[hardlianotion]

Expected index given by

E [X] = \sum_{i \in [2,100)} p(X < i)\prod_{j < i}p (X \ge j)i.

Edit: Sorry rest of reply was wrong. Had to account for not hitting until i^th loop.

[aalhour]

Not sure that adds up, I translated the right side of the equality into a Python statement and it returns a number I am not sure how to interpret:

  1/99 * sum([(i-1)*i for i in range(2, 100)])   # 3266.666666666667

Both the median and mean are around 12 for 10K runs of the loop.

[hardlianotion]

I edited my response with an apology! Thanks for checking!

[aalhour]

No worries, I think your solution coverges with @Someone's above, if I am not mistaken. You're multiplying the probabilities for all values of j up to i for all values of i.

[hardlianotion]

Yes I am and yes I think it does:

edit - actually, the calculations from the other guy are probability calculations, so this is related to the median. I have given you a mean calculation, which should be related to the average. But you already mentioned that they are close ...