Earlier today I was reading a thread where people were talking about some of the more obscure dice, like d30 and d24, and how to replicate them. I hope this was trolling, but there seemed to be a lot of people who just had no idea how random distribution -- one person suggested trying to roll d30 as d10+d20.
What's a good way to explain why different multiple-die-roll methods work to a beginner? The autist in me wants to express it all mathematically, e.g. d100≡(1d10-1)*10+(1d10-1)+1 or d3=modulo(d6, 3)+1, but I know that that's not exactly very intuitive.
anydice.com is a good resource for checking the probability curves of die rolls.
Hudson Watson
There's no helping people who cannot into basic math. Leave them to their diceless freeform hugbox games, eternally arguing about which of two magic ninjas teleports behind the other. But in all fairness, 1d20+1d10 can be the better option depending on what you want to use it for and whether there's an opposition roll.
David Peterson
>d100 as 2d10 >One of the dice is the tens value, the other is the ones value. >If both turn up 0, it counts as 100 instead of zero, so that we have a 100 outcome instead of a zero one. I don't see how that's very difficult.
Juan Reed
>there seemed to be a lot of people who just had no idea how random distributio Yep. Ther's a lot. Math is haaaaarrrrrrd ya know.
Luke Sullivan
And emulating dice with other dice is usually basically that. d36, using 2d6 with one of the dice in the sixes place, and one in the ones place. (heximal numbers) d60, with a d6 in the tens value and a d10 in the ones value.
And similar variations. And for all other values? re-roll when out of range.
Cooper James
>What's a good way to explain why different multiple-die-roll methods work to a beginner? Depends on what you're explaining. Everybody's familiar with 2d6, but it's different than 1d12, even though the possible outcomes are (almost) the same. You just ask them "how many ways is there to roll a 12 on 2d6?" and then "how many ways are there to roll a 7?" and hopefully that should explain it.
Rolling for digits of a large die is probably the best way to express large die in terms of smaller dice. d30 with a d6-as-a-d3 for 0-2 for the tens digit, and a d10 with 0-9 for the ones digit, and then defining a roll of 0-0 as the maximum instead of the minimum because all die rolls on a dn are from 1 to n, instead of 0 to n-1.
Thomas Jones
It may occasionally be more practical to roll a larger die and drop a small number of results as re-rolls, for example, a uniform distribution from 1 to 11 would be easiest to roll with a d12, but with the 12 being a re-roll. You will only need to re-roll less than 10% of the time, and will almost always be done in two rolls.
Luke Wright
>What's a good way to explain why different multiple-die-roll methods work to a beginner?
Write down all possible rolls in their results on a 2d6, like this: 1 1 = 2 1 2 = 3 1 3 = 4 1 4 = 5 1 5 = 6 1 6 = 7 2 2 = 4 2 3 = 5 etc etc
Show him the results. Show that 7 shows up more than the others. Show that 2 and 12 show up the least. At this point he should realize that 2d6 can't replicate 1d12.
Eli James
The other day I was trying to work out which die rolls were possible without resorting to pulling slips of paper out of a bag or re-rolling until you get a number in the right range.
As far as I can tell, there were two main ways to form another evenly-distributed die roll (let's call it dX) from other dice: >Method A: Roll for places. This is how you get 1d100 from 2d10, as well as d60 from d6 and d10 like mentions. You can also use this to get a die roll for any power of the number of faces on the die you're rolling, by using a different number base. Two coinflips becomes d4, three becomes d8, four becomes d16, etc. >Method B: Roll 1dY where Y is a multiple of X, then divide the result by X, take the remainder, and add one. This is how you can get a d3 from a d6 or a d5 from a d10.
I'm writing up some of the ones I found were possible, hold on a sec.
Jordan King
Of course, you can combine these two methods to get more results. "Primitive" means "it's already a real die": >d1: by method B with literally any die >d2: primitive >d3: by method B with d6 >d4: primitive >d5: by method B with d10 >d6: primitive >d8: primitive >d9: by method A with d3 >d10: primitive >d12: primitive >d15: by method A with d3 and d5 >d16: by method A with d4, or with d2 and d8 >d18: by method A with d9 and d2 >d20 primitive >d24: by method A with d2 and d12 >d25: by method A with d5 >d27: by method A with d3 >d30: by method A with d3 and d10, or with d5 and d6 >d32: by method A with d16
The pattern I'm noticing is that you can get any die roll with method B if you have a way to roll for any pair of factors for it.
