Look, you can’t complain about this after giving us so many scenarios involving N locked chests and M unlabeled keys.

https://explainxkcd.com/3015/

  • Dragon Rider (drag)@lemmy.nz
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    17 days ago

    The math is easier if you look for the probability that the arrows are safe. If you do it that way, it’s simply 1/2*4/9, or 2/9.

    • Rentlar@lemmy.ca
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      17 days ago

      That is true. Cueball had asked the question in the way I described, but the check was done in the inverse which is easier to do the math on.