Concerning the Type I and Type II error comparison, a die roll of an unbiased die yields an outcome from a discrete uniform distribution. To test a candidate die for bias, one has to assume a type of bias distribution that is a discrete nonuniform distribution. Since there are several different ways of biasing a die, there is no unique distribution shape for a biased die. The method presented here is perfectly general in that it can be applied to any observable die bias (some bias types can be hidden), indeed, to any discrete problem type, and using binning, to some continuous distribution problems. Also, there are several allowed statistical tests for bias that could be compared for power for the biased die problem. None of these allow for the usual normal distribution assumption to be applied to dice. However, the method here is a data treatment for comparing a known distribution with an unknown one, such that $\beta$ cannot be assigned a priori.
A die has one of two official number patterns with opposite sides adding to 7; (1) the 1,2,3 vertex and the 4,5,6 vertex numbers increase clockwise, (2) numbers$-$same vertices$-$increase counterclockwise. This puts some constraints on the possible shapes for probability mass distribution for a biased die, but the remaining choices are hardly unique, i.e., there are many.
Cheating at craps in the US with loaded dice was common, usually by the casino, until the innovation of placing bets with the house as well as against it in 1907. However, even recently, many Las Vegas casinos' have been referred to their gaming commission for cheating with biased dice. In some jurisdictions, dice in current casino usage have security features including serial numbers.
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However, in some locations at least, a casino may resist attempts to take any die home, unless it has been invalidated by mutilation.
Although it varies depending on the game and type of bet, the house take averages approximately 5%. I cannot do a full cost analysis without knowing who typically sues whom for what, how much it costs to record a number, $n$, of outcomes, how many dice are tested between lawsuits, what the savings would be for a particular level of exclusion of bias, and what level of bias is worth detecting. Depending on the particulars, both type I and II errors could have associated costs to either party in a lawsuit. As one will see, below, the confidence interval width of insignificantly biased dice proceeds as $\sim\dfrac{1}{\sqrt{n}}$, meaning that our cost for increased confidence for bias detection escalates as $\sim n^2$, when the cost of testing is $\sim n$.
Here are some methods of loading a die from the stack exchange gaming site
1) "Shaved" dice, which are not quite symmetrical, but slightly wider or narrower on one axis than on others. A shaved d6 with, say, the 1–6 axis longer than the others will roll those sides less often, making it "less swingy" than a fair d6 should be (but leaving the average roll unchanged). The name comes from cheaters actually shaving or sanding down dice to flatten them, but cheap dice may have this kind of bias simply due to being poorly made. Other similar biases due to asymmetric shape are also possible, especially in dice with many sides.
2) Uneven (concave / convex) faces may be more or less likely to "stick" to the table, favoring or disfavoring the opposite side. The precise effect may depend on the table material, and on how the dice are rolled. Again, cheap plastic dice case easily have this kind of bias, e.g. due to the plastic shrinking unevenly as it cools after molding. Uneven edges can also create bias, particularly if the edge is asymmetric (i.e. sharper on one side).
3) Actual "loaded" dice, i.e. dice with a center of gravity offset from their geometric center, may occur accidentally due to either bubbles trapped inside the plastic or, more commonly, simply due to the embossed numbers on the sides of the die affecting the balance. In fact, almost all dice, with the exception of high-quality casino dice deliberately balanced to avoid this kind of bias, will likely have it to some small extent.
Others
4) Partially melting die in a microwave,
5) Variable loaded die having lead shot in a solid oil filled center that melts in hand heat.
If I were loading a die myself, I might include a "now it's biased", "now it's unbiased" feature (e.g., using magnets).
A quick test is to float a die in salt water, where the concentration of salt is high enough to just float it, and give it a few flicks. If I were doing this myself, I would x-ray any opaque die. In a casino one can drop a die into a glass of water to see which side lands where, and players and dealers spin a die between two opposite vertices held between their fingers to see which side falls down. Also, so called "dice balancing calipers" are used to test for bias, and are available from gambling supply houses. Nothing will test for all possible methods of 'loading' a die. Keeping in mind that there is no simple test for human ingenuity, the most direct method of bias detection is to roll the die to see how it rolls. Under Las Vegas casino conditions, one would need an actual craps (or dice) table to do this, as casino dice have sharp edges and are not very useful for board games. In summary, one would not actually use statistics for a physical die bias testing, but on line casinos are another problem, and for those, there is no alternative to statistical testing.
One might think that the mean is a sensitive test choice, but not really because 3.5 is not an actual result. One can do sign testing on this, but this will be a lower power test, and like other location tests for would not detect numerically balanced biases. Fisher's exact test would be more interesting and possibly more powerful. If I assume that ordinary t-testing can be used, I will be violating assumptions as a discrete uniform distribution is not the same as a normal distribution. If I persist in doing this, the p-values resulting will be non-normally distributed, e.g., a very approximately, a Laplace distribution (it may be a generalized normal distribution, do not know exactly).
(Image of $t$-test probabilities from 10,000 simulated rolls.) What exactly this is is not that interesting because a disadvantage of using the mean value is that it may not detect certain methods of biasing the results. For example, I could make a die so that it would be equally difficult to land on a one and a six, and the average would be unchanged (e.g., shaved die). So, we need something that tests for each of the six number "cells" simultaneously. Chi-square testing is just such a test.
Chi-squared distribution use for die roll testing is routine. This tests the outcome of each result against its expectation for each possible outcome with the least amount of fuss. And a common rule of thumb is to have at least five times as many rolls as there are sides on the die.
