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Say, a bag has 10 balls, in which 9 are red, 1 is black.

Each red ball is worth 1 point, each black is worth 4 points.

I have 8 picks from the bag to start with (the bag refills itself after each pick: returns to 9 red 1 black), once I get a black ball, I will get 8 additional picks, and go on.

Ultimately I would like to calculate the expected total points.

So, for example

        1st pick    red     7 picks remaining
        2nd pick    black   6 + 8 = 14 picks remaining
        ...

One way to solve the problem:

Assume the expected number of red balls is A, black balls is B. $A/B=9$ and $8+8*B=A+B$ So $A = 36,B = 4$

But I am looking for a more formal/generalized way (maybe markov/matrix or wald equation or stopping time) to this kind of problem.

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    "The bag refills..." means it returns to being 9 red and one black?2012-10-19
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    @ThomasAndrews yes it does2012-10-19
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    Is it even true with probability $1$ that you will eventually end this game? That's not at all clear to me.2012-10-19
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    If it is not true, then the expected value is infinite. It might be infinite even if the game will end with probability 1, as in the St Petersburg paradox: http://en.wikipedia.org/wiki/St._Petersburg_paradox2012-10-19
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    Also, when the game ends, if you have $A$ red balls and $B$ black balls, then $A+B$ is the total number of draws, but that is $8+10(B-1)$, so $A+B=8+10(B-1)$ so $A/B = 9-\frac{2}{B}<9$2012-10-19
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    You have different equations for $A+B$; colinfang has $8+8B$, Thomas has $8+10(B-1)$ -- why isn't it $8+10B$? That's the total number of picks we get if we draw $B$ black balls, no?2012-10-19
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    Ah, because I'm brain dead, it is 8+10B. Still, that means $A/B = 10 + 8/B>10$ so when the game is over, we can never have $A/B=9$ or even be close to $9$.2012-10-19
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    @ThomasAndrews sry there was an typo in the question, the additional draws is 82012-10-19
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    Now that you've fixed the question, your calculation is correct and is in fact a good way to calculate the expected point total. I don't see what you don't like about it. It's much better this way than messing with matrices.2012-10-19
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    I still don't like it, because when you've stopped, you've got an over-respresentation of red balls compared to blue. So, while over time, when you might expect $A/B$ to tend to $9$, the game doesn't tend to end when $A/B$ is much less than $9$, so the data selects out that sample, but doesn't select out the sample when $A/B$ is more than $9$.2012-10-19
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    @Thomas If at the end there are $0$ black balls then there are $7$ red balls, if $1$ then $15$, ..., if $4$ then $36$, ..., if $10$ then $78$, etc. and so sometimes the ratio is above $1:9$ and sometimes below.2012-10-19

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Generalised: you have probability $p$ of picking a red ball and $1-p$ of a black ball, and $n$ picks. If you a black ball, you get an extra $n$ picks.

So if $r$ is the expected number of red balls picked and $b$ the expected number of black balls then you can calculate the expectation from one pick (including the values from the possibility of getting an extra set of picks), which you can multiply by $n$: $$r = n(p + (1-p)r),$$ $$b= n(1-p)(1+b),$$ so $$r=\frac{np}{1-n(1-p)},$$ $$b = \frac{n(1-p)}{1-n(1-p)}.$$

For $p=0.9$ and $n=8$ this gives $r=36$ and $b=4$, as you calculated.

If $n \ge \frac{1}{1-p}$ then the denominators are not positive and the expectations are in fact infinite.

Red balls are $\frac{r}{b}=\frac{p}{1-p}$ times as likely as black balls ($9$ times in your example) and this carries through to the expectations. If $B$ is the actual number of black balls then you will have $R=(n-1)B+n$ red balls: while $\frac{R}{B}$ may be less than $\frac{p}{1-p}B$ for small $B \lt b$, it will be more than $\frac{p}{1-p}$ for large $B \gt b$ so long as the expectations are finite.