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In a game a player has a probability of $ \frac{1} {N} $ to win, where $ N \in {\Bbb N} $. It´s possible to calculate the probability of at least one win, after $k$ tries?

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up vote 3 down vote accepted

Yes, the probability of at least one win is 1 minus the probability of no wins: $1-\bigl(1-{1\over N}\bigr)^k$ (assuming independence between games).

In general, you use $P(A)=1-P(\text{not }A)$. The complement of (the "not" of) "at least one" is "none".

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Trying again N times, what is the probability of at least "j" wins D:? I can´t do the same here or yes :S? – August Jan 3 '12 at 2:27

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