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?


  • $\begingroup$ Exact calculation seems like it would take a while. You can get a pretty good approximation out of finding the probability that a particular value (let's say, $1$) appears at least $33$ times (probably using the normal approximation already for this purpose), and then multiplying by $6$, on the supposition that it's very unlikely for it to happen to two or three numbers simultaneously. $\endgroup$
    – Brian Tung
    Mar 20, 2017 at 20:46
  • $\begingroup$ Well, I think we can safely neglect the cases where two of them occur at least $33$ times so we can approximate and say that what you want is just $6$ times the chances that one chosen number occurs at least $33$ times. You can get that exactly from the Binomial distribution or estimate it from the normal approximation. $\endgroup$
    – lulu
    Mar 20, 2017 at 20:47
  • $\begingroup$ Note: The standard deviation is $\sigma = \sqrt {100\times \frac 16\times \frac 56}\approx 3.73$ so you are talking about a $4.5\sigma $ event.....extremely improbable. $\endgroup$
    – lulu
    Mar 20, 2017 at 20:51

1 Answer 1


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.


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