In this specific case, P(redOnKthSample) = (numberOfRedSamples + 1) / (totalNumberOfSamples + 2) = (1+1)/(1+2) = 66% red.
On the second draw, if the ball is green, then you get P(redOnThirdSample) = (1 + 1) / ( 2 + 2) = 50%
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.