Show that probability of drawing a white ball second time from an urn is independent of the additional balls added after the first draw 
Suppose an urn contains 'w' white balls and 'b' black balls and a ball is drawn from it and is replaced along with 'd' additional balls of the same color. Now a second ball is drawn from it. The probability that the second drawn ball is white is independent of the value of 'd'.

My Attempt:
There will be two cases: 
$\text{I}:$ A white ball may be drawn, in that case after the replacement and adding additional balls the total balls will be $=$ b+d+w
Then, the probability of drawing a white ball is $\frac{w + d}{w+b+d}$
$\text{II}:$ A black ball may be drawn, again the total balls have to be $=$ w+b+d
But now, the probability of drawing a white ball is $\frac{w}{w+b+d}$
So, the probability of drawing a white ball on the second draw is $= \frac{2w+d}{w+b+d}$
Which is clearly dependent on d.
But my book says that this statement is true.
Can anybody help me understand this?
 A: 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 $d$.
A: 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) = $w/(w+b)$
A: 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.
A: 
Let us think of a basic Probability concept. Suppose, the original contents of the urn are w white balls and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. Then the probability that the second ball drawn is white is $\frac{w}{w+b}$, i.e. that this probability does not depend on d. Again, the probability that the second ball drawn is Black is $\frac{b}{w+b}$, i.e. this probability also does not depend on d. Please go through the diagram, where, the required probability does not depend on additional balls denoted by d has been established. Therefore, the probability that the 100th or nth ball you pick is black is $\frac{b}{w+b}$ and White is $\frac{w}{w+b}$
