Show that probability of drawing a white ball second time from an urn is independent of the additional balls added after the first draw

Solution 1:

Clearly $$P(X_2=w|X_1=w)=\frac{w+d}{w+b+d}$$ as you say, where $X_i$ is the result of draw $i$.

Also $$P(X_2=w| X_1=b)= \frac{w}{w+b+d}$$ from similar reasoning.

So $$P(X_2=w)=P(X_2=w|X_1=w)P(X_1=w) + P(X_2=w|X_1=b)P(X_1=b)$$

by the law of total probability, and computing this gives

$$\frac{w+d}{b+w+d} \frac{w}{w+b} + \frac{w}{w+b+d}\frac{b}{b+w} = \frac{(w+d)w + wb}{(w+b)(w+b+d)} = \frac{w}{w+b}$$ as the common factor $w+b+d$ cancels out.

So $P(X_2=w)$ indeed does not depend on $d$.

Solution 2:

Imagine that the $w$ white balls are all slightly different shades of white, and same for the $b$ many black balls. So there are $w+b$ balls, each of a different shade.

Now by symmetry, the $w+b$ many possible shades for the 2nd ball are all equally likely, since nothing in he process favors one of the shades over another (having started with the same number [i.e. 1] of each shade).

So the the probability that the 2nd ball is white is

(number of white shades)/(total number of shades) = $w/(w+b)$

Solution 3:

Here's an intuitive (as opposed to algebraic) proof.

The key to this question is that the same number, $d$, would be added in either case. So before the second draw, there are $w+b+d$ balls in there. Imagine you can distinguish the "original" $w+b$ balls vs the "newly added" $d$ balls.

Conditioned on you drawing an original ball, $Prob(white|original) = {w \over w+b}$

Conditioned on you drawing a new ball, $Prob(white|new) = {w \over w+b}$ because all new balls are the same color and that color is white iff the first drawn ball is white, which happens with prob ${w \over w+b}$.

The overall probability is a linear combination of the two terms above, with coefficients depending on $d$ and summing to $1$, but since the two terms are equal, it doesn't matter what $d$ is.

BONUS 1: This proof also shows (inductively) that you can extend the process to adding more balls matching the $2$nd draw, then making a $3$rd draw, then adding even more balls matching the $3$rd draw, then making a $4$th draw, etc., and the unconditional prob of drawing white at the $n$th draw will always be ${w \over w+b}$.

BONUS 2: With a bit more work, this proof also shows that $d$ can be a random variable (e.g. how about $d \sim Poisson(\lambda=13)$), as long as it is independent of the first draw result.