I'm trying to verify that my dice are reasonably fair and balanced. I'm not sure I'm doing it right, though, since I'm basically teaching myself stats without references. To achieve this, I've settled on the chi-squared test. Here's what I'm doing:

I have a sample size $N$ of rolls. On a dC, where $C$ is the number of sides, the expected frequency for each roll 1 thru $C$ is $\frac{N}{C}$. Then I compute the $\chi^{2}$ value for the distribution from $1$ to $C$. Then, I compute the chi squared distribution (in Excel using CHIDIST) with degrees of freedom equal to $C-1$.

Here's an example case:

  • Take a d4, and assume each face was rolled equally 1500 times.
  • Compute the chi-squared value for this, with each face and an expected frequency of 1500
  • Compute the chi squared distribution using chidist and 3 as the degrees of freedom

The problem is, this is returning nonsensical values. For instance, on a d4, if I input 6000 evenly distributed rolls, chidist returns .801252. No matter how high I make $N$, even if there's an even distribution, I'm not getting below .801252. I know I'm doing something wrong here, but I'm not sure what it is.

The chi squared distribution value should be falling to within any sort of significance. Any help?

Edit: Per request, I'm posting a screenshot of my Excel layout and listing the formulas used. I've hidden the irrelevant rows.

enter image description here

Formulas used:

  • C2:C5 = B22/4 (Frequency)
  • B22 = SUM(B2:B5) (Sum)
  • (StDev & StdErr are not relevant)
  • B25 = CHITEST(B2:B5, B91:B94) (Chi squared)
  • B26 = CHIDIST(B25, B1-1) (Chi distribution)

Here's some raw data:

   4  -  6
1| 56    36
2| 45    33
3| 57    33
4| 42    40
5|       29
6|       29
  • 1
    $\begingroup$ If you entered 1500 each for 4 sides of a 4-sided die, why would $p<.05$? $\endgroup$ May 8, 2013 at 3:29
  • $\begingroup$ @gung Well, to be fair, $.05$ was a pretty arbitrary choice. I've edited it - maybe I'm not interpreting the chi distribution properly, though? But it seems to me like it should tend towards 0 for a more uniform distribution. $\endgroup$
    – user25363
    May 8, 2013 at 3:31
  • 1
    $\begingroup$ The p-values you're getting are weird, I think, but in the opposite direction. If you throw a fair 4-sided die 6k times, the expected counts are 1500 each. If you were to enter those numbers into a chi-squared goodness of fit test, the p-value should be exactly $1.0$. $\endgroup$ May 8, 2013 at 4:01
  • $\begingroup$ It may help you to read this thread: chi-squared-always-a-one-sided-test. $\endgroup$ May 8, 2013 at 4:02
  • $\begingroup$ @gung Hmm... I'm not sure then. It's only 1.0 when there's a serious bias in the data. $\endgroup$
    – user25363
    May 8, 2013 at 4:20

1 Answer 1


A chi-square statistic gets bigger the further from expected the entries are. The p-value gets smaller. Very small p-values are saying "If the null hypothesis of equal probabilities were true, something really unlikely just happened" (the usual conclusion is then usually that something less remarkable happened under the alternative that they're not equally probable).

The correct chisquare value for cell counts of 1500 1500 1500 1500 is 0 and the correct p-value is 1:


    Chi-squared test for given probabilities

data:  c(1500, 1500, 1500, 1500)
X-squared = 0, df = 3, p-value = 1

Your formula for the chi square statistic is wrong. One of the images you posted had counts of 1500 1500 0 1500. In that case, the chi-squared value is 1500, and the p-value is effectively 0:


    Chi-squared test for given probabilities

data:  c(1500, 1500, 0, 1500)
X-squared = 1500, df = 3, p-value < 2.2e-16

So you calculated a chi-square of 1 when you should have got 0 and calculated 0 when you should have got 1500.

What formula are you using?

(An additional check - on these four more typical counts

   1    2    3    4 
1481 1542 1450 1527 

you should get a chi-square of 3.5693 )


On the use of chi-square tests to test dice for fairness see this question

I gave an answer there which points out that - if you really want to test dice - you might consider other tests.


As you noted in your response, the CHITEST function in Excel returns the p-value rather than the test statistic (which seems kind of odd to me, given you can get it with CHIDIST).

In the case where all the expected values are the same, a quick way to get the chi-squared value itself is to use =SUMXMY2(obs.range,exp.range)/exp.range.1, where obs.range is the range for the observed values and exp.range is the range of the corresponding expected values, and where exp.range.1 is the first (or indeed any other) value in exp.range, giving something like this:

    1       2       3       4
Exp 1500    1500    1500    1500
Obs 1481    1542    1450    1527

    chi-sq. p-value     
    3.5693  0.31188     

A mildly clunky but even easier alternative is to use CHIINV(p.value,3), to obtain the chi-square statistic, where p.value is the range of the value returned by CHITEST.


