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[-] SHOW_ME_YOUR_ASSHOLE@lemm.ee 7 points 1 year ago

Honestly I'm sure this is the best solution. I get that a d4 is the obvious choice for something that should have a 1/4 chance of happening but a d8 with 4 numbers twice would be the most appropriate.

The only downside I can see is that a d8 and a d8/4 would be easy to mix up at first glance.

[-] fushuan@lemm.ee 4 points 1 year ago

Honestly you only need a d20 and a d6. D4? Divide by 5. D8? D20/5 x d20/10. D12? D6xd10/2

MATH BABY

[-] tissek@ttrpg.network 6 points 1 year ago

D8? D20/5 x d20/10

Am I missing something here? Can this even generate 5 or 7?

D20/5 gives [1...4] and D20/10 [1...2], of course assuming whole numbers. Where to get the factors for 5? 5 can be factored only as 5x1 or 1x5 and the 5 cannot be found either in d20/5 or d20/10. Same is true for 7.

And I don't see it happening either if we allow rational numbers. To get 5 we would get the following expressions
5= d~1~20/5 x d~2~20/10 = d~1~20 x d~2~20/50
or 250= d~1~20 x d~2~20
And two d20 multiplied together cannot give us 250.

Math baby?

[-] Malgas@beehaw.org 2 points 1 year ago

You could do something like ((d6-1)*20+d20)/15.

But that's an awful lot of work just to avoid having a d8.

[-] fushuan@lemm.ee 1 points 1 year ago

You are right, in my mind the d20/2 was some sort of iterator over the d20/5, the correct math would be d20/5+(20/5*(d20/10-1)). To get 5 this expresion would be with a 1-5 in the first one and a 11-20 on the second, the first would be 1 (rounded up) , and the second one 4*(2-1), so 5. The idea is that you use the second one to decide how many batches of the full first batch you add to the first one. As if you were rolling a d100 with two d10 but in base 20/5 instead of base 10. It's not actually base 20/5 but that's the idea, one of the dice is the "tens" dice and the other is the "hundreds" dice.

... math baby

[-] tissek@ttrpg.network 3 points 1 year ago* (last edited 1 year ago)

But then do really need the d8? If we toss that in the bin we can go to the universal d60. This one dice will allow us to get
d2 (even/odd)
d3 (d60/20)
d4 (d60/15)
d5 (d60/12)
d6 (d60/10)
d10 (d60/6)
and d12, d15, d20, d30

Base 60 is cool yo!

[-] fushuan@lemm.ee 1 points 1 year ago

that dice would either be really big, or it would just be a ball that would take too long to stop rolling lol... I want it now.

[-] swab148@startrek.website 1 points 1 year ago

There's a d100

[-] match@pawb.social 3 points 1 year ago

I hate math babies. Least favorite type of baby.

Math baby.

[-] SpacetimeMachine@lemmy.world 1 points 1 year ago

Really it should be just using a d/20 itself divided into 5 parts. For instance, 1-4, 5-8, 9-12, etc.

[-] candybrie@lemmy.world 1 points 1 year ago* (last edited 1 year ago)

To keep the same probabilities, you can only reduce and only to one that is a factor. E.g. d20 can be equivalent to d10, d5, d4 and d2.

Multiplying the rolls messes things up. As an example, for d12 as a d6xd2 you have double the chance to roll 2, 4, and 6 and no chance to roll 7, 9, and 11.

You could make the equation a little more complicated (6×(d2-1))+d6 to make it work.

[-] fushuan@lemm.ee 1 points 1 year ago* (last edited 1 year ago)

You are absolutely right, I was thinking d6d2 as: the D2 rolls 1, it's d6. The D2 rolls 2,its 6+d6. That's not what my math said so my bad!

Edit: your equation is what I had in my mind, which is sorta what we do to roll d100.

[-] Malgas@beehaw.org 4 points 1 year ago

You don't even need a special die for this. Just roll a d8 and subtract 4 if it's 5-8. Just like using a d6 as a d3.

[-] Doug@midwest.social 3 points 1 year ago

I always divide by two and round up for d3

[-] BeefPiano@lemmy.world 2 points 1 year ago

You already have d10 and d100 (d00? What do we call the other one?), so there’s precedent for duplicating shapes.

[-] candybrie@lemmy.world 3 points 1 year ago* (last edited 1 year ago)

But if you roll the d00 on accident, you can easily still treat it as a d10. If you roll the d8/4, you can't.

[-] FinalRemix@lemmy.world 2 points 1 year ago

what do we call the other one?

A golf ball?

[-] Doug@midwest.social 2 points 1 year ago

d% is what I usually see

[-] shuzuko@midwest.social 1 points 1 year ago

I've heard it called a tens die or a percentile die. D100 is usually saved for the actual 100-sided die in my experience.

[-] burble@lemmy.dbzer0.com 2 points 1 year ago

I have a "D3" that's just a D6 with two of each.

this post was submitted on 15 Dec 2023
170 points (93.8% liked)

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