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