D&D General Video on the math behind advantage (and Elven Accuracy)

[EzekielRaiden]

Man 13.8 is lower than you’d think reading folks talk about advantage being a big deal.

As noted, this is an issue of considering mean difference, rather than typical difference. If we think of advantage from the perspective of the lowest die, then if you rolled a 1 advantage will (on average) give you +9.5. If the "lowest" die is 20 (meaning you double-crit), your average gain is zero. Over the range of values most likely to occur, however, the net change in probability is about 20-30 percentage points, hence, most people gloss the benefit as roughly +5.

Wow- he’s really charismatic! That was a great video. I lasted until hypercubes and then had to bail. I’m barely functional in a 3D world. Try as I might, I can’t grok 4D!

Oh yeah, Matt's great. I'm a big fan of both of the presentations he's given at the Royal Institution, on his two books, Things To Make And Do In The Fourth Dimension and Humble Pi: A Comedy of Maths Errors. Sadly I haven't read either book yet but someday, if I have actual money, I might snag a copy of each.

I love Matt Parker. He's awesome.

Indeed.

 

[Olaf the Stout]

His stand-up is pretty good, too. If you’re into spreadsheets.

I have spreadsheets for so much of my stuff!

 

[overgeeked]

Interesting stuff, but I think we can all agree that the most valuable takeaway here is that the average result of rolling an infinity sided die with advantage is 2/3 (roughly 0.667).

I watched that earlier today, at least up until he started talking about roll 3, keep 1.

It is really interesting that the average result of roll 2, keep the highest 1 will always be 2/3 of the max result of that die type.

 

[Charlaquin]

I watched that earlier today, at least up until he started talking about roll 3, keep 1.

It is really interesting that the average result of roll 2, keep the highest 1 will always be 2/3 of the max result of that die type.

Well, 2/3 if the max result plus 0.5. And If you’d kept watching, it turns out the average result of roll 3, keep the highest is always 3/4 of the max result plus 0.5. And the average of roll 4, keep the highest is always 5/6 of the max result plus 0.5. And we know that the average result of a single die roll is always 1/2 of the max result plus 0.5. So, he conjectures (though doesn’t prove) that the average result of rolling an n sided die m times and keeping the highest is always n*(m/(m+1)+0.5)

 

[Paul Farquhar]

That's assuming the dice is numbered with the complete set of integers from 1 to n.

But it is provable, it was probably just too long for video.

 

[Charlaquin]

That's assuming the dice is numbered with the complete set of integers from 1 to n.

Yes.

But it is provable, it was probably just too long for video.

Oh, cool!