# fair die or not, from 3D printer

I made a 3D printed die today, but depending on the heat applied, it may or may not be a "fair" die (i.e. have an equal chance of landing on each face).

I have just tried rolling it 150 times. The frequency results came out to:

$$\begin{array}{c|c} \hline 1's & 21 \\ \hline 2's & 30 \\ \hline 3's & 23 \\ \hline 4's & 31 \\ \hline 5's & 21 \\ \hline 6's & 24 \\ \hline \end{array}$$

How would I calculate the chance that this die is fair?

• I'm not familiar with the term "Laplacian die." Do you perhaps mean a "fair die" (one in which each side is equally likely to appear when tossed)? Otherwise, could you please define your terms? – JMoravitz Dec 16 '15 at 21:08
• What is a Laplacian die? Is it a fair die? If so, you can't, if you're a Bayesian; you need a prior. Otherwise, you try to refute the null hypothesis that the die is fair, and evaluate the probability of obtaining the results you did with something like the chi-squared test. – Brian Tung Dec 16 '15 at 21:09
• The $\chi^2$-value here is $3.92$, which is absolutely non-significiant. There is no reason to doubt the fairness of the die. – Peter Dec 16 '15 at 21:41
• @Peter But there is little reason to believe in fairness of the die either... (given that it's the first ever dice produced by this printer and fairness of a die is very sensitive to small variations in shape). – A.S. Dec 18 '15 at 6:39
• @cel I don't see how you can consider "probability of getting a 6" as an inherent property of a die since the outcome of the throw depends not only on the die itself but on mechanics of the throw - which automatically leads to a limit in how well you can differentiate different dice. Basically, no number of throws will ever distinguish $1/6$ from $1/6+10^{-100}$ due to fluctuations in the throw itself and unclear relationship between die bias, throw distribution and outcome. – A.S. Dec 18 '15 at 7:39

Goodness-of-fit test. Computation in R. Results agree with Comment by @Peter.

 obs = c(21, 30, 23, 31, 21, 24)
chisq.test(obs)

Chi-squared test for given probabilities

data:  obs
X-squared = 3.92, df = 5, p-value = 0.561


If observed counts are $X_i$ and expected counts are $E = 150/6 = 25,$ Then the chi-squared goodness-of-fit statistic is $Q = \sum_{i=1}^6 (X_i - E)^2/E,$ which is approximately distributed as $Chisq(DF = 5).$ The critical value for a test at level 5% is 11.07. We fail to reject the null hypothesis that all six faces are equally likely because $Q = 3.92 < 11.07.$

Power. However, I'm wondering if 150 rolls is enough. Suppose your die is markedly biased so that faces 1, 2, and 3 each have probability 5/36 and faces 4, 5, and 6 each have probability 7/36. Then the following simulation shows that only about 27 in 100 tests with 150 rolls would reject the hypothesis of fairness. That is, the power of the goodness-of-fit test against this particular degree of bias is about 27%. More modestly biased dice would fail the test at an even lower rate.

 m = 10^5;  q = numeric(m); E = 150/6
for(i in 1:m) {
faces = sample(1:6, 150, repl=T, prob=c(5,5,5,7,7,7)/36)
x = table(faces);  q[i]=sum((x-E)^2/E)}
mean(q > qchisq(.96, 5))
## 0.2702


The histogram shows values of $Q$ for 100,000 tests, each using 150 rolls of such a biased die. The vertical line is the critical value for a test at level 5%. The curve is the density of $Chisq(5)$.

Prompted by a comment, I ran a slight modification of the R code that shows 89% power for the same biased die as above, but using 600 rolls for the test. The corresponding graph is shown below.

Note: I have posted a Bayesian analysis of the data for face 1 separately.

• How many rolls would i roughly need to to have a 90% chance that the test is right an a die that is biased only 5% so 5.6 or 6.4. Also what would happen if there are only 2 sides that are different e.g. 1 and 6 with 5/36 and 7/36 respectively and the rest is 1/6? – Eumel Dec 18 '15 at 8:48
• It would require several more simulations to answer such questions. The R code is easy to modify, but each run takes a couple of minutes on my ancient computer. For a 150 rolls, the power against the alternative distribution (1/9, 1/9, 1/9, 2/9, 2/9, 2/9) is about 90%. But that is an extremely biased die. You are right to suppose that power depends on degree of biasedness and number of rolls. – BruceET Dec 18 '15 at 17:03
• @Eumel As a ball-park, consider that to distinguish $B(n,p)$ and $B(n,p+\epsilon)$ you need $n\gg p(1-p)\epsilon^{-2}$, so you need at least $750$ tests. – A.S. Dec 18 '15 at 17:39