EBOK at Thu, 24 Oct 2024 09:22:18 UTC No. 16446929

The epitome of today's science

Anonymous at Thu, 24 Oct 2024 09:22:18 UTC No. 16446930

>>16446928
You can roll multiple dice by the way if you choose.

Anonymous at Thu, 24 Oct 2024 09:26:41 UTC No. 16446931

Use the d8, if you get 8 just roll again until you get something between 1-7

Anonymous at Thu, 24 Oct 2024 09:28:21 UTC No. 16446932

>>16446931
Yeah I thought of that but is there a way to do it without having to re-roll if a certain thing happens?

Anonymous at Thu, 24 Oct 2024 09:35:41 UTC No. 16446938

>>16446932
I don't think so, you can't make a multiple of 7 with these sides, nor can you do something like taking the sums of multiple dice cause then the distribution won't be uniform

EBOK at Thu, 24 Oct 2024 09:37:05 UTC No. 16446939

Derp. I'm posting in a retarded onions thread.

Anonymous at Thu, 24 Oct 2024 09:40:17 UTC No. 16446942

>>16446939
Quit playing grab-ass and say something of substance genius.

EBOK at Thu, 24 Oct 2024 09:40:25 UTC No. 16446943

MOVE ALONG with this gay shit, stop making us focus on such wackery

Anonymous at Thu, 24 Oct 2024 09:41:56 UTC No. 16446946

My idea / train of thought is if you rolled a D20 7 times and calculated the total value and have 7 brackets. Having got to thinking of the math /probability of it though

for example 1-20 = 1
21-40 = 2
41-60= 3

etc.

Anonymous at Thu, 24 Oct 2024 09:43:55 UTC No. 16446949

>>16446946
However it wouldnt work because the minimum roll would be 7 obviously but thats just my idea. Not sure how to solve this problem

Anonymous at Thu, 24 Oct 2024 09:49:12 UTC No. 16446953

Perhaps if you roll a certain number of a specfic or multiple different dice x times, there is a value for example the minimum, maximum sum, chance of a range of values etc equal to 1/7 or 0.142857 repeating

Anonymous at Thu, 24 Oct 2024 10:06:14 UTC No. 16446970

>>16446949
It wouldn't be uniform I don't think.

Anonymous at Thu, 24 Oct 2024 10:10:26 UTC No. 16446973

Just roll a normal die and add 1

Anonymous at Thu, 24 Oct 2024 10:23:13 UTC No. 16446985

>>16446970
Yeah well the solution for 9 which is another prime number is easy. Roll 2 d6, you get 36 which is divisible by 9. As for 7 idk.

Anonymous at Thu, 24 Oct 2024 10:32:03 UTC No. 16446991

>>16446985
9 is not a prime number...

Anonymous at Thu, 24 Oct 2024 10:35:43 UTC No. 16446994

>>16446991
>.
true its not. idk w.hat i was thinking, either way i meant for 9. 430am here

Anonymous at Thu, 24 Oct 2024 10:37:44 UTC No. 16446995

>>16446994
You under some kind of time constraint to solve this?

I've been probing Claude with questions about this for the last 30 minutes and we can't figure it out, even doing funny stuff like multiplying the outcomes of dice roles adding then up and taking modulos

Anonymous at Thu, 24 Oct 2024 11:00:16 UTC No. 16447026

Roll two d6, assign the pairs of values through (1,1) to (1,5) in groups of five to the seven outcomes, discard (6,6).

Anonymous at Thu, 24 Oct 2024 11:04:26 UTC No. 16447031

>>16447026
Ok, but is it possible without rejection sampling?

Anonymous at Thu, 24 Oct 2024 11:16:08 UTC No. 16447041

>>16447031
Pretty sure that's a no, boss. You can get arbitrarily close, but the only tools you have are multiplying things by numbers that aren't seven, and you ultimately want to get to a number that's divisible by seven.

Anonymous at Thu, 24 Oct 2024 11:20:35 UTC No. 16447045

>>16446995
No not under any time constraint just can't sleep this week.
I just dont think rolling an 8 is a satisfactory answer since you can technically do that for any single number.

Is there no multiple of 7 that can be created using 4,6,8,12,20 in any combination ?

