_Average_ number of rolls to roll a '1'?
College probability (not that I ever learned anything) and calculus (in which I did learn something) is decades in the past. If I have an n-sided die, n > 1, what is the average number of rolls it will take to roll my first '1'? (I typically would be using n = 100.)
It's 1/6 to succeed on the first try, 5/6 * 1/6 of succeeding on the second try, 5/6 * 5/6 * 1/6 of succeeding on the third try, etc. It looks like it might be a limit calculation, but 'm not sure how to write a formula for that—and not sure I would be able to do that calculation any more. I'm not sure the magnitude of that formula doesn't indicate that it will go to infinity, which would mean, in my mind, there is no average.
A related, but secondary, question would seek to calculate the average number of rolls it would take to roll a number in the range of 1 to x, x < n.
Posted Apr 6, 2022, 3:58 PM UTC
The average is n. Like you say, you can compute this as a series. But here is a less computational argument: suppose you roll this die many times in a row, like 6000 times for your 6-sided die. You will get on average 1000 sixes, yes? By thinking of your 6000 rolls as a large number of trials of rolling until you get a 1, that means you made on average 1000 trials. The sum of the lengths of those trials is of course 6000, so the average length is 6000/1000=6.
By a similar argument, it should be n/x.
Very clever! Thanks