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.
From a play session perspective? If the GM is that good that they can mental math it, I would much rather be given one roll than a series of rolls. Ask anyone about their horror stories about grappling in 3e about how much that kills the game flow.
Also: The verbiage is ambiguous (less so if you have the context of how many attacks per round a player has and what feat they are using) but I think you can represent “I grabbed two at once” and “I grabbed one and then one” with a binomial coefficient. Been more than a minute but poking chatgpt to remember the notation (nCk) and it is likely representable as (5C2)/(10C2) which is approximately 22.2%.
As for the dice? I forget if the type of die meaningfully impacts this but 3d6+1d4=4-22. Whether a 16 maps to that 22.2% range is beyond my brain right now as this comment was mostly because I forgot the difference between nCk and nPk and felt like googling that.
The type of dice used can meaningfully impact this. The chance of a 2 or 12 rolling 2d6 is 1/36, the chance rolling 1d8+1d4 is 1/32. The chance of rolling 7 on 2d6, the most common result, is 1/6. The chance of rolling a 5, 6, 7, 8, or 9 on a 1d8+1d4, all equally likely, is 1/8 each.
Unlike you I can’t begin to remember the elegant way to find this. I also assume Randall would have it at least close to right.
Anydice.com can handle this stuff easily. As already pointed out in another comment, it does perfectly match. What it will not tell you is if you grabbed one or 2 arrows, though presumably a roll of 1-x could be used to say you got one, and x+1-15 means you got two.
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.
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.
Yeah, d10 (1-5 cursed, 6-10 normal), then repeat with 1 or 10 being a reroll. I’m curious what the intent of the character was, whether they’re looking for a simple answer or something contrived like the DM’s answer.
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.
Its xkcd so I assume Randall et al did the math.
From a play session perspective? If the GM is that good that they can mental math it, I would much rather be given one roll than a series of rolls. Ask anyone about their horror stories about grappling in 3e about how much that kills the game flow.
Also: The verbiage is ambiguous (less so if you have the context of how many attacks per round a player has and what feat they are using) but I think you can represent “I grabbed two at once” and “I grabbed one and then one” with a binomial coefficient. Been more than a minute but poking chatgpt to remember the notation (nCk) and it is likely representable as (5C2)/(10C2) which is approximately 22.2%.
As for the dice? I forget if the type of die meaningfully impacts this but 3d6+1d4=4-22. Whether a 16 maps to that 22.2% range is beyond my brain right now as this comment was mostly because I forgot the difference between nCk and nPk and felt like googling that.
The type of dice used can meaningfully impact this. The chance of a 2 or 12 rolling 2d6 is 1/36, the chance rolling 1d8+1d4 is 1/32. The chance of rolling 7 on 2d6, the most common result, is 1/6. The chance of rolling a 5, 6, 7, 8, or 9 on a 1d8+1d4, all equally likely, is 1/8 each.
Unlike you I can’t begin to remember the elegant way to find this. I also assume Randall would have it at least close to right.
Anydice.com can handle this stuff easily. As already pointed out in another comment, it does perfectly match. What it will not tell you is if you grabbed one or 2 arrows, though presumably a roll of 1-x could be used to say you got one, and x+1-15 means you got two.
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.
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.
Yeah, d10 (1-5 cursed, 6-10 normal), then repeat with 1 or 10 being a reroll. I’m curious what the intent of the character was, whether they’re looking for a simple answer or something contrived like the DM’s answer.
or: repeat with the already rolled number being a reroll
Sure, but I usually prefer high/low success/failure