Is my six sided dice weighted?

Question

I was playing a game with my son, and it seemed that we rolled two or five about half the time. He was wondering if the die was weighted.

So I re-rolled it a bunch of times and jotted down the results:

1: 4 times
2: 12 times
3: 13 times
4: 4 times
5: 13 times
6: 8 times

I realise that this is a small sample (only 54 datapoints), but to my very untrained eye, the 1 and 4 seem very low. I was wondering how I would go about determining the probability that the die is fair. I was expecting about 9 each. I did a ChiTest in excel, which I am not sure how to calculate manually and got 0.069

Does this mean that there's a 6.9% chance that the die is unweighted, and therefore a 93.1% chance it's weighted, or have I misinterpreted the results entirely?

Answer 1

This is not the right interpretation of the $0.069$ number. Indeed, this wrong interpretation is extremely common, and people from doctors to judges to scientists to policy makers routinely get this wrong (and go on to make poor decisions as a result).

The $0.069$ number means this: if your die is fair, then the probability of getting a result at least as extreme as the one you observed is $0.069$. In other words, if you do sets of 54 trials over and over again, you'll see distribution as extreme or more in 6.9% of the trials.

You can use this informtaion however you want to make deductions about whether the die is fair or unfair. However, it makes no sense to talk about "the probability that the die is fair" unless we have an underlying space of all possible events as reference (all dice in the world, for example, or all ways of manufacturing dice). The result from a chi-square test simply does not provide that type of information.

Answer 2

It would be better at least 5 obs but also 4 is fine.

$$\chi_{(5)} ^2=\frac{(4-9)^2}{9}+\frac{(12-9)^2}{9}+\frac{(13-9)^2}{9}+\frac{(4-9)^2}{9}+\frac{(13-9)^2}{9}+\frac{(8-9)^2}{9}=10.22 $$

The critical $\chi_{(5)}^2$ at 5% is 11. Your dice is fair. To calculate the exact p-value you must use a calculator.

You interpreted well your results

Answer 3

I would not consider your die as weighted, we can see this as a binomial distribution problem $$P(X=r)={{n}\choose{r}}P^{r}(1-P)^{n-r}$$

This basically models your case, $r$ is the number of succeses, $n$ would be the number of trials and $P$ the probability of succes. Replacing and computing for rolling number 1 four times in 54 trials

$$P(X=4)\approx0.0268$$

Which would be 2.68%, since it is not incredibly low, considering all the posibilities (one 1, two 1s,..., 54 ones) I would guess it is not. But you could always apply the large number rule, meaning the more tries, the more the real possibility asimilates to the theorital possibility.