Five points are dropped independently and uniformly on [0,1], cutting it into six intervals. What is the expected length of the longest interval?

Five points are dropped independently and uniformly on [0,1], cutting it into six intervals. What is the expected length of the longest interval?

Approach: Use the exchangeability of the six spacings, write the probability that all of them are below x by inclusion-exclusion over which ones exceed x, then integrate the survival function.

49/120. The five points give six spacings between consecutive order statistics, and they are exchangeable. Inclusion-exclusion gives P(all spacings <= x) = sum_j (-1)^j C(6,j)(1 - jx)^5, where terms with jx > 1 are dropped. Integrating the survival function collapses the sum into the harmonic number: E[longest] = (1/6)(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) = (1/6)(49/20) = 49/120 = 0.4083. For n points the answer is H_{n+1}/(n+1), so the longest gap shrinks like log(n)/n while the average gap shrinks like 1/n. Here the longest piece runs about 2.45 times the average piece of 1/6.

Follow-up: What is the expected length of the shortest of the six intervals?

Key concepts: spacings, inclusion-exclusion, harmonic number, order statistics.