Where does the biggest of five random numbers land?

Question

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

Accepted Answer

The maximum of the five is below a level t exactly when all five numbers are below t. Since each is below t with probability t, P(M \le t) = t^5. The median is the level where this equals one half: m^5 = 	frac12 \quad\Rightarrow\quad m = \left(	frac12ight)^{1/5} = 2^{-1/5} \approx 0.871. So the largest of five uniforms is a coin flip to land above or below about 0.871. Note the median 0.871 sits above the mean of the maximum (5/6 \approx 0.833): the distribution of a maximum is skewed toward the top of the range. Takeaway. The maximum of n uniforms has CDF t^n, so its median is (1/2)^{1/n…

Where does the biggest of five random numbers land?

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