FiveThirtyEight: D&D Riddle

[Telonius]

This week's "Riddler Classic (https://fivethirtyeight.com/features/can-you-find-the-best-dungeons-dragons-strategy/)" feature on fivethirtyeight is D&D-related (5th ed, in fact). Have at it, all your crazy mathematicians!

[Ibrinar]

About the Riddler Express I can do a sum of 6 (use 4 1s to limit the possible places for the one in a square to the middle and put a 2 in the middle) is there a lower option?

About the d&d one disadvantage of advantage is the better one for average values, here https://statmodeling.stat.columbia.edu/2014/07/12/dnd-5e-advantage-disadvantage-probability/ this give the probabilities for advantage/disadvantage. for advantage you have an 75.1% chance of getting 11 or higher, which give you an 56.4% chance to have neither of two advantages be below 11. Other way around for advantage of disadvantage. (I guess to do this properly mathematically I would have to show that it isn't weirdly skewed but I did simulate it and got a 13.8 avg for dis of ad, and 7.17 for ad of dis.)

About the extra if I am not doing it wrongfor 14 and higher normal is better than disadvantage of advantage, and for 8 and lower ad of dis is better than normal.
I don't think adof dis is better than dis of ad for any value.

[Khedrac]

About the Riddler Express I can do a sum of 6 (use 4 1s to limit the possible places for the one in a square to the middle and put a 2 in the middle) is there a lower option?
I don't think your proposal would have a unique solution which I understood to be a requirement or this.

About the d&d one disadvantage of advantage is the better one for average values, here https://statmodeling.stat.columbia.edu/2014/07/12/dnd-5e-advantage-disadvantage-probability/ this give the probabilities for advantage/disadvantage. for advantage you have an 75.1% chance of getting 11 or higher, which give you an 56.4% chance to have neither of two advantages be below 11. Other way around for advantage of disadvantage. (I guess to do this properly mathematically I would have to show that it isn't weirdly skewed but I did simulate it and got a 13.8 avg for dis of ad, and 7.17 for ad of dis.)

About the extra if I am not doing it wrongfor 14 and higher normal is better than disadvantage of advantage, and for 8 and lower ad of dis is better than normal.
I don't think adof dis is better than dis of ad for any value.

For the D&D one I brute-forced it in Excel, and you are not doing it wrong.
Adv of Dis and Dis of Adv are indeed mirror distributions
And needing a 14 or better is indeed the point where Dis of Adv becomes better than a single dice.

There is another option he did not consider for having both Advantage and Disadvantage at the same time - roll three dice and discard the highest and lowest. I haven't run the numbers, but it will make for a very centered distribution - both high and low rolls being very unlikely.

[jayem]

About the Riddler Express I can do a sum of 6
That's my easy solution too upper bound, that 3 is insufficient I think is equally trivial.
Anything doable in 5 has to involve "5*1" or "3*1+2" which I think don't work but there might be something clever.

Ah, it's the success we're changing for the AoD DoA dice. I couldn't see why my intuitive feel was dice size dependent. Although not really thought about it.


For any underlying probability of success P:
The probability of succeeding under advantage is PA=2*P - P 2 (You win if you succeed on either dice)
The probability of succeeding under disadvantage is PA=P2 (You win if you succeed on both dice)

by composition of the functions.
PAoD=2*PD - P D2=2*(P2) - (P 2)2 = 2P2[/sub]-P[sup]4
=P (2P-P3)

while
PDoA=PA2=(2*P - P 2)2 = 4P2-2*2*P3+P4
=P (4P-4*P2+P3

Sanity check if P=0 both are trivially 0. if P=1 then Paod=1, Pdoa=1

So PAoD>P if (2P-P3)>1 which graphing is when P>0.618
And similarly for PDoA, when P>0.382

Which I think almost makes sense with the 14 given earlier, when I remember that a D20 starts at 1. I like the symmetry, but feel I might have a sign wrong.

Finally
PAoD > PDoA if:
(2P-P3)>(4P-4*P2+P3
(-6P-2P3+4*P2)>0
(-6-2P2+4)>0 Which is never the case if P is between 0 and 1

[KillianHawkeye]

Does a statistics question based on a fake D&D rule somehow qualify as D&D trivia? Why would anyone care? :smallconfused:

[Peelee]

Does a statistics question based on a fake D&D rule somehow qualify as D&D trivia? Why would anyone care? :smallconfused:

A.) There is no fake D&D rule presented, it's explicitly listed as a hypothetical scenario in which an existing rule is changed.
2.) Who called it D&D trivia? I can't find that claim in either the article or in this thread.
iii.) People interested in puzzles and/or math/statistics would care because it fits their interest.

[Telonius]

The answer (https://fivethirtyeight.com/features/somethings-fishy-in-the-state-of-the-riddler/)is in:



As the puzzle’s submitter, Emma Knight, observed, you didn’t need to work out all the cases. Both advantage of disadvantage and disadvantage of advantage (try saying that 10 times fast!) required a total of four rolls. We can label those four rolls in (nonstrictly) increasing order 1, 2, 3 and 4. At this point, there are three equally likely pairings of these numbers, where each pair represents the two rolls of an advantage or disadvantage: (12)(34), (13)(24) and (14)(23).

For these three scenarios, advantage of disadvantage produced results of 3, 2 and 2, for an average of about 2.33. Meanwhile, disadvantage of advantage produced results of 2, 3 and 3, for an average of about 2.67. Finally, rolling a single die would result in the average of all four numbers, or 2.5. That meant you would have the highest expected roll with disadvantage of advantage.

But for Riddler Nation, this solution was just the beginning...


Much math follows.