (Hk|Ek−1I) = 1− pe − ²P (Hk|Hk−1I) = 1− pe + δIf we write the probabilities for the kth trial as a vector,Vk ≡(P (Ek|I)P (Hk|I))Then this Markov process can be described by the matrix equationVk =((pe + ²) (pe − δ)(1− pe − ²) (1− pe + δ))k−1V1The general solution becomes(?),P (Ek|I) = (pe − δ)− (²+ δ)k−1(pe²− (1− pe)δ)1− ²− δThe center-of-mass argument becomes the limit of ²→ 1− pe and δ → pe: once the first bounce occurs, thestate remains constant forever.In this model, there doesn’t seem to be any obvious criterion to determine the number of bounces. Thisparameter, n, should be treated as a nuisance parameter, between zero and a handful of bounces in realisticsituations. Otherwise, we could the calculated version above.74 Statistical Model of Bouncing• simulation– choose an angle, θ– if E > ECM(θ) then choose another angle– if E < ECM(θ) then stuck in that well• recursive formulation– given a uniform distribution over angle– given one particular well, with minimum m– probability of obtaining a depth of l − ECM(θ) less than an energy ² is approximatelyF (²) =0 ² < 0√²m 0 < ² < m1 ² > mthis is also the probability of escape, given energy ², and a particular well– define Fe and Fh as the wells for edge and heads respectively. me = l −R and mh = l − h/2– the probability for finally landing on edge on the ith bounce, given a sequence of bounce energies asbefore En = γnEo, ispedge(i) = pe(1− Fe(Ei)) + peFe(Ei)pedge(i+ 1) + phFh(Ei)pedge(i+ 1)= pe(1− Fe(Ei)) + pedge(i+ 1)(peFe(Ei) + phFh(Ei)pedge(i+ 1))– Implemented something likeif (i>i_max) {p=pe;} else {p=pe*(1-F(E,me))+p_edge(gamma*E,h,R,gamma,i+1)*(pe*F(E,me)+pht*F(E,mh));}5 Data5.1 PVC Solid Stock, Flipped by Hand5.2 Murray Teare 1993, Brass Nuts5.3 Murray 2005, Automatic Flipping6 Model ComparisonP (Mi|D, I) = P (D|Mi, I)P (Mi|I)P (D|I)P (Mi|D, I)P (Mj |D, I) =P (D|Mi, I)P (Mi|I)P (D|Mj , I)P (Mj |I)if all of the models are a priori equally likely, and we look at the logarithm, we havelogP (Mi|D, I)− logP (Mj |D, I) = logP (D|Mi, I)− logP (D|Mj , I)8Figure 5: Theories for the Probability of an Edge-falling Coin: Simple Bounce model compared to Murray andTeare, 1993 simulations. Note that the simple bounce model value of γ should be compared to the Murray andTeare, 1993 value of γ2.For models with extra parameters,= we need to marginalize over those parameters, such asP (Mi|D, I) =∫dξP (D|ξ,Mi, I)P (ξ|I)this has the benefit of automatically penalizing more complex models unless they can be justified with a particularlybetter fit (i.e. a quantitative Occam’s razor).7 The Problem with Flipping• rolling8 Conclusions9 Bibliographydaveweb: http://users.frii.com/davejen/coin3.htmReferencesEdward Pegg, J. (1997). A Complete List of Fair Dice. Masters Thesis.Levin, E. (1983). Experiments with loaded dice. American Journal of Physics, 51(2):149–152.9Figure 6: All of the Data For Coin FlippingFigure 7: Subset of the Data.10Figure 8: Subset of the Data. Different γ.Murray, D. B. and Teare, S. W. (1993). Probability of a tossed coin landing on edge. Physical Review E, 48(4):2547–2552.11

Teaching Bayesian Model Comparision with the Three-Sided Coin

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Teaching Bayesian Model Comparision with the Three-Sided Coin

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