[bonanova]

On average, how many times must I roll a fair six-sided

die before all the numbers 1, 2, 3, 4, 5, 6 show up?

[Prime]

It could be considered this way...

To determine simple averages, (forget the die for a moment) you add up all possible outcomes and divide by the number of outcomes. In theory, you could roll a die any number of times without getting all six numbers. You could roll forever and only get 3's! With infinity (as well as every integer) in the averaging, on average you would never roll all six. It really depends on your definition of "average" in the question.

Infinite series may add up to a finite number (converge). See my topic Dice Gamblers' course for discussion and relevant math.

[Guest]

That formula is incorrect.

Formula for the number of times at least one number is not rolled is more complex and yields smaller number of combinations.

Oh, right. I'm counting some combinations up to six times...

[Prime]

I used a different approach, and thought I would get the same answer as Chuck. I did not....

I calculated the number of times rolling a k sided die n times does, and does not, yield all k values. I then solved for the value of n that yields equal probability of either outcome.

...

And there may be another flaw with that approach. The 50% probability for an event to occur and the average number of trials for an event to occur are not the same thing. For example, it takes on average 6 rolls to roll number "6". However, "6" comes up with a 50% probability in less than 4 rolls. See my preceding discussion with Bonanova. (His winning an actual gambling match notwithstanding.)