Do all the red draws beat all the blue draws?

Question

Draw 3 "red" numbers and 3 "blue" numbers, all 6 independent and uniform on [0, 1]. What is the probability that every red number is larger than every blue number? Then generalize to m reds and n blues.

Accepted Answer

Because all 6 numbers are drawn from the same continuous distribution, they are exchangeable: every ordering of the 6 values is equally likely, and ties have probability zero. The colors are just labels slapped on afterward. "Every red beats every blue" means that when we rank the 6 values from smallest to largest, the three blues occupy the three lowest ranks and the three reds occupy the three highest. Out of all \binom{6}{3} = 20 equally likely ways to choose which three of the six positions are blue, exactly one puts all blues at the bottom. So P = \frac{1}{\binom{6}{3}} = \frac{1}{20} = …

Do all the red draws beat all the blue draws?

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