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A two-person method to simulate die rolls (2023)

(blog.42yeah.is)
76 points Fraterkes | 2 comments | 05 Dec 25 19:32 UTC | HN request time: 0s | source
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xg15 ◴[07 Dec 25 21:47 UTC] No.46185507[source]▶
> We can prove that in an ideal situation, the die roll will be fair. Assuming both parties can come up with unbiased random numbers ranging from [0;12)...

Doesn't that assumption remove the entire problem though? I thought the whole reason for the method was that people can't easily think of an unbiased random number.

Or put differently, if that's your starting point, what's stopping you from simply doing (A mod 6) + 1?

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jstanley ◴[07 Dec 25 23:11 UTC] No.46186338[source]▶
Because you can rig the answer if it's just one person. But if two of you use the method from the post, and both commit to your answers before revealing them, then neither of you can rig it.
replies(1): >>46193981 #
1. IAmBroom ◴[08 Dec 25 16:09 UTC] No.46193981[source]▶
That assumes "rigging" is the only non-random component. There's also unintentional bias; when asked to name a random number between 1-100, multiples of ten are underrepresented due to our bias.
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2. jstanley ◴[08 Dec 25 16:32 UTC] No.46194312[source]▶
It solves the rigging problem, it doesn't necessarily make claims as to whether that is the only problem that exists.