• Artisian@lemmy.world
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    11 hours ago

    They don’t say that the random answer is chosen uniformly (though that is the norm in the field). If we relax that, then we’re putting a distribution on these where we want:

    P(correct with distribution (a,b,c,d)) = some value shown on A,B,C,D

    I don’t see any assumption that we will pick using that distribution, so I think this avoids the recursion.

    Unfortunately this has too many solutions. If you put a total of 0.25 weight on A and D, then the rest does not matter. If you put 0.5 weight on C, again the rest is irrelevant.

    • Triumph@fedia.io
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      10 hours ago

      You’ve added details that aren’t in the question. It’s like asking what are the odds of rolling a “one” on a 1d4, and then saying “Well, if it’s not a fair 1d4, then …”

      • Artisian@lemmy.world
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        10 hours ago

        … I… I literally talked about this. It’s the first words.

        They don’t say that the random answer is chosen uniformly (though that is the norm in the field). If we relax that,…

        What more was needed?

        • Triumph@fedia.io
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          9 hours ago

          If we relax that, …

          The question is as posed. We have no indication that we should assume a different distribution of “random”.

          • Artisian@lemmy.world
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            9 hours ago

            My favorite indication that someone is using a word in an unusual way is that their question has no answer if you interpret it as usual. I reply to you because you argued very nicely that, if this makes sense at all, it must be with a different use of language than we expect.