Anonymous at Thu, 24 Oct 2024 11:45:00 UTC No. 16447057

>>16446932
I don’t think so. There’s no regular polyhedron with a multiple of seven sides. Mixes of different shapes like can get multiples of seven; ex. Concave polyhedrons like truncated octahedrons or cuboctahedron have 14 sides, and the convex tetrahemihexahedron has 7 sides, but because all the sides are different shapes they’re going to have different probabilities of being landed on.

The only thing I could think of would be maybe there’s some weird fucking convex shape you could make where the probabilities of the rolls work out that if you make so many of the sides 1s, 2s, 7s, etc. it works out that the probabilities are the same, but I’d say it’s very unlikely.

Anonymous at Thu, 24 Oct 2024 11:46:34 UTC No. 16447059

You could roll a d12, count 1 or 2 as "one", 3 or 4 as "two", etc. Then whenever you get 11, or 12, roll 2d4 and count 1 or 2 as "six" and 3 or 4 as "seven".

Anonymous at Thu, 24 Oct 2024 11:49:47 UTC No. 16447062

>>16447045
You could roll 7d4 and reassign the value based on the total:
1-4: 1
5-8: 2
9-12: 3
13-16: 4
17-20: 5
21-24: 6
25-28: 7
You could also do the same with 7 of any other regular die, but it can’t be done with one die without rerolls.

Anonymous at Thu, 24 Oct 2024 11:51:57 UTC No. 16447066

>>16447062
rolling 7 d4 the minimum value would be 7, so this wouldnt work

Anonymous at Thu, 24 Oct 2024 11:57:47 UTC No. 16447070

>>16447066
shit, yeah, i didn't think that through, and come to think of it the different sums wouldn't have equal probabilities either

fuck it's too early for this shit

Anonymous at Thu, 24 Oct 2024 12:04:56 UTC No. 16447075

>>16446928
Flip a coin seven times. Increment count if and only if you flip heads. There you go..any other questions

Anonymous at Thu, 24 Oct 2024 12:06:19 UTC No. 16447078

>>16446928
The problem as you've phrased it is trivially solvable with 2 d6s:

1. Roll the first d6
2. Add the other d6 mod 2.

This will give a maximum value of 7 and a minimum value of 1. However I think there's unstated constraints to this problem, as you've shown with your responses here:

1. Each integer should have a probability of 1/7 occuring; and
2. There are no rerolls based on the outcome of a previous roll.

Ultimately I think this is impossible because ultimately 7 is a prime number, but I'm thinking about ways to prove this.

Anonymous at Thu, 24 Oct 2024 12:09:55 UTC No. 16447081

>>16447075
This has a normal distribution

Anonymous at Thu, 24 Oct 2024 12:23:45 UTC No. 16447090

>>16447059
Then the probability of getting "one" is 1/6...

Anonymous at Thu, 24 Oct 2024 12:25:15 UTC No. 16447092

>>16447078
I mean obviously the problem is totally trivial if we don't care about uniform probabilities, just roll a d8 and map 7,8 to 7.

Anonymous at Thu, 24 Oct 2024 12:26:31 UTC No. 16447094

>>16447078
Clever thought, but unfortunately not a constant distribution; 1 and a 7 will each have a 3/36 probability, while 2-6 will have a 6/36 probability.

Anonymous at Thu, 24 Oct 2024 13:08:00 UTC No. 16447121

sum(roll d6 7 times) % 7 + 1

Anonymous at Thu, 24 Oct 2024 13:12:45 UTC No. 16447127

Here's a leftfield idea: there are 14 "e"s if you write out the numbers one to twelve in English. Maybe you can do something with that.

Anonymous at Thu, 24 Oct 2024 14:11:44 UTC No. 16447164

>>16447121
Not uniform unfortunately

Anonymous at Thu, 24 Oct 2024 14:12:50 UTC No. 16447165

>>16447045
>Is there no multiple of 7 that can be created using 4,6,8,12,20 in any combination
your range of outcomes is the cartesian product of the dice you roll

Anonymous at Thu, 24 Oct 2024 14:52:43 UTC No. 16447226

>>>>16446928
Flip three coins.
000 -> 1
001 -> 2
010 -> 3
011 -> 4
100 -> 5
101 -> 6
110 -> 7
111 -> Reroll all three coins.

Anonymous at Thu, 24 Oct 2024 14:54:25 UTC No. 16447230

>>16447090
Welp don't mind me, I'm retarded.

