Since it kept me up, I'll update..
Let's simplify and look at three balls, whose possible permutations are
1. rrr 2. rrg 3. rgr 4. rgg 5. grr 6. grg 7. ggr 8. ggg
There are two approaches that are intuitive and wrong, and then the right approach. Specifically it depends which question you're answering.
1. What are the odds when picking two random balls over and over from the same permutation.
- You might assume the odds of each ball are completely independent, making the odds of a first red n/100, and then a second red slightly lower at n-1/99. This is what happens when you pick two random balls over and over from the same permutation.
- You might assume that if you've selected a red ball first you must be in permutations 1-4, making the odds of a second red ball 50% (permutations 1-2). This is what happens when choosing the first (and second) ball in order from a random permutation.
- This is what we're actually being asked to do.
- In this case the odds of picking red first are proportional to the number of reds in the permutation. Specifically, picking red is guaranteed when in permutation 1, and 67% likely when in permutations 2,3, and 5, but only 33% likely when in scenarios 4, 6, and 7, and impossible in scenario 8. While it is equally likely to be in permutations 4 or 5, unlike scenario 2, you're more likely to choose a red ball when in 5 than 4, and in general, you're more likely to pick a red ball first when in a permutation that has more red balls to choose from. And because it does, the second draw is more likely to be red as well.
- Like in the Monty Hall problem, the color of the first ball is, while random, unintuitively not arbitrary, but rather an indication that there are more balls of that color to choose from in the urn, which gives you an advantage in knowing what the second draw will be.