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xkcd #3015: D&D Combinatorics (xkcd.com)

submitted 2 months ago by [email protected] to c/xkcd

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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.

https://explainxkcd.com/3015/

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[–] Ostrakon 28 points 2 months ago* (last edited 2 months ago) (8 children)

You have a 1 in 2 chance of pulling a cursed arrow the first time.

If you pulled a cursed arrow the first time, the second arrow has a 4 in 9 chance to be cursed. Otherwise, it's 5 in 9.

Personally I'd have resolved this as a single d10 once, and rerolled a 10 on the second arrow. I haven't done the math to know if 3d6+1d4 <16 yields the same probability though.

[–] [email protected] 12 points 2 months ago* (last edited 2 months ago) (1 children)

I dm Call of Cthulhu, so simply roll a luck check.
The chance doesn't follow maths, it follows the whims of That Which We Do Not See.
And Randall has pushed his luck with them too far already.

[–] ggppjj 4 points 2 months ago* (last edited 2 months ago)

I was gonna say, sounds like a great use-case for quantum statistics. Until the roll, each arrow is in a superposition where it can be said to be simultaneously cursed and normal. Luck check for each shot until all of either are fully gone.

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