Probability that any outcome of a dice roll happens more than X times out of Y trials

I'm trying to determine the probability that a person experiences a "lucky number" when rolling a single, fair, 6-sided dice over a set of rolls in a single trial. A "lucky number" in this case is any face of the die that occurs visibly more common than one would normally expect. If you roll a six-sided die 100 times, you expect the outcome to occur with ~16.6 results of 1, 2, 3, 4, 5, and 6 ea, on average.

For example, you roll a six-sided dice in 100 independent trials, what is the probability that the occurrence of rolling any side of the dice happens at least 33 times over the course of the 100 independent trials? It doesn't matter if the roll was 1, 2, 3, 4, 5, or 6, just that the same result happened at least 33 times out of the 100 trials.

How would I calculate this?

Thanks.

Answer

The chance that $1$ comes up exactly $33$ times in $100$ comes from the binomial distribution. The chance of success is $\frac 16$ and failure is $\frac 56$ so it is ${100 \choose 33}(\frac 16)^{33}(\frac 56)^{67}\approx 0.00003$ If we sum from $33$ to $100$ we get the chance of at least $33\ 1$s, which is about $0.0005$ per Alpha. You can multiply these by $6$ to get the chance for any number, as it is very unlikely we doublecount by having at least $33$ of two different numbers. So the chance of a "lucky number" happening by chance is about $0.0003$ or one in $3300$. Pretty unlikely, but rarer things happen all the time.