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The probability of winning a game is 0.6 and losing it is 0.4. Then how many games should one play, so that overall probability of winning that many games exceeds 0.8?

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    I don't see how the title relates to the question.2012-10-04

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As noted in the comments, the question is not clear. When you have clarified the question, you may find this useful: if you play exactly $n$ games, then the probability of winning exactly $r$ games is the coefficient of $x^ry^{n-r}$ in the expansion of $(.6x+.4y)^n$. By the Binomial Theorem, $(.6x+.4y)^n=\sum_{r=0}^n{n\choose r}(.6)^r(.4)^{n-r}x^ry^{n-r}$