# Tag Info

4

The pigeonhole principle The probabilistic generalisation of the pigeonhole principle tells us that for $n$ pigeons randomly assigned to $m$ pigeonholes with equal probability, the chance that one hole will contain at least 2 pigeons is: $$1-{{(m)_n}\over{m^n}}$$ where $(m)_n$ is the falling factorial. For us, our pigeons are the number of dice and ...

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You're going to have a fairly limited range where success (of some level) is possible, but isn't guaranteed. For 1d6, it's obviously impossible to roll doubles. 7d6 guarantees at least one double by the pigeonhole principle. So only from 2d6 to 6d6 do you have any uncertainty at all. Because my conditional probability skills have atrophied, I just brute-...

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For a numerical answer, we can use the AnyDice code I just recently wrote for another answer, which calculates the highest number of identical values in a roll of dice: function: highest number of matches in ROLL:s { MAX: 0 loop I over ROLL { COUNT: ROLL = I if MAX < COUNT { MAX: COUNT } } result: MAX } output [highest number of matches ...

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I thought I would have a go at trying to make something as this seems like a fairly interesting problem. I looked into it and if you only want to roll 2d20 it is impossible to get a distribution perfectly between no advantage and advantage but you can at least in theory make something really close, even if the dice are indistinguishable. I originally made ...

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TLDR: Roll regular advantage, then also roll a d4, if d4 rolls 1 then flip the value (20->1, 1->20, 19->2, 11->10, etc) Inpired by the answers I saw... So after I posted the question I continued experimenting and did manage to come to the realization that to get the distribution I was looking for, the simplest way was probably just to modify the ...

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Advantage when the dice Match I like Eric's answer, and agree that "The most straightforward way to add "half Advantage" is to only make it work half of the time." However, that particular solution involves the addition of a third die. Granted, a pretty low increase in complexity, but strictly speaking unnecessary. An even simpler 'half ...

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I would suggest rolling one or two d12 and a d20 for partial advantage. Here are the odds for "at least" results using Anydice. a straight d20 (black) highest of a d12 and a d20 (orange) highest of two d12 and a d20 (blue) normal advantage, highest of two d20 (green) Analysis: Likelyhood of getting 13 or more is exactly same as with straight ...

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Erik's suggestion of effectively flipping a coin to choose between a normal roll and a roll with advantage is pretty much the ideal "half advantage" roll, as it produces a distribution exactly halfway between normal and advantage. Its only down side is that you end up having to roll an extra die (or flip an actual coin, I guess). If you'd instead ...

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The most straightforward way to add "half Advantage" is to only make it work half of the time. You can do this by using different colored D20s and throw in another d6 to decide whether you're allowed to choose or not. For example, a red D20, a green D20, and a green d6. If you roll 4-6 on the d6 you pick the highest die, if you roll 1-3 on the d6 ...

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