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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?

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  • $\begingroup$ The recursive approach sounds like a good approach to me, but I don't understand where the values you are using come from. Think about the terminal case first: if there is only one red ball and two blue balls, what is the expected number of balls needed to draw the red ball? Now step up to your original case where there are two red balls, and use this first result. $\endgroup$
    – JKL
    Commented Jun 22, 2022 at 1:47
  • $\begingroup$ Values - e.g. first 2 are red have probability - 2/3 * 1/3 = 2/9. Sorry i dont quiet follow your reasoning. So terminal case with one red ball and 2 blue balls = 1/3 + 2/3 (E+1)? $\endgroup$
    – Chase
    Commented Jun 22, 2022 at 2:08
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    $\begingroup$ I'm not sure where you are getting $E + 1$ and $E + 2$ from. The recursive case should really look at what happens if you draw one ball at a time; this is the one described in the answer below, which you should try to understand. $\endgroup$
    – JKL
    Commented Jun 22, 2022 at 3:35
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    $\begingroup$ @Chase : That is (almost) the calculation for the expected time until two consecutive draws of red balls from the urn when the balls are returned to the urn. Partitioning on the next two draws.$$\begin{align}E(T)&=\tfrac 49{\sf E}(T\mid R_1,R_2)+\tfrac 13{\sf E}(T\mid B_1)+\tfrac 29{\sf E}(T\mid R_1,B_1)\\&=\tfrac{8}{9}+\tfrac 13(1+{\sf E}(T))+\tfrac 29(2+{\sf E}(T))\end{align}$$However, that is not what you want to do. The issue you have to consider it that red balls are replaced by blue balls as they are drawn, and you seek the expected draw count until both have been replaced. $\endgroup$
    – Graham Kemp
    Commented Jun 22, 2022 at 4:08
  • $\begingroup$ Yes, you are right @GrahamKemp. I was calculating expected time until two consecutive draws of red balls (not a correct approach). $\endgroup$
    – Chase
    Commented Jun 22, 2022 at 21:13

1 Answer 1

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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?

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  • $\begingroup$ I follow your solution. Whats the intuition for adding 1 at front in both E2 and E1? $\endgroup$
    – Chase
    Commented Jun 22, 2022 at 23:05
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    $\begingroup$ @Chase When you have $2$ red balls, the expected extraction is one (the first extraction) plus the expected extractions after that first one. $\endgroup$
    – leonbloy
    Commented Jun 23, 2022 at 12:22

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