this post was submitted on 24 Nov 2024
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xkcd
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2d10 would be used if each arrow had a 10% chance of being cursed. But that's not the case. There are 10 arrows, five are cursed, and 2 are selected. Therefore the first arrow would have a 5/10 chance of being cursed, while the second selection would have either a 4/9 or 5/9 chance of being cursed depending on whether or not the first arrow was cursed.
To solve this, requires using combinatorics. There are 10 choose 2 (45) ways to choose two arrows, of which there are 5 choose 2 (10) ways to choose 2 arrows that are non-cursed. This works out to be 2/9 odds to pull two safe arrows. Which means you need to get funkier with the dice.
If they pulled one, checked if it was cursed, and then pulled another, you'd be right
But they pulled two out of ten at the same time.
So roll two d10s, and say odds are cursed and even would regular. And that's good enough.
I mean, maybe I'm missing something and I didn't spell it out exactly what I meant in the first comment, but that should be the exact same odds as the action.
But what if I roll 2 times 1? I can't pull arrow 1 twice!
Re-roll on a repeat
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.
you will lose. the first arrow has 50% of being cursed the second 4/9 or 5/9.
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
the odds of what exactly are 2/10?
This is where a LLM might come in handy. Just tell it the parameters and say roll random. I think D&D could really benefit from the LLM. Shouldn't be too hard to just let it be the DM. That way everyone can be in the party 😂
Except an LLM has no way to roll anything random, it will just predict the most likely text for a random roll, which isn't remotely the same thing.
Why couldn't they be paired with a true rng? They can reference outside sources.
Why use dice and simple math to solve a problem, when we can use an enormous pile of circuitry, electricity and vector calculus to get an algorithmically determined string of text that contains a probabilistically likely description of the correct dice to roll?