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When random is not actually random enough

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Hacker News | Oct 6, 2026 | steveklabnik

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Assuming that our random choice function has being chosen. The previous weights would be represented with (4, 3, 3) instead. 4 + 3 + 3 = 10 and so 3 / 10 = 0. 3 as we expect. Then simply draw a random number choice = random_between(0, 29).

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