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The larger a group of people is, the more probable it is that they have to wait for at least one person before they can finally leave for the next pub. Why?

Assume the probability that one person delays is 3%. Then, for, say, 30 persons, it is almost certain -- 90%, or $3 \cdot 30\%$. Right?

Well, of course not. For $3\% =: \alpha$ and $30 =: n$, it is more along the lines of $1 - (1-\alpha)^n$. Had that at school. But still, for a fixed $\alpha$ there should be an $n \neq 0$ so that the "approximation" holds exactly. (Is that true anyway?)

$$ 1 - (1-\alpha)^n = \alpha n $$

I believe that there's no algebraic solution for this equation. But how to prove this (and perhaps similar problems)?

EDIT: After plotting the difference $1- (1-\alpha)^n-\alpha n$, it seems clear that the only values for $n$ where the equation holds are $0$ and $1$. This should be provable using elementary calculus, so I'm closing the question.

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Let $n$ be an integer $\ge 2$. For $0\lt x\le 1$, let $f(x)=\left(1-(1-x)^n\right) -nx$. Note that $f'(x)=n(1-x)^{n-1}-n=n\left((1-x)^{n-1}-1\right)$. Since $0\lt x\le 1$, $f'(x)$ is negative in this interval, meaning that $f(x)$ is decreasing in that interval.

Thus we have $1-(1-x)^n =nx$ only at $x=0$.

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You actually don't need calculus to resolve this problem--using the binomial theorem, $(1-\alpha)^n = \sum^n_{k=0} (-\alpha)^k {n \choose k}$.

Expanding the first few terms yields:

$(1-\alpha)^n = 1 - n\alpha + {n \choose 2}\alpha^2-{n \choose 3}\alpha^3+...$

If $\alpha$ is sufficiently close to 0, the first two terms should be a good approximation (a little calculus is hidden in this line).

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It may be worth noting that the reason $\alpha n$ is the incorrect answer is because it counts scenarios with multiple successes ("delays" here) multiple times, so it follows then that it will overestimate the real probability whenever $n\geq2$.

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