This answer should go to the top.

Except it doesn’t give much intuition for why?

I find it hard to remember a ton of different rules, so I just count configurations, and I'm right on these problems basically every time. You count configurations by enumerating all possible combinations of unknowns, then "score" them by likelihood. The configuration's probability is its score divided by the total score. Then take the weighted sum of your test value, with the weights being the configuration probabilities.

There are 101 configurations (the different values of N). We know our first ball was red. So N == 0 has a score of 0 (it's impossible since we found a red), N == 6 has a score of 6 (there are six different ways we could draw a red), N == 100 has a score of 100. The sum of all the scores is 5050, so the probability that N == 100 is 100/5050 = 1.98%.

The value we're interested in is p(2nd ball == red). In the N == 100 configuration, that's 100%: it contributes 100% * 1.98% to the final "score". In the N == 6 configuration, that's 5/99 (we already drew one red), so it contributes 5/99 * 6/5050 to the final score. Add up the final scores and you get 66.67%.

As long as you can feasibly enumerate all possible configurations and score them accurately, this approach basically never fails.