# How can you test to see if a dice is weighted?

I was browsing Etsy today and came across this.

What tests are there to see if the dice are usable, ie, if one side isn't favored over another, and if all sides are balanced?

Would this just be to roll the dice a large nubmer of times and collect the data? Or are there other tests that could be done without doing that?

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## 1 Answer

A statistical test would require a large number of rolls of the dice. For a simpler statistical example, to test whether a coin is fair by tossing a coin $N$ times would result in approximately $\frac{N}{2}$ heads but the standard deviation of that count would be about $sqrt(N)$ so to get to a level of significance where the standard deviation is only 0.01 on the 0.50 value that you are trying to measure would require $N = (0.01)^{-2} = 10000$ tosses of a coin.

Another test you could try would be to try to accurately measure where the center of mass of the die is. You could do this by seeing if the die can approximately balanced on a knife edge which is bisecting one of the die's faces. Do this for 3 of the orthogonal faces of the die and if they all are balanced, then the center of mass is in the geometric center of the die. However, I don't know how to determine what error in this measurement would be achievable and how much an off-center center of mass would affect the fairness of the die.

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The stats part of this is wrong. There are deterministic ways that you can tell if you are dealing with weighted dice/coins, and using a far fewer number of trials that what you posit. I would think that an easy way to tell whether the dice was weighted is to suspend it in some medium where it floats, and then set it spinning. If it is weighted, there will be a skew in which side faces upward. –  Jen Nov 5 '11 at 1:18
I like the idea of floating the die. What is wrong with the stats? The flips of a coin are like a random walk in 1 dimension and the standard deviation of your distance from 0 grows like $sqrt(N)$ AFAIK. - just looked it up, the distance is $sqrt(\frac{2 N}{\pi})$ –  FrankH Nov 5 '11 at 7:50
My internet connection keeps cutting out so I can't post a proper response, but look up the chi-square test. You can determine with reasonable accuracy pretty quickly if a dice (or coin) is weighted based on the disparity between the expected and observed concurrences. –  Jen Nov 5 '11 at 10:41
Good dice has some imbalance, because the dots are carved, and six pits make that side lighter than the on pit on the opposite side, etc. –  Georg Nov 5 '11 at 10:48