# Probability of a fair die appearing to be biased

I'm interested in the probability of a die appearing to be biased when it is, in fact, fair. I'm trying to derive a result given, without proof, on YouTube: http://youtu.be/6guXMfg88Z8?t=1m29s

The basic idea is this: you suspect a fair 20 sided die to be biased. The video claims that, if you roll your dice 100 times, there is a 1-in-50 chance of you getting an excess of threes by pure chance.

I've tried to count the cases, but I'm getting into trouble. Let's say there are $k$ threes, then there are

$$\frac{100!}{k! \ (100-k)!}$$

ways of distributing the threes amongst the 100 throws. The next part is where I'm getting stuck. I need to count the number of ways of distributing the other 19 numbers amongst the remaining $100-k$ throws. I suspect that this might be related to the number of partitions of $100-k$.

-
The probability of exactly $k$ threes if the die is fair is $\frac{100!}{k!(100-k)!}(1/20)^k(19/20)^{100-k}$. –  André Nicolas Dec 19 '12 at 22:48

The probability of getting a three in any one roll is $1/20$. Hence the number of $3$'s in $n=100$ rolls follows a $\operatorname{Bin}(100,1/20)$ binomial distribution, that is $$P(k\;3's) = \binom{100}{k}(1/20)^k(19/20)^{100-k}.$$ You would expect to see, on average, $100/20=5$ 3's in 100 rolls, and so the probability of getting an excess of 3's would be $$\sum_{k=6}^{100}{P(k\;3's)}=38.4\%.$$
If the question just asks for the probability that $3$ is the number you see most often in your 100 rolls, then the answer is obviously $1/20$, by symmetry. This is because every number 1,...20 is equally likely to be the one which is rolled most often.
Maybe, they don't really say what they mean (I only glimpsed at it). Remembering that the standard deviation of the number of 3's is $\sqrt{100\times(1/20)\times(19/20)}=3.1$, one might consider an excess of $3's$ to be any number larger than $8$. This would give a probability of $6.3\%$. For $10$ or more 3's one would get $2.8\%$. –  Eckhard Dec 19 '12 at 23:04