Ok, so I think I have answered my own question.

Now, the initial problem I was trying to solve, is what is the average number of draws you have to make to get marble A out of a bag of 100 (with replacement if you don't get A)?

(I simplified the problem above to 2 marbles to establish a method - a good tip for problem-solving!)

Pr(1 draw) = \(\displaystyle \frac{1}{100}\)

Pr(2 draws) = \(\displaystyle \frac{99}{100} * \frac{1}{100}\)

Pr(3 draws) = \(\displaystyle (\frac{99}{100})^2 * \frac{1}{100}\)

Pr(n draws) = \(\displaystyle (\frac{99}{100})^{n-1} * \frac{1}{100}\)

So average number of draws = \(\displaystyle \sum\limits_{n=1}^{infty}n*(0.99)^{n-1}*0.01\)

Wolfram Alpha says:

View attachment 39714
So, on average, you'd have to take 100 draws before you get marble A.

And if there are n marbles in the bag, then the average number of draws would be n.

Intuitively, I thought it would be less. I can visualise the distributions and sort of see why. Is this really a simple concept that I am overthinking?? Any thoughts would be appreciated.