xkcd

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submitted 1 month ago by [email protected] to c/[email protected]

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

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[–] [email protected] 10 points 1 month ago (1 children)

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.

[–] [email protected] 3 points 1 month 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.