The derivation of Pearson Chi-squared from the central limit theorem appears here and press the [show] link. An example for d6 testing appears here. An analysis of a simulation in Mathematica of 10,000 rolls (die) with testing appears next.
nt = 10000; die =
RandomVariate[DiscreteUniformDistribution[{1, 6}], nt]; h =
Histogram[die, Automatic, ChartStyle -> {Opacity[0.3]}];
Print["The number of 1's through 6's for 10000 simulated die rolls is ",
bc = BinCounts[die, {1, 7}], " having Chi-squared p = ",
bcdie = BinCounts[die, {1, 7}]; s = Sum[(nt/6 - bc[[j]])^2/(nt/6), {j, 1, 6}];
p = 1 - N[CDF[ChiSquareDistribution[5],s]],
". The Bonferroni corrected Poisson distribution confidence interval is from ",
low = InverseCDF[PoissonDistribution[10000/6], 0.025/6], " to ",
high = InverseCDF[PoissonDistribution[10000/6], 1 - 0.025/6]]
lh = Plot[{low, high}, {x, 1, 7},
PlotRange -> {{1, 7}, {0, Automatic}}];
Show[lh, h]
The number of 1's through 6's for 10000 simulated die rolls is
{1679,1696,1619,1651,1689,1666} having Chi-squared p = 0.78885. The Bonferroni corrected Poisson distribution confidence interval is from 1560 to 1775.
The confidence intervals from 0.4167% to 98.583% for individual cells are equivalent to 95% CI's for every cell.
This amounts to a $+6.5$% and $-6.4$% spread using a Poisson counting model, suggesting that, a $12.9$% total range around 1/6th of the number of rolls would not detect a loaded die on all faces, and any die face showing a count beyond that would suggest significant bias.
Now the number ($n$) of rolls we need to perform to reduce our 95% confidence interval to a given fraction $y\sim\dfrac{12.8173}{n^{0.499151}}$ as per the following plot,
is a power function suspiciously decreasing proportional to approx. $1/\sqrt{n}$. Note that the Poisson confidence interval itself, in blue, ratchets for low count rates, not unexpectedly as Poisson counting is a whole number result.
Calculation via simulation of Type I error (false positive rate, FPR) and Type II error (false negative rate, FNR) requires 1) models for biased and unbiased dice 2) varying amounts of bias 3) selection of alpha 4) multiple random outcomes to build up a table.
The model used was to simulate extra weight placed by drilling holes under the dots of the six to make two bias models, one for which the six shows 4/10$^{\text{ths}}$ as frequently a one or six result (a $\frac{3}{2}$ odds ratio of seeing a one versus a six) and one for 1/3$^{\text{rd}}$ of the time (an odds ratio of twice). As the six is opposite the one, we should roll a one 6/10$^{\text{ths}}$ of the time, and 2/3$^{\text{rds}}$ of the time, that we see either number, respectively.
The biased die on the left panel is one trial of 1000 rolls of the die with a 1/3$^{\text{rd}}$ bias. The 95% and 99% confidence intervals are shown which represent the choices of $\alpha$ of 0.05 and 0.01 respectively. We then repeat this experiment 1000 times to accumulate histogram information, which is next displayed as probability histograms.
This shows the biased die results in orange and the unbiased die results in blue. The histogram categories are conveniently displayed at 0.05 and 0.01 probability widths and scaled for probability so that we can read power (1-FNR = 1 - type II error, in orange) and type I error (FPR, in blue) from the result in the first histogram category. Doing this repeatedly allows us to build up a table for 1000 trials of each level of bias for each type of error and each $\alpha$, shown next. This is done using the Chi-square distribution calculation, as it is more accurate than Poisson for this problem.
$$\left(
\begin{array}{ccccc}
\alpha & \text{Bias} & \text{n} & \text{FPR} & \text{FNR} \\
0.05 & \text{4/10} & 200 & 0.054 & 0.797 \\
0.05 & \text{4/10} & 1000 & 0.043 & 0.199 \\
0.05 & \text{4/10} & 4000 & 0.052 & 0. \\
0.01 & \text{4/10} & 200 & 0.008 & 0.918 \\
0.01 & \text{4/10} & 1000 & 0.008 & 0.375 \\
0.01 & \text{4/10} & 4000 & 0.019 & 0. \\
0.05 & \text{1/3} & 200 & 0.039 & 0.486 \\
0.05 & \text{1/3} & 1000 & 0.044 & 0.001 \\
0.05 & \text{1/3} & 4000 & 0.045 & 0. \\
0.01 & \text{1/3} & 200 & 0.011 & 0.728 \\
0.01 & \text{1/3} & 1000 & 0.017 & 0.004 \\
0.01 & \text{1/3} & 4000 & 0.012 & 0. \\
\end{array}
\right)$$
Notice that the recovered type I errors are on the same order as those specified a priori, and are fairly accurate. Most notably, there is a variable not specified in the question. That is, the more subtle the bias, the more difficult it is to detect, i.e., the power then decreases.
Specifying a given ratio of type II and $\alpha$, as is requested in the question is ambiguous as it does not relate to accuracy of testing. What is really desired is high accuracy of testing, i.e., $\text{Accuracy} =1-(\text{FPR}+\text{FNR})$, which is achieved by minimizing the sum $(\text{FPR}+\text{FNR})$.
The above method is general, and applies to electronic results with a single die outcome of bias, like an all-too-frequent one, something that is not possible with a physical die which we could not bias for one's without reducing the frequency of outcomes of six's.
My advice is, if you want to gamble, buy shares.