In an urn with $b$ blue and $r$ red balls, each time (call it a trial) a ball is chosen at random and then put again in the urn along with $c$ extra balls of the same color. Now probability of getting a blue ball in the 1st trial = $\frac{b}{b+r}$. Surprisingly, I also see that probability of getting a blue ball in the $n$-th trial is also $\frac{b}{b+r}$. I don't understand the intuition behind this.
Another question is the intuitive argument to prove that the number of red balls in the first $n$ trials follow an uniform distribution between $0$ and $n$ when $b=r=c$.
Answer
It's roughly like the following.
Easy case: Suppose there were equal numbers of each colour. Then how could the probability - assuming you don't know what's come before - ever be anything other than a half? For every eventuality that you end up picking a blue ball, there's an exactly complementary possibility in which the colours are all reversed. The contents of the bag will typically be skewed, but the probability of it being skewed either way is the same.
General case: Essentially a restatement of a proof by induction. The first turn is easy. What about the second turn? Think about it like this: either you draw a ball which was in the bag at the start, or you draw a ball which was added after the first turn. If it was the first case, then the probability is just the original one. On the other hand, if you drew a new ball, it is necessarily of the colour drawn in the previous (first) turn. But the probability this is blue is just the same as the probability the first ball drawn was blue again!
In fact, this is a conceptual proof: the ball drawn on the $n$th turn is from a distribution over colours which we can break down according to when each ball entered the bag. But the probability distributions for each turn are determined to be exactly those of all previous turns and so nothing can ever change!
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