probability of risk of ruin

Hello everyone. This is my first post. But trust me it wont be my last. I realized to get help with math i should associate with people who have already solved my problem instead of doing it on my own.

If you have Mathematics of poker and look at page 281 he goes into detail about the risk of ruin model.. b=bankroll, R(b) is the risk of ruin function. If the game has negative expectation, then R(b)=1 for all b. If there is no probability of a negative outcome, then R(b)=0 for all b.

Where the risk of ruin function for the sum of two bankrolls is equal to the product of the individual risk of ruin values for each bankroll.

R(a+b)=R(a)R(b) . He than goes on to talk about rolling die and if it comes up 1 or 2 we lose $100. If the die comes up 3-6 we win $100. We win more than $33 per toss. Easily understood. But suppose we have only $100. What is our chance of eventually losing this $100 if we play the game indefinitely? U

On the first roll we have a 1/3 chance of going broke and a 2/3 chance of growing our bankroll to $200. Suppose we call $100 one unit. Then the risk of ruin of a bankroll of $100 is R(1), and the risk of ruin of a bankroll of $200 is R(2).

R(1)=1/3+2/3 R(2)

Now we know from the above formula that R(a+b)=R(a)+R(b), so:

substituting and solving:

R(1)=1/3+2/3 R(1)2( the 2 is to the power of R, or squared, I dont know how to write it in computer form I am learning)

2R(squared)-3R+1=0

(2R-1)(R-1)=0

He then goes to say this gives us two possible risk of ruin valus, 1/2 and 1. The R(1) for this game is 1/2

Im lost on the first part of the formula R(1)=1/3+2/3 R(1)2 . How did they change that to step 2? From there I know how to get the problem. It becomes a Coefficient on x2 Term, with a Positive Constant I know this is simple algebra. I just need help with simplifying the first part of the equation and than when he simplied it how he came up with a probability of 1/2 for risk of ruin.

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