Three-sided Cylindrical Dice

Authors: Matthew, Rebekah, Deepu, Brian, Fiona, Miracle, Carsten, Zoe

I. State the problem.

Is this a fair die? The Riverbend gang is bound to find out. We are testing cylindrical dice. Will the die land on all the sides the same amount of times?

II. Talk about the Riverbend Math Center and who worked on the project.

Here we do a lot of math and Amanda makes it easier for us to learn math related things. The people who participate are Fiona, Rebekah, Matthew, Deepu, Brian, Travis, Victoria, Valeria, Laura, Zoe, Jennifer, Martha, and Jimmy.

III. Conjectures

Height = diameter.
Some people thought the height would be the same as the diameter. Miracle conjectured, "Because the sides would be the same, I thought it would be fair and land on both sides."
Height = radius
Also, some thought the height would equal the radius.
Equal area on all three sides.
Another conjecture was that the area would be equal on all the sides.
Show diagram (Travis' conjecture)
Basically, this conjecture relies on he idea that the coin will land on its side if a at the time that the coin touches the grounline from the center of gravity to the ground passes through theside d. Assuming a no velocity, the coin will move such that the center of gravity gets closer to the ground, which would be such that the ide hits the ground. After this, the coin doesn't have enough energy to tip any further (to do so would require more energy than the coin originally had as potential). viewing a cross-section of the coin, it becomes clear that for the probabilites of the coin landing on the sides or the top or the bottom to be equal, there must be the same probability that a random line would go through each one of these parts (this again assumes that little or no rotaton is present). for this to be possible, the arcs created by circumscribing the cross-section must be equal. thus the sum of the arc lengths for the side (2x) must be equal to the arc lengths for each of the top and bottom. That makes 6X total, which means x must be 60 degrees. With this, we compute that the height must be 2*3^(1/2)/3 times the radius of the coin.
Show diagram (Von Neumann's conjecture)
1/phi x diameter = height
Another person thought that the height divided by the diameter should equal 1 / phi, or 0.618.

IV. Explain how we tested this

  • Matthew made a LEGO NXT robot that threw the dice to double-check our candidates.
  • Explaining the Formula for Standard Deviation:
  • 68% of the time, a sample will fall within 1 standerd deviation of the mean.
  • 95% of the time, a sample will fall within 2 standerd deviation of the mean.
  • 99.7% of the time, a sample will fall within 3 standerd deviations of the mean.
  • mu= 0.333
  • sig= 0.067
  • mu+sig= 0.4
  • mu+2sig= 0.467
  • mu+3sig=0.534
  • mu-sig=0.266
  • mu-2sig=0.199
  • mu-3 sig=0.132
  • If our proportion is NOT between 0.199 and 0.467 then we are 95% confident that it is not a fair die.


We made our dice, then we tossed them in the air about 50 to 100 times. On a piece of paper we recorded how many times the die landed on each side. For example if you tossed the die in the air 50 times and it came up 16 times on the side,you would write the fraction 16/50 to represent the proportion of times it landed on its side. Then you would turn the fraction into a decimal. To find the decimal you have to divide the numerator by the denominator. In this case the answer would be 0.32. If the decimal was between 0.199 and 0.467 it would be a candidate or else it would be a reject.

Our dice.

  • Our materials were: PVC pipe, washers, wood, and/or felt pads. The materials the dice were made might of made a difference in the ratios. We also used metal washers.

VI. How we determined which dice were fair, and which conjectures were the best.

  • We had 12 candidates and 19 rejects.
  • A total of 31 dice.
  • 39% candidates and 61% rejects.
  • So that means

VII. Our results.

  • Some of our dice were candidates and some were rejects.
  • If the height is bigger then the diameter the die is a reject.
  • Matt's conclusion: If the height is equal to the diameter the die is a reject.
  • Zoe rolled a dice that had dementions of 1 inch in height and 2 and 1/2 in diameter, and it worked two times. She used one felt pad and a block of wood.