Anonymous at Thu, 24 Oct 2024 15:27:49 UTC No. 16447283

>>16447226
Halting Problem

Anonymous at Thu, 24 Oct 2024 15:45:08 UTC No. 16447306

>>16446928
https://leetcode.com/problems/implement-rand10-using-rand7/description/

Anonymous at Thu, 24 Oct 2024 20:05:08 UTC No. 16447785

>>16446928
Make a 7-sided dreidel.

Anonymous at Thu, 24 Oct 2024 20:11:32 UTC No. 16447791

>>16447785
>answer the question with a retarded non sequitur about imagnary jew shit
thanks retard

Anonymous at Thu, 24 Oct 2024 20:16:42 UTC No. 16447795

>>16447031
It is not possible without rejection sampling.

If you roll any number of platonic solids, you 4^A*6^B*8^C*12^D*20^E equiprobable outcomes. No number 4^A*6^B*8^C*12^D*20^E is a multiple of 7. So no matter how you assign the different outcomes to 7 different results, you will never get an equiprobable distribution of 1-7. You can either approximate it with enough entropy that the nonuniformity of your result distribution is small enough to not matter, or you can use rejection sampling like >>16446931.

Anonymous at Thu, 24 Oct 2024 20:24:54 UTC No. 16447807

>>16447795
You can roll one d6 and give the answer as max(7[today is sunday], roll)

Anonymous at Thu, 24 Oct 2024 20:38:57 UTC No. 16447818

>>16447164

it is

map(n -> count(x -> sum(rand(1:6, 7)) % 7 + 1 == n, 1:10^9), 1:7)

142860978
142854898
142864106
142846428
142847119
142845049
142838835

Anonymous at Thu, 24 Oct 2024 20:43:09 UTC No. 16447819

>>16447818
Sum one d6 7 times mod 7 is also correct.

Anonymous at Thu, 24 Oct 2024 20:49:10 UTC No. 16447826

>>16447819
It's not. 6^7 is not a multiple of 7, so the distribution cannot be uniform.

Anonymous at Thu, 24 Oct 2024 20:54:57 UTC No. 16447837

>>16447826
True, per >>16447165

Anonymous at Thu, 24 Oct 2024 20:58:56 UTC No. 16447842

>>16447819
You get 0 39990/279936 of the time. You get 1-6 39991/279936 of the time

Anonymous at Thu, 24 Oct 2024 22:02:44 UTC No. 16447903

>>16446928
Roll a d4 3 times.
Let T be the number you're drawing.

Start with T=0.
If the first roll gives 3 or 4, add 1 to T
If the second roll gives 3 or 4, add 2 to T
If the third roll gives 3 or 4, add 4 to T.

Restart the procedure until you get T>0.

Anonymous at Thu, 24 Oct 2024 22:28:46 UTC No. 16447948

>>16447791
OP asked if fair odds for a single seven-sided polyhedron are doable - the answer is yes, but only by imposing selection rules (ex. 1d8, re-roll 8s)

OP later stipulated whether there was a way to do it with a single die without re-rolls, which is where the conversation is now, and so far the answer seems to be no (all of the modulus solutions break the condition for constant probability distributions).

Using a heptagonal pyramid instead of a regular polyhedron presents a possible solution to the problem - instead of a die being thrown, the 'top' is spun, and (if properly balanced), will land on one of the seven sides with a constant probability distribution.

Only other option would be to find combination of die rolls where the number of microstates can be divided into seven equal multiplicities without any overlap of resulting macrostates.

Anonymous at Thu, 24 Oct 2024 22:30:41 UTC No. 16447950

>>16447948
or >>16447807

Anonymous at Thu, 24 Oct 2024 23:19:10 UTC No. 16447976

>>16446928
:D

Anonymous at Thu, 24 Oct 2024 23:24:09 UTC No. 16447977

>>16447791
kek, this guy always seethes if you mention the small hats

Anonymous at Thu, 24 Oct 2024 23:26:57 UTC No. 16447984

>>16447977
>i'm afraid say jew but you're the one seething
nice cope

Anonymous at Thu, 24 Oct 2024 23:36:08 UTC No. 16448000

>>16447984
>i'm afraid to say kike, but i'll have a fucking shit fit over someone saying dreidel
nice seethe

Anonymous at Fri, 25 Oct 2024 00:17:15 UTC No. 16448053

>>16448000
>t. crying kike seething because no one cares about your fake dreidel hate crimes
wassup rabbi

Anonymous at Fri, 25 Oct 2024 01:13:33 UTC No. 16448142

>>16447795 has the right idea - because 7 is prime there's no combination of dice where the multiplicity will be a factor of 7, which means there's no way of working out a fair roll system without invoking *some* kind of selection rule

Anonymous at Fri, 25 Oct 2024 01:21:06 UTC No. 16448148

>>16446973
Fucking vaccines.

