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/

  • @NuXCOM_90Percent@lemmy.zip
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    297 days ago

    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.

    • @godot@lemmy.world
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      7 days ago

      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.

      • @AEsheron@lemmy.world
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        37 days ago

        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.