I've been trying to work through this in my head and can't figure out what I'm missing. Here's the problem:

A 20 sided die is rolled 100 times and we count the number of 20's we roll. Well, it turns we rolled zero 20's. That's odd! But how odd is it? My first instinct is to say the probability of not getting a single 20 in 100 rolls is $$\frac{19}{20}^{100} = 0.0059$$

That makes sense to me, but I wanted to reframe it in the context of the binomial distribution with n=100 and p=1/20. I set up my z test as: $$Z = \frac{X-np}{\sqrt{np(1-p)}} = \frac{0-100(\frac{1}{20})}{\sqrt{100(\frac{1}{20})(\frac{19}{20})}} = -2.294$$ which equates to a p-value of ~0.011 or 1.1%. That's a big difference between the two methods and I can't spot a mistake with either. Where'd I go wrong?

  • 1
    $\begingroup$ The $p$-value is Til probability, is based on an asymptotic approximation to the distribution of $X$, and is particularly approximate in the tail $X=0$. $\endgroup$
    – Xi'an
    Sep 13, 2022 at 6:21
  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$
    – Galen
    Sep 13, 2022 at 6:21
  • $\begingroup$ It's not so much the binomial distribution and common sense not matching up, but the binomial distribution and the normal approximation not doing so. $\endgroup$ Sep 13, 2022 at 7:23
  • $\begingroup$ An alternative approximation would be to use a Poisson distribution to suggest $e^{-100/20}\approx 0.0067$, which is closer even if not good $\endgroup$
    – Henry
    Sep 13, 2022 at 8:28

1 Answer 1


The problem is that what you are referring to is not binomial distribution, but normal approximation of it. As you can learn from the Wikipedia article on binomial, the normal approximation does not always work well and it works mostly for approximating the area of the distribution close to the mean, that's not your case.

Binomial distribution has a probability mass function

$$ {n\choose k} p^k (1-p)^{n-k} = 1 \times 1 \times \frac{19}{20}^{100} $$


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