Chance the largest of four random numbers beats 0.9

Question

Draw 4 independent numbers uniformly from [0,1] and let M be the largest. What is the probability that M exceeds 0.9?

Accepted Answer

The maximum stays below 0.9 only if all four numbers are below 0.9, which happens with probability 0.9^4. So P(M 0.9) = 1 - 0.9^4 = 1 - 0.6561 = 0.3439. Even though any single draw clears 0.9 only 10\% of the time, having four chances lifts the odds that at least one (hence the maximum) does so to about 34\%. Takeaway. For the maximum of n uniforms, P(M t) = 1 - t^n; extra draws sharply raise the chance of touching the top of the range.

Chance the largest of four random numbers beats 0.9

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