Ask HN: We just had an actual UUID v4 collision...

Re:Ask HN: We just had an actual UUID v4 collision... by susam on Saturday May 09, 2026 @01:19AM (#48072030)

And for people who like equations, here is my attempt at explaining it.

Assume each flip is independent and the bias remains same in each flip.

Let

  P(H) = p,
  P(T) = 1 - p.

Then

  P(HH) = p^2,
  P(HT) = p(1 - p),
  P(TH) = (1 - p)p,
  P(TT) = (1 - p)^2.

Therefore

  P(HT or TH) = 2p(1 - p).

Now calculate

  P(HT | HT or TH) = p(1 - p) / (2p(1 - p)) = 1/2,
  P(TH | HT or TH) = (1 - p)p / (2p(1 - p)) = 1/2.

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Re:Ask HN: We just had an actual UUID v4 collision... by taegee on Saturday May 09, 2026 @03:23AM (#48072715)

You don't need conditional probability here, as the flips are independent.

It's just p(H)p(T).

And p(H)p(T) = p(T)p(H), thus 2*p(H)p(T) = 2p(1-p).

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Re:Ask HN: We just had an actual UUID v4 collision... by cdaringe on Saturday May 09, 2026 @11:40AM (#48075845)

Thats how i noodled thru it internally

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Re:Ask HN: We just had an actual UUID v4 collision... by clickety_clack on Saturday May 09, 2026 @02:02PM (#48076887)

That wasn’t what he was trying to prove, but the proof could be done without conditionals like this:

If: p(H)p(T) = p(T)p(H)

And: p(H)p(T) + p(T)p(H) = 1

Then: p(H)p(T) = p(T)p(H) = 0.5

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Re:Ask HN: We just had an actual UUID v4 collision... by susam on Saturday May 09, 2026 @09:14PM (#48079988)

Independence tells us how to compute the probability of a sequence like HT or TH:

  P(HT) = P(H)P(T) = p(1 - p)

But the question I am addressing is not just "what is the probability of HT?" It is "given that the two flips are different, what is the probability that the order was HT rather than TH?"

That is a conditional probability:

  P(HT | HT or TH)

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