Balls you expect to draw to finish red balls.

Question

There are 2 red balls and 1 blue ball in the urn. After one ball is drawn another blue will go in.

How many balls you expect to draw to finish the red balls?

My attempt:

I tried to use recursive technique (not sure if it is the right way to go about it). Let E be the expected number of balls drawn before we finish the red balls. Therefore:

E = 2/9 * (2) (first 2 are red) + 1/3 * (E+1) (first is blue) + 4/9 * (E+2) (first is red and second is blue).

Is this the correct way to approach it?

Answer 1

Let's be more precise. Let $E_k$ be the expected number of remaining balls to extract, given that we have $k$ red balls in the urn. Then, of course, $E_0=0$. We want to find $E_2$. But

$$E_2 = 1 + \frac{2}{3} E_1 + \frac{1}{3} E_2$$

and $$E_1 = 1+ \frac{1}{3} E_0 + \frac{2}{3} E_1 = 1 +\frac{2}{3} E_1$$

Can you go on from here?