“Bring Out Your Polyhedrons!”

Maybe contributor Jeff Partridge has too much time on his hands, or maybe he is a Dungeons and Dragons fan. Either way, he poses the following:

If you want fun with dice, start working on whether you get the same odds of a number turning up if you roll two 20-sided dice as opposed to one 40-sided die. He goes on to add that the 20-sided dice he plays with are numbered from 0-19, and asks what is the relative probability of throwing a number, for instance, 13 with either the single 40-sided die or the pair of 20-sided dice. What do you think?

Before reading my thoughts on his problem, give it a few minutes of thought. You might have a different answer than mine.

First off, the two conditions are not identical because the combination of 20-sided dice cannot generate an answer of 39 as the single 40-sided one can. Re-numbering one of the 20-sided dice to go from 1-20 doesn’t help because then you can’t throw a 0. However, if you decide 39 isn’t all that important, then you can quickly show how the probabilities of the two conditions are quite different for the remaining possible numbers. This should be obvious to anyone who has played craps and realized there are more ways to shoot a 7 than either a 2 or 12. The probability of occurrence is certainly not even for all possible numbers from 0-38.

Just as in craps played with 6-sided dice, the 20-sided dice Jeff uses have a special number that have the highest probability of being thrown. That number is 19. Why? How is that related to the probability of getting a 7 in craps? What would be the special number if the 20-sided dice were numbered from 1-20 as normal 6-sided dice are?

Assume that one wishes to make a dice-based randomizing system that generates 40 possible numbers with equal probability; is that possible using a pair of 20-sided dice?

One way is rather simple. Paint the dice different colors so they can be distinguished. Say one is red and the other blue. Throw the dice and record the value of the red die as modulo 5 and multiply it by ten. Record the value of the blue die as modulo 2 and add it to the first result. The result will have a value in the range 0-39 with equal probability.

However, since presumably these dice are being thrown as part of a game, then maybe we can enliven the game a bit. Assume the dice are thrown as above, but they are indistinguishable. The thrower has the option of deciding which one is the “tens” die and which is the “ones.” For instance, assume the throw comes up 12 and 5. Then the player has the choice of calling the final number 21 or 16 depending on which one is the “tens.” Would that add spice to the game? I don’t know.

Without going to the complication of dice that are not based on 6, 8, 12, or 20 sides (why are these numbers special?), what equally probable ranges of numbers can you generate with “simple” algorithms and throwing no more than three dice at a time? For instance, throwing any single die and assigning even or odd to 1 or 0 gives an equal probability range of 0-1. Can you think of a way to generate an equal-probability range of 0-6 (i.e. 7 values)?

Thanks, Jeff, for the inspiration.

For those who wish to delve further into decision theory without wading through a lot of equations, I have posted a tutorial on elementary decision theory. It shows examples of faulty physicians’ diagnoses (important for those considering surgery) and how to evaluate anti-terrorist activities (important for everyone). That tutorial can be found here.