What is the average number of times it would it take to roll a fair 6-sided die and get all numbers on the die? The order in which the numbers appear does not matter.

I had this questions explained to me by a professor (not math professor), but it was not clear in the explanation. We were given the answer (1−(56)n)6=.5 or n=12.152

Can someone please explain this to me, possibly with a link to a general topic?

**Answer**

The time until the first result appears is 1. After that, the random time until a second (different) result appears is geometrically distributed with parameter of success 5/6, hence with mean 6/5 (recall that the mean of a geometrically distributed random variable is the inverse of its parameter). After that, the random time until a third (different) result appears is geometrically distributed with parameter of success 4/6, hence with mean 6/4. And so on, until the random time of appearance of the last and sixth result, which is geometrically distributed with parameter of success 1/6, hence with mean 6/1. This shows that the mean total time to get all six results is

6∑k=16k=14710=14.7.

*Edit:* This is called the coupon collector problem. For a fair n-sided die, the expected number of attempts needed to get all n values is

nn∑k=11k, which, for large n, is approximately nlogn. This stands in contrast with the mean time needed to complete only some proportion cn of the whole collection, for some fixed c in (0,1), which, for large n, is approximately −log(1−c)n. One sees that most of the nlogn steps needed to complete the full collection are actually spent completing the last one per cent, or the last one per thousand, or the last whatever percentage of the collection.

**Attribution***Source : Link , Question Author : eternalmatt , Answer Author : Did*