You are on the right path in solving the problem, however, there is a nice technique for evading the calculations, which is finding the expected value using indicator functions.
First, number the red balls from $1$ to $r$ and define $X_i$ to be the function indicating whether ball number $i$ is drawn before the first blue ball or not.
So $X_i$ is $1$ if the i'th ball is drawn before the first blue ball and zero otherwise.
Now, it is easy to see that $X$ is the sum of indicators, i.e.
$$X = 1 + \sum_{i=1}^r X_i $$
so the expected value of both sides are equal, and using the expected of the sum property and the symmetry between $\mathbb{E}[X_i]'s$, we get:
$$\mathbb{E}[X] = 1 + \sum_{i=1}^r \mathbb{E} [X_i] = 1 + r\mathbb{E}[X_1] = 1 + r\mathbb{P}[X_1] $$
the problem now boils down to finding the probability that the first red ball is drawn before the first blue ball,
which is just $\frac{1}{b+1}$, since out of b+1 different ways of putting 1 red ball and b blue balls next to each other, only the one which the first red ball comes first is valid. you can then arrange other red balls where ever you want.
Finally, the expected value is given by:
$$\mathbb{E}[X] = 1 + \frac{r}{b+1} $$
There are three blue balls, all equally likely to be the first ball chosen. In two cases, the second ball chosen will also be blue, and in one case, it will be red.
EDIT What you say in your objections is true, but it has nothing to do with the case. Instead of balls, let us says that we have one urn with two gold coins, one with a gold coin and a silver coin, and one with two silver coins. The coins are the same size and cannot be distinguished by touch. You draw a coin at random from a randomly chosen urn, and it's gold. What is the probability that the other coin in the urn you have chosen is gold? According to your logic, it $1/2.$
Now I tell you that the gold coin in the urn with the silver coin is dated $1900,$ and the two gold coins in the same urn are dated $1901$ and $1902.$ You draw a gold coin, and place it on the table, and tails is showing, so that date cannot be seen. Now, what is the probability that the other coin in the urn is gold? This is the question being asked. The answer is that the other coin is gold if when we flip the coin we've drawn, the date is $1901$ or $1902$ but not if it's $1900.$
Do you really want to maintain that it's equally likely to be $1900$ as it is to be $1901$ or $1902?$
Best Answer
Let $E(n)$ be the expected number of turns to turn the balls blue, assuming there are $n$ red balls left and $5-n$ blue balls. Thus, the problem is to find $E(5)$. Clearly, $E(5) = E(4) + 1$, as a red ball will definitely be turned into a blue ball. Then, $E(4) = 1+\frac{4}{5}E(3)+\frac{1}{5}E(4)$, as there is a $\frac{4}{5}$ probability of picking a red ball and turning it blue, and a $\frac{1}{5}$ probability of picking a blue ball and keeping it blue. $$$$ Similarly, $E(3) = 1+\frac{3}{5}E(2)+\frac{2}{5}E(3)$, $E(2) = 1+\frac{3}{5}E(1)+\frac{3}{5}E(2)$, $E(1) = 1+\frac{1}{5}E(0)+\frac{4}{5}E(1)$. $E(0) = 0$, as if there are no red balls left, no more turns are needed. Solving this system of equations yields an answer for $E(5)$, which is what you want. $$$$ You could also use a Markov chain/matrix to find the expected value - this is pretty much the same thing.