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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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For $n$ games, we need $(0.6)^n>0.8\implies n<1$, but $n\ge 1$, so no solution. – lab bhattacharjee Oct 4 '12 at 4:52
Is the problem that you want to win $n$ games and are asking how many games $m$ should be played to have 0.8 chance of winning $n$ of them? – Ross Millikan Oct 4 '12 at 4:55
@labbhattacharjee: in your reading, which is what the question asks, we clearly need $n=0$. Then we have greater than 80% chance of winning no games. – Ross Millikan Oct 4 '12 at 4:56
I don't see how the title relates to the question. – joriki Oct 4 '12 at 7:48

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, $$(.6x+.4y)^n=\sum_{r=0}^n{n\choose r}(.6)^r(.4)^{n-r}x^ry^{n-r}$$

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+1, a powerful rule very explained very nicely. – NoChance Oct 4 '12 at 7:18

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