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[-] match@pawb.social 28 points 11 months ago

d8s with duplicate sides gang

[-] BleakBluets@lemmy.world 22 points 11 months ago* (last edited 11 months ago)

Flipping a coin two times and reading the result as binary gang. (Don't actually do this, coins aren't as fair)

[-] ryven@lemmy.dbzer0.com 23 points 11 months ago

They're only off by about 1% and the bias depends on which side was up, it's not that bad. I wouldn't expect most inexpensive dice to be substantially fairer than that.

[-] PapaStevesy@midwest.social 2 points 10 months ago

Would spinning the coin get rid of that bias?

[-] ryven@lemmy.dbzer0.com 2 points 10 months ago

I have not read any stats on coin spinning, so I don't know!

[-] PapaStevesy@midwest.social 3 points 10 months ago

I suppose it's basically flipping turned 90 degrees, so it might still be affected but which face is forward when being spun.

[-] milkisklim@lemm.ee 11 points 11 months ago

D12 with numbers in triplicate appreciators over here

[-] Khrux@ttrpg.network 6 points 11 months ago

D100 with 25 of each option lover right here

[-] SHOW_ME_YOUR_ASSHOLE@lemm.ee 6 points 11 months 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.

[-] Malgas@beehaw.org 4 points 11 months 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 10 months ago

I always divide by two and round up for d3

[-] fushuan@lemm.ee 4 points 11 months 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 11 months 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 11 months 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 11 months 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

[-] match@pawb.social 3 points 10 months ago

I hate math babies. Least favorite type of baby.

Math baby.

[-] tissek@ttrpg.network 3 points 10 months ago* (last edited 10 months 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 10 months 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 10 months ago

There's a d100

[-] SpacetimeMachine@lemmy.world 1 points 11 months 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 11 months ago* (last edited 11 months 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 11 months ago* (last edited 11 months 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.

[-] BeefPiano@lemmy.world 2 points 11 months 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 11 months ago* (last edited 11 months 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 11 months ago

what do we call the other one?

A golf ball?

[-] Doug@midwest.social 2 points 10 months ago

d% is what I usually see

[-] shuzuko@midwest.social 1 points 11 months 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 11 months ago

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

[-] MyTurtleSwimsUpsideDown@kbin.social 2 points 11 months ago

I’d actually like to see a d8/d4 hybrid. Basically take a caltrop d4, snip a bit off the ends to make a truncated tetrahedron. You’ll then have 4 large hexagonal faces and 4 small triangular ones. Put the numbers on the triangles. If it lands upside down, then it is just house rules whether to use the bottom face or to reroll. Or just number the large faces too.

It’s a similar concept to the round safety d4s; just less… round.

[-] Khrux@ttrpg.network 2 points 11 months ago

Wilder option which is definitely hard to mass produce:

A truncated tetrahedron like this but with wireframe corners, so it rolls like a D4 but there is a clear upward face.

[-] RizzRustbolt@lemmy.world 1 points 10 months ago

Altered Magic 8-ball Syndicate.

[-] Aremel@lemmy.world 14 points 10 months ago

Just use a dreidel, ya dingus.

For your health.

[-] SHOW_ME_YOUR_ASSHOLE@lemm.ee 7 points 10 months ago

I've never actually had to roll a d4 on a table. Baldur's Gate handles it for me.

[-] NkdFstZoom@ttrpg.network 2 points 10 months ago

This is the way

[-] gmtom@lemmy.world 9 points 10 months ago

I don't get what this means, roll on top or bottom of what?

[-] TheGreenGolem@lemm.ee 11 points 10 months ago

There are 2 kinds of D4 dice. The result can be on the bottom or on the top of the "piramid".

[-] myrrh@ttrpg.network 3 points 10 months ago

...well, there are more than two kinds of d4s, but there are two popularly-styled tetrahedral d4s...

[-] mobius_slip@beehaw.org 4 points 10 months ago

GURPS for the win, bell curves are the way to go.

[-] acockworkorange@mander.xyz 1 points 10 months ago

FATE will still get ya!

[-] TQuid@beehaw.org 2 points 11 months ago

Arch d4 gang

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

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