Limiting chance the max stays below 1 minus 2/n

Question

Draw n independent numbers uniformly from [0,1] and let Mn be the largest. As n grows large, what is the limiting probability that Mn stays below 1 - 	frac2n?

Accepted Answer

The maximum is below t when all n draws are below t, so P(Mn \le t) = t^n. With t = 1 - \frac2n: P\!\left(Mn \le 1 - 	frac2night) = \left(1 - 	frac2night)^n \longrightarrow e^{-2} \approx 0.1353. Pushing the target down by two units of \frac1n instead of one squares the limit, from e^{-1} to e^{-2}. Takeaway. The rule \left(1 - 	frac{c}{n}ight)^n 	o e^{-c} shows the shortfall of the maximum below 1 follows an exponential law at scale 1/n.

Limiting chance the max stays below 1 minus 2/n

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