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

submitted 2 days 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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[–] givesomefucks -4 points 2 days ago (1 children)

Specific number doesn't matter.

We have a binary result: cursed or regular. You probably wouldn't lose anything by flipping a coin twice as long as the distribution of the arrows in the quiver is truly random.

Like, if you're looking closer than that, you might as well account for when the arrows were added to the quiver, if they were added at the same time, how much the quiver has been jostled. The line has to be drawn somewhere, which I think is literally the joke of the comic....

There's a very very simple solution but the DM is about to overthink and come up with the same result as the easy way, which they'll realize after taking the long way around.

I know I've lost a lot of time by thinking "it's worth being exact" and then I found out, no it really wasn't worth it.

[–] SwordInStone 4 points 2 days ago (1 children)

you will lose. the first arrow has 50% of being cursed the second 4/9 or 5/9.

[–] givesomefucks -4 points 2 days ago (1 children)

That is if he pulled two individually, one after another.

Not two at the same time.

The odds are two out of ten.

There is no simpler way to explain this, I'm sorry if it still isn't working

[–] SwordInStone 4 points 2 days ago

the odds of what exactly are 2/10?