# Is there such a thing as a fair die?

Is there such a thing as a fair die? On dice where the number is represented by a scooped out dot, surely that makes a difference? Has anyone done any research?

In fact thinking about it, why would a coin flip be fair? the physics on each side is completely different.

• With respect to fair dice, yes, casinos have a huge monetary interest in having dice be very very close to fair. The randomization comes in large part from bouncing off the floor and walls of the area where you throw them, and I suspect the dots play an insignificant role in that. – jbowman Apr 28 '12 at 13:46
• For the coin, see Andrew Gelman & Deborah Nolan's article in The American Statistician, You Can Load a Die, But You Can’t Bias a Coin. – onestop Apr 28 '12 at 13:48
• – nico Apr 29 '12 at 10:55

I think the concept of 'fair' is hard to define. Since a given roll of the die will produce a deterministic result (in other words, physics determines what the result is) we can't really say that there is a certain 'probability' of rolling a one. This relates to the mind projection fallacy, which essentially says that probability is a property of one's state of information of a phenomena, not a property of the phenomena itself. Relating to the roll of a dice, the result is not just based upon the die, but also the method in which it is rolled. If we 'know' enough about a given roll (the die's material composition, it's initial orientation, the forces applied to it, the environment it will land in, etc.) we can (theoretically) model all of the motion that occurs in that roll with arbitrary accuracy and instead of finding a 1/6 'probability' of landing on a given side, we will be near certain that it will land on some side.

This all is very unrealistic of course, but my point is that the method of rolling is as important as the die's physical makeup. I think a good definition of a 'fair' die would be one in which under reasonable constraints (on computing power, time, accuracy of measurements) it is not possible to predict the result of a roll with some level of confidence. The specifics of these constraints would be dependent on the reasons you are checking if the die is fair or not.

Aside: Suppose I tell you I have an 'unfair coin' and I will give you a million dollars if you can correctly guess what side it will land on. Do you choose heads or tails?

• The first paragraph of this answer displays an almost prototypical Laplacian view of randomness. – cardinal Apr 29 '12 at 3:03
• This reminds me of the Eudaemonic Pie, where some students tried to predict Roulette based on a shoe-computer :-) – thias Apr 29 '12 at 8:30
• @cardinal I disagree very much. This is basically the exact view espoused by E.T. Jaynes in his 2003 book, which is a decidedly un-Laplacian view in favor of a much more objective Bayesian view. – ely Apr 29 '12 at 14:57
• @EMS: P.S. Laplace (1814), Essai philosophique sur les probabilités, Courcier, pp. 2-3: Nous devons donc envisager l'état présent de l'univers, comme l'effet de son état antérieur, et comme la cause de celui qui va suivre. Une intelligence qui pour un instant donné, connaîtrait toutes les forces dont la nature est animeé, et la situation respective des êtres qui la composent, si d'ailleurs elle était assez vaste pour soumettre ces données à l'analyse, embrasserait dans la même formule, les mouvemens des plus grands corps de l'univers et ceux du plus leger atome: ... – cardinal Apr 29 '12 at 23:26
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A little Googling reveals a Wikipedia (gasp!) article on dice. It includes remarks on the precision of dice that mentions the problem of scooping out the dots (they are refilled with material of the same density). Are these going to be exactly fair? How will you define that? How close to 1/6 does each result have to be to qualify?

• No die is fair, but to test out whether any given one is biased is a matter of the number of rolls (i.e., time). If during a realistic lifespan of a die and, say, a million rolls, you don't have enough power to detect differences from 1/6, as well as independence of the outcomes, then for all practical reasons it is a fair die. This is the same question as to how many replications one should use in Monte Carlo to detect a small sample bias of an asymptotic estimator: you KNOW there is a bias, but you may fail to find it with 1000 or 10000 Monte Carlo samples, so you conclude it's OK. – StasK Apr 28 '12 at 14:35
• I am thinking in terms of Pearson $\chi^2$ of $H_0: p_1 = \ldots p_6 = 1/6$. What significance level should be attached to this test may be open to discussion. So how the "distance" from fair is $z_{1-\alpha/2}\sqrt{1/6\cdot5/6\cdot1/n}$ where $n$ is the number of times you rolled the die. – StasK Apr 28 '12 at 15:55
• I am failing to see the point, indeed. Pearson $\chi^2$ is not a test of independence, and there is a multitude of ways to break it. In Pearson testing paradigm, the choice of the critical level reflects how much trust you put into the null (or how strong the evidence in favor of the alternative should be to reject the null). I am not inclined to go into philosophical discussions though. In Bayesian paradigm, you would have to construct a weird nonregular prior with point mass at the fair point and some absolutely continuous distribution elsewhere, and I just don't know how well it works. – StasK Apr 29 '12 at 3:46
• I'm not sure why you bring up independence; none of my criticisms deal with that. I'm saying that you're working with $P(Data | H_{0})$ when what matters epistemically is $P(H_{0} | Data)$. In the Pearson setting, you also have to assume a weird non-regular prior, and in fact it would be more buried in assumptions and less accessible than setting it up in a Bayesian paradigm, since you're implicitly embedding all of that into the term $P(Data | H_{0})$ by virtue of what that is equal to under Bayes' theorem. – ely Apr 29 '12 at 4:59
• I disagree @PeterFlom, look at Danial Johnson's answer above. It's about your own mind's state of ignorance about the die, not about anything empirical regarding the die. This is why $P(Fair | Data)$ and your prior beliefs about the die do matter, but the more frequentist approach of basing it on test statistics with $P(Data | Fair)$ doesn't matter. It's absolutely not about mere tolerances, because one can always concoct a totally unfair die that satisfies any computable test statistic to within whatever accuracy you want. You really must use the idea of priors and states of knowledge. – ely Apr 29 '12 at 15:01