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If I've played the lottery a certain $N$ number of times (and I didn't look into the results of each game), what this $N$ number must be to get me a $50\%$ chance of have already won at least one of my games? (imagine I 'll look up at all the results at once, after acheiving this chance) and the chance of winning this particularly lottery is $1$ in $1$ million.

I was discussing this with my colleague and I particularly think that there's no exactly number $N$ of getting $50\%$ of chance of already have won. I think that when you ve played zero games the chance is $0\%$ and if you play an infinite numbers of games the chance is $100\%$ that you have won, but there s no ascending curve.

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The number $X$ of successes in $N$ trials is binomially distributed with parameters $N$ and $p=\frac{1}{1000000}$. So, by complementarity the probability of at least one win, i.e. $P(X\ge 1)$, is equal to $1$ minus the probability of $0$ wins, i.e. $P(X=0)$. By the formula of the binomial distribution we get $$P(X\ge 1)=1-P(X=0)=1-\dbinom{n}{0}p^0(1-p)^N=1-(1-p)^N$$ Therefore, if you want at least $50\%$ chance to have won the lottery, then solve $$0.5\le P(X\ge 1)=(1-p)^N \iff N \ln(1-p)\le \ln 0.5 \iff N \ge \frac{\ln 0.5}{\ln(1-p)}$$ For $p=\frac{1}{1000000}$ this gives $$N\ge 693146.834 $$ and since $N$ is an integer it must be at least equal to $693147$ trials.

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If $p$ is the probability of winning the lottery in a single try (so here $p=\frac 1{10^6}$) then the probability of not winning (in a single try) is $1-p$, and the probability of not winning in $n$ independent trials is $(1-p)^n$. It follows that the probability of winning at least once in $n$ independent trials is $1-(1-p)^n$. Thus you want to solve $$1-(1-p)^n=.5\;\implies (1-p)^n=.5\implies n=\frac {log(.5)}{log(1-p)}$$

Easy to check that with $p=\frac {1}{10^6}$ we have $n\sim 693,147$ . It's true, of course, that $n$ is not integral, but I'm not sure that's terribly significant. If we use that (integral) value we get a win probability of about $.50000008$

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