>captcha: 255 (i.e. 2^8-1)
Robert Lopez
Op being accurate is good and all but most people don't play dnd to do math problems. Its better that its slightly weighted than being a pain in the ass to calculate for every die roll.
Nolan Green
So you should be able to simulate dX for any X whose prime factorization has only 2s, 3s, and 5s. Any multiple of 7, 11, 13, 17, 19, ... can't be emulated using these methods.
Grayson Garcia
Also, anything you can get with a full set of d2/d4/d6/d8/d10/d12/d20 could in theory be replicated by just d6s and d20s.
Blake Edwards
>some nerds are actually bad at math I feel truly sorry for these people
Nathan Edwards
d6es and d10s, even.
Connor Torres
>So you should be able to simulate dX for any X whose prime factorization has only 2s, 3s, and 5s. Pretty much. I'd try and prove that if I weren't terrible at proofs.
>Any multiple of 7, 11, 13, 17, 19, ... can't be emulated using these methods. Geometrically fair d14s do exist, and d22s apparently. This means that multiples of 7 and 11 are also possible.
John Campbell
>Everybody's familiar with 2d6, but it's different than 1d12, even though the possible outcomes are (almost) the same. You just ask them "how many ways is there to roll a 12 on 2d6?" and then "how many ways are there to roll a 7?" and hopefully that should explain it.
This. I used to use this technique to teach basic probability to elementary kids.
Caleb Reed
what the fuck is going on in that picture
Grayson Parker
To immediately prove that 2d6 can't fairly emulate 1d12, ask them how many ways there are to roll a 1 with 2d6.
Juan Harris
crushing fetish?
Jaxon Peterson
basic math is the best way to do it, but you can make it easier to see by showing that something like d6+d6 is not d12 because you can never, EVER roll a 1, and then going into how many combinations there are for numbers like 6, 7 and, 8 while there's only one way to ever get a 12 (and still NO way to get a 1)
> and how to replicate them. theres no excuse in this day and age. 3d printing services are everywhere and they're all over school campuses and many hobbyists own them. it's one of the few parts that even the most fervent hater of the technology would readily admit is a good use of it (since the cost in filament for just one die is probably gonna be less than shipping+handling alone for a single d30 or d24, and even the least precise printer could probably handle making it)
Isaac Anderson
d30 can be done with a d3 and a d10 can't it?
Thomas Adams
Why they keep arguing on mathematical subjects then?
Adam Perez
>Just make snake eyes count as one, bro. It's not like 2 is ever gonna be a success anyway What now?
Benjamin Robinson
Dice "problems" aren't hard. The solution to the D30 problem was given within 3 or so posts. After that, it was just trolls and idiots rolling in and throwing out their ideas.
Works perfectly fine. Just need a D3.
Evan Johnson
so how do you roll a 2 now?
Carson Allen
>What's a good way to explain why different multiple-die-roll methods work to a beginner? Why would you need to? All the beginner needs to know is that you need a number between X and Y, and he needs to roll these dice to give you that number. If you're feelign generous, you tell him, "it needs to be above/below this number."
That's all a beginner needs. To much information scares them. If they ask for more, THEN you give them more.
Noah Lewis
>random distribution I find it worse how many people can't grasp that 2 chances at at 75% roll it is not the same as 1 chance at 150%, it is not guaranteed to happen.
Liam Campbell
>lol just make 2 and 1 count as two bro
Jonathan Kelly
>Why they keep arguing on mathematical subjects then? Because every mouth-breathing neanderthal who's thrown dice and bullshitted houserules for a few months decides that somehow makes them experts at statistics.
Eli Bennett
You know a d6 is a d3 right?
Sebastian Perry
Using the methods defined by you can somewhat easily (if in a convoluted way) make a fair d12 with 2d6. You can use method A to get a d36 then use method B and divide by 3 to get a d12.
Then if you roll a 2 or a 3 that's a 1 (since that gives 3 ways out of 36 to roll a 1), if you roll a 4 that's a 2, etc.
You need to be able to distinguis between the two die for that though (so that 3 4 and 4 3 aren't the same, for example), so you'd need two different colors or something similar. And a chart ideally.
Carson Garcia
Here's an idea gujs
To roll a d20 with a bonus, you just roll and take the face value... unless the value you roll is equal to or less than the value of the bonus, in which case you add 20 to that value.
This means that 1d20(10)+4 would count as 10, whereas 1d20(5)+4 would count as 5, but 1d20(2)+4 would count as 22.
There's an equal number of ways to roll each possible result, and it's less math to do.
Caleb Robinson
I't's also confusing and convoluted, and not actually simpler than just adding the value in the first place.
