<s>[INST] Chance of winning simple dice game Tossing two fair dice, if the sum is 7 or 11, then I win; if the sum is 2, 3 or 12, then I lose; if the sum is one of rest of numbers then I toss the two dices again.  What is probability of winning?
What I have tried: every tossing, p(win)=$\frac{2}{9}$, and p(lose)=$\frac{1}{9}$, p(tossing again)=$\frac{2}{3}$, so the probability of winning is $\frac{2}{9}+\frac{2}{3}\frac{2}{9}+(\frac{2}{3})^2\frac{2}{9}+(\frac{2}{3})^3\frac{2}{9}+.....=\frac{2}{9}\sum^{\infty}_{n=0}(\frac{2}{3})^n=\frac{2}{3}$
Does anyone could tell me why my idea is wrong? What is the correct solution? Thanks!
 [/INST] With your edited solution i think you are now correct with 2/3.  
Since $$\frac{\frac{2}{9}}{\frac{2}{9}+\frac{1}{9}}=\frac{2}{3}$$
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