vase with blue and red balls At first I hope this is not a duplicate post. I tried to find it but I have not found it. 
I hope that someone could help me with understanding the exercise.
This question is about a vase with r red balls en b blue balls.
I have the following two questions:
In each of the questions we have different vase with r red balls and b blue balls. So everything you will do at each question does not matter for the other question.


*

*You will pick randomly balls from the vase, till you pick one blue ball. Proof that the total amount of balls that is picked is equal to
$$ \frac{b+r+1}{b+1}  $$

*What is expected amount of remaining balls if you keep removing balls from the vase till all the balls in the vase has the same color?
I have trouble of understanding how I can do this question. I know that the +1 in the formula is because you have b-1 balls, so you know that 1 blue ball is picked.
My thoughts about question 2 is as follows:
Fist I calculated the probability that the only remaining balls are red.
I defined X for the remaining red balls after the last blue has been picked. 
Then I numbered every red ball 1 < j < b, to find the probability that ball j has been picked before all the blue balls.
We need to calculate the value of P. This means that I have to calculate what the probability is that every ball has been picked except for the blue ball j has been picked. So what is the chance that blue b j is picked at last?
This is $$ \frac{1}{1+r}  $$
Now I can calculate the expected value of only remaining red balls left, which is:  $$ EX[x] = \frac{b}{r+1}  $$ 
Now I only have to calculate the expected amount of the only remaining blue balls and add this to the expected value of only remaining red balls and divide that by 2. Like this: 
$$ EX[X] = \frac{\frac{b}{r+1}+ \frac{r}{b+1}}{2} $$
I am not sure if any of this is correct. Both of the questions are without replacement.
if anything is unclear, please ask.
 A: Question 1: It is easiest to prove this by induction.  If $r = 0$ (there are no red balls) and $b \geq 1$, then clearly the first ball will be blue.  That is, the number of balls $N$ that we need to draw in order to draw a blue ball has expected value $E(N) = 1$.  Note that we can write this as
$$
E(N) = \frac{b+1}{b+1} = 1
$$
in accordance with the desired formula.  Now, consider any other value of $r$, and $b \geq 1$ still.  Suppose that we already have shown that for $r-1$ red balls and $b$ blue balls, the expected number of balls until we draw a blue ball is $\frac{b+r}{b+1}$, in accordance with the desired formula.  We now draw a ball from an urn with $r$ red balls and $b$ blue balls.  With probability $\frac{b}{b+r}$, the first ball is blue, and $N = 1$.  With probability $\frac{r}{b+r}$, the first ball is red, and we now have an urn with $r-1$ red balls and $b$ blue balls.  By the premise, we already know what the expected number of additional balls that need to be drawn in order to produce a blue ball; it is $\frac{b+r}{b+1}$.  Therefore, for the urn with $r$ red balls and $b$ blue balls, we have
$$
\begin{align}
E(N) & = \frac{b}{b+r} \cdot 1
       + \frac{r}{b+r} \cdot \left(1+\frac{b+r}{b+1}\right) \\
     & = \frac{b}{b+r} + \frac{r}{b+r} + \frac{r}{b+1} \\
     & = 1 + \frac{r}{b+1} = \frac{b+r+1}{b+1}
\end{align}
$$
and we are done.
Question 2: By symmetry, the expected number of remaining balls left when only balls of one color (either red or blue) remain is equal to the number of balls of the same color we draw at the beginning.  (To see this, draw all the balls out from the urn, and lay them out on a table, in sequence.  The sequence from left to right is exactly as probable as the sequence from right to left.)
With probability $\frac{r}{b+r}$, the first ball is red.  We are then left with $r-1$ red balls and $b$ blue balls, and we want to know how many balls we draw before (not until) we draw the first blue ball.  This is one less than the result from Question 1.  To this we have to add the first red ball.  So when the first ball is red, our answer is
$$
N_{red} = \frac{b+r}{b+1}-1+1 = \frac{b+r}{b+1}
$$
With probability $\frac{b}{b+r}$, the first ball is blue, and by symmetry, the number of consecutive blue balls at the start is
$$
N_{blue} = \frac{b+r}{r+1}
$$
Thus, the expected number of consecutive balls drawn of the same color (which is also the answer to the original Question 2) is
$$
E(N) = \frac{r}{b+r} \cdot N_{red} + \frac{b}{b+r} \cdot N_{blue}
     = \frac{r}{b+1} + \frac{b}{r+1}
$$
