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
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    2 个月前

    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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      2 个月前

      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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        2 个月前

        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.