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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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