Die code Candidate/Reject Material Description Height (Inches) Diameter (Inches) Height / Diameter Times Tested # of 1s # of 2s # of 3s (Lateral surface) Fraction of 1s Fraction of 2s Fraction of 3s Notes
B.004Reject1 felt padheight = 1/4 of radius0.12510.1350282200.560.440
A.015Rejectwood dowelsee Measurements >>0.251.1250.2250222710.440.540.02
A.014Rejectwood dowelequal-area0.1880.750.2550242240.480.440.08
A.016Rejectwood dowelequal-area0.2510.2550232700.460.540
C.005RejectPVC pipeTravis's conjecture0.3751.0630.355136??0.71??Information partly illegible; also, testing suspicious
C.001CandidatePVC pipe sandwiched between metal washersVon Neumann's conjecture0.43810.44501517190.30.340.38
A.017Reject3/4" wood dowel (1 1/16" high) sandwiched between felt padsVon Neumann's conjecture0.43810.442610250.0400.96Testing suspicious
B.005Reject4 felt padsVon Neumann's conjecture0.87520.4450221990.440.380.18
A.001Candidatewood dowelradius = height0.3750.750.5501719140.340.380.28
A.005Candidatewood dowelTravis's conjecture0.510.5501617170.320.340.34
B.001Candidatemetal washer sandwiched between felt padsradius = height0.510.5551819180.330.350.33
B.002Candidate5 felt padsradius = height120.5501910210.380.20.42
A.008Rejectwood dowelradius = height0.3750.750.5531626110.30.490.21
A.004Candidate5/8 wood dowel, 3/8 felt padradius = height0.50.9380.53501517180.30.340.36
A.004Candidate5/8 wood dowel, 3/8 felt padradius = height0.50.9380.53501616180.320.320.36
B.003Candidate3 felt padsradius = height0.56310.56501814180.360.360.28
C.003RejectPVC pipesee Measurements >>0.6251.0630.59501413230.280.260.46
A.020Candidatewood doweldiameter x (1 / phi) = height0.7731.250.62501315220.260.30.44
A.019Candidatewood doweldiameter x (1 / phi) = height0.4640.750.62501713200.340.260.4
A.002Candidatewood dowelTravis's conjecture0.62510.63541920150.350.370.28
C.006RejectPVC pipeTravis's conjecture0.62510.635097340.180.140.68
A.003Candidatewood dowelTravis's conjecture0.6881.0630.651032931430.280.30.42
C.002CandidatePVC pipe sandwiched between metal washerssee Measurements >>0.7510.75501515200.30.30.4
A.007Rejectwood dowelsee Measurements >>0.9381.0630.885073400.140.060.8
A.006Rejectwood doweldiameter = height1.0631.06315055400.10.10.8
A.009Rejectwood doweldiameter = height0.750.7515043430.080.060.86
A.010Rejectwood doweldiameter = height0.750.7515014450.020.080.9
A.012Rejectwood doweldiameter = height1.0631.06315078350.140.160.7
A.013Rejectwood dowelsee Measurements >>1.1251.0631.065157390.10.140.76
A.011Rejectwood dowelsee Measurements >>1.5631.0631.4751911310.180.220.61
C.004RejectPVC pipesee Measurements >>2.18812.19500050001Unevenly cut - smallest height listed
A.018Reject3/4" wood dowel (2" high) sandwiched between 4 felt pads on each endsee Measurements >>3.56313.56500050001

VIII. Remaining questions.

  • Does the weight of the dice make a difference in our results?
  • Is the diameter of the cylinder always higher than the height if the cylinder is a candidate?
  • Does it matter what material it was?

IX. What we would do differently if we did this experiment again.

We would remake our dice out of different materials to see if the different materials really did make a difference in the ratios.

Riverbend Community Math Center
(574) 339-9111
This work placed into the public domain by the Riverbend Community Math Center.