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> In order to get a collision with probability 1% you need to produce about 4e37 hashes

Does "get a collision" here mean finding a particular collision with a known hash, or finding any collision between any two of the generated hashes?

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The latter. I used the formula from the Wikipedia page on the birthday attack: p ~= 1 - e^{-n^2/2H} where n is the number of items you've picked and H is the size of the universe

I see, thanks!

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