What is the likelihood of any side of a die appearing more than 200 times after 1000 dice rolls?

Question

What is the likelihood of any side of a die appearing more than 200 times after 1000 dice rolls? It seems like I need to figure out the standard deviation for this but not sure how to do this. I know the expected amount of times a number pops up is 1000/6, but how do I calculate the standard deviation and variance?

Answer 1

For a cumulative outcome, I use TI-83's binomcdf. For more than 200 repeats, the input will be $$1 - \text{binomcdf } (1000,\frac{1}{6},200)$$ This is one minus the cumulative probability of getting 0 to 200 repeats.

Answer 2

Consider the complementary problem, i.e. the probability that each number from $1$ to $6$ appears $200$ times or less.

The result of rolling a die $1000$ times is a sequence of $1000$ integers, each in the range $1$ to $6$. There are $6^{1000}$ such sequences, all of which we assume are equally likely. We would like to count the sequences in which each number appears $200$ times or less. To do this, we consider the more general problem in which the die is rolled $r$ times, and let $a_r$ be the number of sequences in which each number appears $200$ times or less. Let $f(x)$ be the exponential generating function of $a_r$, i.e. we define $$f(x) = \sum_{r=0}^{\infty} \frac{a_r}{r!} x^r$$ Since there are $6$ possible numbers and each appears $200$ times or less, $$f(x) = \left( 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots +\frac{1}{200!}x^{200} \right)^6$$ The probability that each number from $1$ to $6$ is rolled $200$ or fewer times in $1000$ rolls is then $$p = \frac{a_{1000}}{6^{1000}} = \frac{1000!\; [x^{1000}] f(x)}{6^{1000}}$$ where $[x^{1000}] f(x)$ denotes the coefficient of $x^{1000}$ when $f(x)$ is expanded.

I do not know a feasible paper-and-pencil method of finding this coefficient, but it can be computed by a computer algebra system such as Mathematica. The result is $[x^{1000}] f(x) = 3.46799 \times 10^{-1790}$, so $p = 0.985082$.

The answer to the original problem, the probability that at least one number appears more than $200$ times, is $1-p=0.0149184$.