Where does the smallest of five random numbers sit?

Question

Draw 5 independent numbers uniformly from [0,1] and let L be the smallest of them. Find the value m for which L is equally likely to fall below or above m (the median of the minimum).

Accepted Answer

The minimum is above a level x exactly when all five numbers are above x, so P(L x) = (1-x)^5. The median is where the minimum is a coin flip to be above or below: P(L m) = 	frac12 \quad\Rightarrow\quad (1-m)^5 = 	frac12 \quad\Rightarrow\quad 1 - m = 2^{-1/5} \approx 0.8706. Hence m \approx 1 - 0.8706 = 0.129. By mirror symmetry this is exactly 1 minus the median of the maximum of five uniforms. Takeaway. The minimum of n uniforms has survival function (1-x)^n, so its median is 1 - (1/2)^{1/n}, hugging the bottom of the range.

Where does the smallest of five random numbers sit?

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