Edit: Here's a couple of plots for the two sets of data you posted.

The solid grey lines are the expected numbers of times each face comes up on a fair die.

The inner pair of dashed lines are approximate limits between which the individual face results should lie 95% of the time, if the dice were fair. (So if you rolled a d20, you'd expect the count on one face - on average - to be outside the inner limits.)

The outer pair of dotted lines are approximate Bonferroni limits within which the results on all faces would lie about 95% of the time, if the dice were fair. If any count were to lie outside those bounds, you'd have some ground to suspect the die could be biased.

plot of d4 results, with bounds plot of d6 results, with bounds

Edit to add some explanation --

The basic plot is just a plot of the counts. For a die with $k$ sides, the count of the number of times a given face comes up (an individual count) is effectively distributed as a binomial. If the null hypothesis (of a fair die) is true, it's binomial$(N,1/k)$.

The binomial$(N,p)$ has mean $\mu=Np$ and standard deviation $\sigma=\sqrt{Np(1-p)}$.

So the central grey line is marked at $\mu=N/k$. This much you already know how to do.

If $N$ is large and $p = 1/k$ is not too close to 0 or 1, then the standardized $i^\text{th}$ count ($\frac{(X_i-\mu)}{\sigma}$) will be well approximated by a standard normal distribution. About 95% of a normal distribution lies within 2 standard deviations of the mean. So I drew the inner dashed lines at $\mu \pm 2\sigma$. The counts are not independent of each other (since they add to the total number of rolls they must be negatively correlated), but individually they have approximately those limits at the 95% level.

See here, but I used $z=2$ instead of $z=1.96$ - this is of little consequence because I didn't care if it was a 95.4% interval instead of a 95%, and it's only approximate anyway.

So that gives the dashed lines.

Correspondingly, if the toss counts were all independent (which they aren't quite, as I mentioned), the probability that they all lie in some bound is the $k^\text{th}$ power of the probability that one does. To a rough approximation, for independent trials if you want the overall rate (across all faces) of falling outside the bounds to be about $5\%$, the individual probabilities should be about $0.05/k$ (there are several stages of approximation here, not even counting the normal approximation; to work really accurately you need lots of faces and small probabilities of falling outside the limits).

So with $k$ faces and an overall $5\%$ rate of being outside the limits, I want the probability of the individual ones being above each limit to be roughly half that or $0.025/k$. With 4 faces that's a proportion of $0.025/4 = 0.00625$ above the upper limit and $0.00625$ below the lower one.

So I drew the outer limits for the four-sided die (d4) case at $\mu \pm c\sigma$, where $c$ cuts off $0.00625$ of the standard normal distribution in the upper tail. That's about $2.5$ (again, it doesn't pay to be overly accurate about the limit, since we're approximating this thing all over).

The d6 works similarly, but the limit for it cuts off an upper tail area of $0.025/6$, which roughly corresponds to $c=2.64$ at the normal distribution.

For the d4, if the null hypothesis were true, the actual proportion of cases with any of the counts falling outside the outer bounds of 2.5 standard deviations from the mean would actually be about 4.2% (I got this by simulation) - easily close enough for my purposes since I don't especially need it to be 5%. The results for the d6 will be closer to 5% (turns out to be around 4.5%). So all those levels of approximation worked well enough (ignoring the existence of negative dependence between counts, the Bonferroni approximation to the individual tail probability, and the normal approximation to the binomial, and then rounding those cutoffs to some convenient value).

These graphs are not intended to replace the chi-squared test (though they could), but more to give a visual assessment, and to help identify the major contributions to the size of the chi-square result.

  • $\begingroup$ Interesting. The chi squared test I'm using (see edited above) returns 0.31188? $\endgroup$
    – user25363
    May 8, 2013 at 20:40
  • $\begingroup$ Actually, this lead me directly to the solution. I'm going to post an answer, but thank you so much for your help! $\endgroup$
    – user25363
    May 8, 2013 at 20:44
  • $\begingroup$ Hi Telthien - as you no doubt already figured out, 0.31188 was the p-value corresponding to the test statistic of 3.5693. If you have some actual die-roll results (unless it's more than about ten I guess) you're looking at, would you be able to post the counts in your question? Real data like that can be very useful, but it takes a long time to produce. $\endgroup$
    – Glen_b
    May 8, 2013 at 22:56
  • $\begingroup$ By 'more than about ten' I meant "results on more than ten dice"; sorry if that was confusing. $\endgroup$
    – Glen_b
    May 9, 2013 at 0:32
  • $\begingroup$ Alright! I've posted some of my raw data for a D4 and a D6. Let me know if there's any other way I can help! $\endgroup$
    – user25363
    May 9, 2013 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.