Anonymous at Fri, 25 Oct 2024 01:33:48 UTC No. 16448168

>>16447283
Not relevant in practice:
Since the wave functions describing the positions of all your particles is not compact, you could pop out of existence just like that, with all your particles spread all over the universe. Is not impossible, is just extremely improbable as keep hitting 111 forever.

Anonymous at Fri, 25 Oct 2024 02:54:07 UTC No. 16448266

>>16448168
You can do better.

Anonymous at Fri, 25 Oct 2024 03:11:11 UTC No. 16448295

>>16446928
Roll a 6 sided die. If it comes up 6 then roll again. 1-3 is 6 and 4-6 is 7. Boom seve sided die with equal weights.

Anonymous at Fri, 25 Oct 2024 06:29:27 UTC No. 16448482

>>16446938
I am exhausted so I don't know if this will work, or if I got the odds right.

Use a 3 element array that will be populated with 1's and 0's to represent the binary numbers from 0-7. There are 12 1's and 10 0's that need to be randomly distributed to randomly generate 0-7.

The first roll determines the position we start in. If the first roll is 6 it is a 1 to balance the 1's and 0's. From there move through the array populating it with 1/2 rolls. If the binary number is 0 do another 1/3 roll to replace a 0 with a 1.

Might fuck around in python tomorrow to see what this actually converges to.

Anonymous at Fri, 25 Oct 2024 06:31:37 UTC No. 16448486

>>16448482
nevermind, same number of 1's and 0's all the population rolls are 1/2
am retarded

Anonymous at Fri, 25 Oct 2024 08:18:08 UTC No. 16448600

Roll a 7 million sided die coloured in 7 colors (because it looks pretty), ignore any non-uniform outcomes past the 6th significant digit.

Anonymous at Fri, 25 Oct 2024 09:03:53 UTC No. 16448637

>How to generate a random number between 1/7 using only platonic solids/platonic solid dice
Get a large cube. Draw a circle with 7 equal subdivisions on the top of it, labeled 1-7. Spin the cube around the circle. Drop a tetrahedron in the center of it. Whichever section the tetrahedron slides off the cube in is your number.

You're thinking in too many dimensions.

Anonymous at Fri, 25 Oct 2024 09:04:56 UTC No. 16448638

How hard is it to construct a seven sided shape?

Anonymous at Fri, 25 Oct 2024 09:26:17 UTC No. 16448656

>>16446928
Roll a d8. Discard an 8 (roll again). Everything else is just German engineering.

Anonymous at Fri, 25 Oct 2024 09:52:47 UTC No. 16448669

>>16446928
You need a 4 sided, a 2 sided and a 1 sided die.

Anonymous at Fri, 25 Oct 2024 10:41:10 UTC No. 16448708

>>16446928
roll the d4 7 times and divide by 4.

Anonymous at Fri, 25 Oct 2024 10:45:10 UTC No. 16448710

>>16448708
Smallest number you can get is 7

Anonymous at Fri, 25 Oct 2024 10:48:56 UTC No. 16448713

>>16448710
ups sorry, I'm a bit tired, take the remainder mod 7 + 1

Anonymous at Fri, 25 Oct 2024 10:50:48 UTC No. 16448716

>>16446928
Roll d4
Roll d6 ceiling[#/2]
How the fuck did nobody say this yet

Anonymous at Fri, 25 Oct 2024 11:30:53 UTC No. 16448741

>>16448713
Is this uniform?

Anonymous at Fri, 25 Oct 2024 12:08:15 UTC No. 16448778

>>16448716
Because it excludes a roll of 1 and it’s not a constant distribution.

Anonymous at Fri, 25 Oct 2024 12:24:41 UTC No. 16448796

>>16448638
>How hard is it to construct a seven sided shape?
Not at all
>How hard is it to construct a seven sided shake where each side has an equal probability of being landed on?
Very. There’s only like three regular polyhedrons with faces that are a multiple of seven, and none of them have equal numbers of differing polygons which means you can’t distribute the numbers in a way that’s likely to get a constant distribution. And because seven is prime you can’t just break up the odds on any single die or even any combination of dice to get constant odds of 1/7.

