Is that true? You'd still have evidence that the distribution of balls tilts one way, wouldn't you?
If you change their simulation formula to pull from a binomial instead of uniform... it looks 50/50.
Since the coins are independently flipped, you can assume that it’s still just a binomial distribution of size n-1 and the same p.
Let’s ignore RRG because it becomes a 50-50 anyway.
RRR has a 100% chance of yielding Red. But is 3X less likely than RGG.
RGG has a 33% chance of yielding Red but was 3X more likely.
Edit: so still 50/50 Given a draw of red, you have exactly the same probability of having been in RRR and RGG. So you were equally likely to have 100% chance or a 0% chance.
Someone picks a coin without you knowing which one, then starts flipping.
The first toss shows heads.
Does that mean subsequent flips from the same coin will be more likely to be heads?
Yes.
Answer is only no if all coins are the same with no bias, which they are not.
You now have 101 coins, with biases ranging from 100 percent heads (or urn with only red balls) to 100 percent tails (or urn with only green).
Someone chooses one of the 101 coins randomly (each coin has equal probability of being chosen) and starts flipping it.
First flip shows heads, what is the probability the next flip is also heads?
Answer is 2/3.
The point is that knowing what was the first flip gives information about what biases a coin may have.
The original problem has a minor technical twist that balls are drawn without replacement, but it’s an irrelevant detail for N>10 total balls.
You’ve replaced it with a binomial of a uniformly selected p, which ends up being, amusingly, the same as a uniform.
It’s mostly irrelevant what kind of distribution this bag of coins comes from (except for some degenerate cases). What matters is that selecting coins from this bag gives coins of varying biases.