#### overgeeked

##### B/X Known World

I tried to goof around with anydice to answer this myself, but it didn't work out.

So about what kind of bonus that is? Anyone know?

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- Thread starter overgeeked
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I tried to goof around with anydice to answer this myself, but it didn't work out.

So about what kind of bonus that is? Anyone know?

It's actually advantage on the tens only, then. The tens die is 0-9. Which means it's modeled pretty well by "output [highest 1 of 2d10]-1" on anydice, which winds up giving you mean of 6.15 and SD of 2.35. Given the mean is usually 4.5, the advantage is about +1.65; given that it's the tens die, it would be about a +16 on average.

But how about in that middle of the range where it's most effective? Well, as per anydice you have a 84% chance of rolling at least a 4, versus 60% for a straight roll, whereas for a 5 it's 75%, versus 50% for a straight roll, and for a 4 it's 64%, versus 40% for a straight roll. So it winds up being about a +25% bonus in the middle of the range.

Awesome, thanks.

It's actually advantage on the tens only, then. The tens die is 0-9. Which means it's modeled pretty well by "output [highest 1 of 2d10]-1" on anydice, which winds up giving you mean of 6.15 and SD of 2.35. Given the mean is usually 4.5, the advantage is about +1.65; given that it's the tens die, it would be about a +16 on average.

But how about in that middle of the range where it's most effective? Well, as per anydice you have a 84% chance of rolling at least a 4, versus 60% for a straight roll, whereas for a 5 it's 75%, versus 50% for a straight roll, and for a 4 it's 64%, versus 40% for a straight roll. So it winds up being about a +25% bonus in the middle of the range.

It's interesting that it's about the same benefit. Advantage on a d20 is a +3 to a +5, or about +15-25%. Advantage on the tens digit of a d100 roll is about +15-25%.

A d20 is just a d100/5.It's interesting that it's about the same benefit. Advantage on a d20 is a +3 to a +5, or about +15-25%. Advantage on the tens digit of a d100 roll is about +15-25%.

That is a good point, and I wondered about that.

If you roll a d4 with advantage (1e DM being generous with wizard HP?), you have a 100% chance of rolling at least a 1 (no change), a 94% (rounded) chance of rolling at least a 2 (versus 75), a 75% chance of rolling at least a 3 (versus 50), and a 44% chance of rolling at least a 4 (versus 25). Here it ranges from 20-25%. So it looks like advantage is somewhat stronger for smaller dice, but the effect is small.

*slightly* different though--the d100 is more granular. Also the mean of a d20 is 10.5, the mean of a d100 is 10.1=50.5/5.

If you roll a d4 with advantage (1e DM being generous with wizard HP?), you have a 100% chance of rolling at least a 1 (no change), a 94% (rounded) chance of rolling at least a 2 (versus 75), a 75% chance of rolling at least a 3 (versus 50), and a 44% chance of rolling at least a 4 (versus 25). Here it ranges from 20-25%. So it looks like advantage is somewhat stronger for smaller dice, but the effect is small.

The probability distribution'sA d20 is just a d100/5.

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Yes, I'm aware. That wasn't very helpful, NPC Thom.A d20 is just a d100/5.

Right. Which is why I assumed advantage would be different, mathematically, on the d10 vs the d20. I wonder if that range of benefit holds for the rest of the dice. At a guess I'd say yes it's probably in that same range as it only moves to +15-25% when you get to a d20. Weird.That is a good point, and I wondered about that.

If you roll a d4 with advantage (1e DM being generous with wizard HP?), you have a 100% chance of rolling at least a 1 (no change), a 94% (rounded) chance of rolling at least a 2 (versus 75), a 75% chance of rolling at least a 3 (versus 50), and a 44% chance of rolling at least a 4 (versus 25). Here it ranges from 20-25%. So it looks like advantage is somewhat stronger for smaller dice, but the effect is small.

The probability distribution's slightly different though--the d100 is more granular.

D&D 5E has an "advantage" concept where instead of rolling 1d20, you roll 2d20 and take the higher. Likewise, disadvantage means rolling 2d20 and taking the lower. How does this affec...

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P(rolling at least n)= 1-((n-1)/20)^2 = 1-(1/400)(n^2-2n+1) = 399/400+(n/200)-(n^2/400).

Compare that with a regular d20 roll, where

P(rolling at least n)= 1-(n-1)/20 = 1-(1/20)(n-1) = 19/20-n/20.

Theoretically we should be able to sub in x for 20, giving us

P(rolling at least n on dx with advantage) = 1-((n-1)/x)^2 = 1-1/x^2 + (2n/x^2)-(n^2/x^2)

P(rolling at least n on dx) = (1-1/x)-n/x.

So the advantage from advantage is [1-1/x^2 + (2n/x^2)-(n^2/x^2)]-[(1-1/x)-n/x].

You can simplify a little to (1/x-1/x^2) + (2n/x^2)-n/x -(n^2/x^2) = ((x-1)/x^2) + ((2n-nx)/x^2) -(n^2/x^2) =

((x-1 + 2x-nx-n^2)/(x^2)) = ((3-n)x-1-n^2)/(x^2).

Where you go with that is your own best guess.

So, for example: 20...

For a given roll of X+, just sum the values for X to sides.

(the visual proof is on an episode of mathologer for this week, tho' it's part of a proof in a different algorithm.

For the d6, the advantage results look like this:

6 B(

5 B(

4 B(

3 B(

2 B(

1 B(

And the visual proof

Advantage | 6 | 5 | 4 | 3 | 2 | 1 |

6 | 6 | 6 | 6 | 6 | 6 | 6 |

5 | 6 | 5 | 5 | 5 | 5 | 5 |

4 | 6 | 5 | 4 | 4 | 4 | 4 |

3 | 6 | 5 | 4 | 3 | 3 | 3 |

2 | 6 | 5 | 4 | 3 | 2 | 2 |

1 | 6 | 5 | 4 | 3 | 2 | 1 |

For the odds of, say, 3+ on advantaged d6 (5+7+9+11)/36 or 32/36, or 8/9; the 2- is (3+1)/36. or 1/9.

The same methodology works for any arbitrary 2 dice.

So, the 10 on d10 is 19/100 Advantage, and 1/100 disad

The 9 is 17 & 3

the 8 is 15 and 5

the 7 is 13 and 7

and so on.

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