The only options I can think of stretch the definition of rolling a die - you could construct a seven-sided top and spin it, or you could construct a long seven-sided prism, determining the minimum length to ensure that it always tips over when perturbed if it happens to land on either of the non-rectangular faces.

Anonymous at Fri, 25 Oct 2024 12:40:59 UTC No. 16448821

>>16448741
Doesn't look like it

Anonymous at Fri, 25 Oct 2024 12:47:24 UTC No. 16448834

>>16448821
What I thought.

I think this problem is impossible without accepting rejection sampling or approximation.

Anonymous at Fri, 25 Oct 2024 12:50:56 UTC No. 16448842

>>16448834
Yeah, 7 throws of a d20 work much better

Anonymous at Fri, 25 Oct 2024 13:23:13 UTC No. 16448899

>must have even number of sides
bro just pick a 14 sided dice
map the numbers such that each of the 14 represents 2 of the 7 values
1 - 1
2 - 1
3 - 2
4 - 2
5 - 3
6 - 3
....
13 - 7
14 - 7

I don't see how this is a hard problem at all

Anonymous at Fri, 25 Oct 2024 13:31:05 UTC No. 16448916

>>16446928
Since nobody as far as I can see has offered a proof that this is impossible* (though many have surmised that it is), I’ll attempt to do so. These dice have uniform outcomes of all 1/4, 1/6, 1/8, 1/12, and 1/20, and for each outcome you can roll another die of your choice (or not) to subdivide that outcome into 4,6,.. equally likely outcomes. You can also group these outcomes together as in your example, adding the probabilities together. The problem then amounts to multiplying these probabilities together and adding them to get 1/7, which is clearly impossible as in the addition of rational numbers, the denominator of the sum (after simplification) has to divide the product of the denominators of the summands. In this case, the outcomes are always a reciprocal of a product of 4,6, etc, so the denominator of a sum of some outcomes must be a factor of a product of these numbers. Since 7 is prime, if it divides a product of these numbers then it must divide one of them, but it doesn’t, so you can’t generate any number p/7 for p that isn’t divisible by 7.

However, this is only if you require that you can only make finitely many subdivisions of the outcomes; otherwise, as was pointed out here >>16446931 it is possible to use platonic solid dice to generate a uniform distribution with 7 outcomes, but only if you’re okay with having no limit on how many times you have to roll dice. I believe the expected number of dice rolls for that option is about 1.143, so I suspect that this is the most efficient way to do this in terms of number of dice rolls.

Anonymous at Fri, 25 Oct 2024 13:38:19 UTC No. 16448932

>>16448899
There is no regular 14-sided polyhedron with an equal number of different polygon faces, which means your probabilities aren’t going to be the same.

Anonymous at Fri, 25 Oct 2024 13:46:09 UTC No. 16448949

>>16448932
The faces don’t have to be regular polygons for the die to be fair, there just has to be symmetry between faces—that is, it need only be isohedral. If you look up “14-sided dice,” the shape where each face is a kite is what anon is talking about. Granted, it breaks OP’s requirement of the dice being Platonic solids, but otherwise it works just fine.

Anonymous at Fri, 25 Oct 2024 13:55:44 UTC No. 16448964

>>16448949
This. A heptagonal bipyramid is the closest thing to a conventional die that would meet OPs requirements.

Anonymous at Fri, 25 Oct 2024 13:59:41 UTC No. 16448970

>>16448669
This gives 8 outcomes not 7

Anonymous at Fri, 25 Oct 2024 14:21:49 UTC No. 16449000

>>16448964
A 14-sided bipyramid worka except for a tiny boggling detail that it can’t land with any one face pointing up because of the odd axial symmetry, you’d have to either base the result of the roll on the side that landed face down or else make it a 28-sided bipyramid and divide the faces into groups of four adjacent faces such that the common vertex of those four surfaces points up when the die lands on a face opposite it.

Anonymous at Fri, 25 Oct 2024 15:19:59 UTC No. 16449057

>>16446928
Interesting problem.

I believe, thanks to >>16448916, that I have a proof that it is impossible.

First, though, I've found a way to get pretty damn close to a completely random number, however I'm getting an itch that it's impossible to do this, though I haven't proved it yet.

Roll 4 d6 and add their values. Call the sum r, which has a min and max value of 4-24.

