this post was submitted on 15 Dec 2023
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[–] [email protected] 28 points 11 months ago (6 children)

d8s with duplicate sides gang

[–] BleakBluets 22 points 11 months ago* (last edited 11 months ago) (1 children)

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

[–] [email protected] 23 points 11 months ago (1 children)

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.

[–] [email protected] 2 points 11 months ago (1 children)

Would spinning the coin get rid of that bias?

[–] [email protected] 2 points 11 months ago (1 children)

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

[–] [email protected] 2 points 11 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.

[–] [email protected] 10 points 11 months ago (1 children)

D12 with numbers in triplicate appreciators over here

[–] [email protected] 5 points 11 months ago

D100 with 25 of each option lover right here

[–] [email protected] 4 points 11 months ago (3 children)

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.

[–] [email protected] 4 points 11 months ago (2 children)

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

MATH BABY

[–] [email protected] 6 points 11 months ago (2 children)

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?

[–] [email protected] 1 points 11 months ago (2 children)

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

[–] [email protected] 3 points 11 months ago

I hate math babies. Least favorite type of baby.

Math baby.

[–] [email protected] 3 points 11 months ago* (last edited 11 months ago) (1 children)

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!

[–] [email protected] 1 points 11 months ago (1 children)

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.

[–] [email protected] 1 points 11 months ago

There's a d100

[–] SpacetimeMachine 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 1 points 11 months ago* (last edited 11 months ago) (1 children)

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.

[–] [email protected] 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 2 points 11 months ago (4 children)

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

[–] candybrie 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 2 points 11 months ago

what do we call the other one?

A golf ball?

[–] [email protected] 2 points 11 months ago

d% is what I usually see

[–] [email protected] 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.

[–] [email protected] 1 points 11 months ago

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

[–] [email protected] 2 points 11 months ago (2 children)

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.

[–] [email protected] 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 1 points 11 months ago

Altered Magic 8-ball Syndicate.

[–] Aremel 15 points 11 months ago (2 children)

Just use a dreidel, ya dingus.

For your health.

[–] [email protected] 7 points 11 months ago

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

[–] [email protected] 2 points 11 months ago

This is the way

[–] gmtom 10 points 11 months ago (1 children)

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

[–] [email protected] 13 points 11 months ago (1 children)

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

[–] [email protected] 4 points 11 months ago

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