Bear in mind the default behaviour if there were just two balls and you never added any more. Then the proportion of red picks vs blue picks would tend to 1/2. So there's naturally a tendency for the proportion to concentrate in the middle.
As you say, the way in which new balls are added tends to push the proportion towards the extremes.
The uniform distribution is the result of these two tendencies exactly cancelling out.
This is not correct as stated. It does not “tend towards extremes” as you might expect intuitively. That would be the case if half the time it approached 100% blue and half the time it approached 100% red, which is precisely not what happens.
For me as well. And when my stochastic probability professor posed this question to the class by way of hands, it was nearly unanimous in favor of the 0/100% end behavior.
But the problem is symmetric, and even if pick a red, you end up with two reds and one blue, so not so much imbalanced.
And even if the mix becomes really imbalanced, say 7 red and one blue, picking the rarest color will have more effect than picking the most common one. So you could consider that the system tries to balance itself naturally, hence avoiding huge swings in some direction or the other.
Usually 'this sort of thing' would go to the extremes: colours that are already prevalent have a bigger chance of getting more added to them.
It's interesting that they don't.
Normal distribution would be extremely weird and unexpected (and not even really possible): we know for sure that the proportion in the end has to be between 0% to 100%. Normal distributions don't have such cutoffs.
As you say, the way in which new balls are added tends to push the proportion towards the extremes.
The uniform distribution is the result of these two tendencies exactly cancelling out.