Then the value ((r-4) mod 7) + 1 is a value from 1-7

However, the probabilities of getting each value are not all equal. It is pretty damn close, however, assuming the program I wrote to verify is correct

For all values except 4, there are 185 rolls each that give the respective value. However, for 4, there are 186 rolls, so the probability is very slightly skewed towards 4

So that gets pretty damn close. But it gives me an idea to prove that it's impossible to get equal probabilities for all values

Notice that (185 * 6) + 186 = 1296, where 6^4 = 1296

In other words, it seems apparent to me that if we are to have an equal distribution of probabilities across the 7 values, the number of possible rolls must also be divisible by 7

So to disprove this, take k1,k2,...,k5 to be the number of dice rolled for each platonic solid respectively

If the product (4^k1)(6^k2)(8^k3)(12^k4)(20^k5) can be shown to NOT be divisible by 7, then we can prove that there is no possible selection of dice/rolls that could give an equal chance of getting any value 1-7 via any modification

I did some work on paper and (unless I made an algebra mistake) the product reduces to (2^m)(3^n)(5^o) where

m = 2(k1) + k2 + 3(k3) + 2(k4) + 2(k5)
n = k2 + k4
o = k5

The final part I realized only after >>16448916 said:
>Since 7 is prime, if it divides a product of these numbers then it must divide one of them

Since (2^m)(3^n)(5^o) is a product of 2, 3, and 5, and 7 divides none of them, then there is no possible combination of platonic dice that can produce an equal probability of getting values 1-7

Q.E.D. (I hope)

Anonymous at Fri, 25 Oct 2024 15:21:10 UTC No. 16449058

>>16449057
Ignore the part where I said
>however I'm getting an itch that it's impossible to do this, though I haven't proved it yet.

I forgot to edit it out after I completed the proof, again thanks to >>16448916 for the nugget of wisdom about 7 not dividing a product of numbers it cannot divide.

🗑️ Anonymous at Sat, 26 Oct 2024 14:45:14 UTC No. 16450658

>>16446928
(d20+1&0 faced coin) mod 7

Anonymous at Sat, 26 Oct 2024 19:48:58 UTC No. 16451129

>>16446928
It's too simple.
1.Roll a dice twice.
2.If the sum is not 36, divide the sum by 7 and add 1 to the residue.
3.If the sum is 36, return to 1.

Anonymous at Mon, 28 Oct 2024 05:31:33 UTC No. 16453222

>>16446931
It's not an answer because you can keep rolling 8.

Anonymous at Mon, 28 Oct 2024 05:48:06 UTC No. 16453230

>>16446928
There's no easy solution

Anonymous at Mon, 28 Oct 2024 06:19:37 UTC No. 16453262

>>16446928
It's impossible without rerolls
I don't recall the proof, but it is provable

Anonymous at Mon, 28 Oct 2024 07:00:53 UTC No. 16453290

Throw a d4. Draw that number from a stack of d6,d8 and 2d12, and then throw those. You end up throwing d6 or d6+d8 or 6+8+12 or 6+8+12+12. This process has 84=7*12 outcomes. They aren't equally likely, so this b84 (b for biased) may need to be repeated a few times.

Anonymous at Mon, 28 Oct 2024 07:51:34 UTC No. 16453319

>>16453222
Yes but practically the probability of doing so is miniscule.

Do not respond to me with some infinitist mathematical jibber jabber chicanery boy

Anonymous at Mon, 28 Oct 2024 07:54:24 UTC No. 16453323

>>16446928
Roll two four-sided die. One numbered 1-2-3-4, the other labelled 0-1-2-3, then sum them.

Anonymous at Mon, 28 Oct 2024 08:48:59 UTC No. 16453350

>>16453323
1 = 0&1
2 = 1&1, 0&2
3 = 0&3, 1&2, 2&1
...
You see the problem?

Anonymous at Mon, 28 Oct 2024 08:51:53 UTC No. 16453351

>>16446928
Put a dice in a lava lamp and have it float to the top.

Anonymous at Mon, 28 Oct 2024 15:42:43 UTC No. 16453626

>>16451129
>2.If the result is not (1 -> 6)*.
>3.If the result is (1 -> 6)*.

Anonymous at Tue, 29 Oct 2024 05:15:56 UTC No. 16454473

>>16446928
roll a d8 and a d6. divide the total by 2.

Anonymous at Tue, 29 Oct 2024 07:42:23 UTC No. 16454555

>>16454473
Did you read the thread at all?