>>1557818
>Two of the dice are guaranteed to roll 31 each time (chosen randomly) and the remaining four can be any number 0 through 31.
This means you can disregard those 2 dice and just rephrase the questions as such:
>What are the odds that I can roll one of the dice with a 31?
>Two dice?
>Three dice?
>All four?

>>1557829

Doing the easiest (rephrased) one first:
Chance of all four getting 31 is: (1/31)^4 = 0.00000108
>>1557818
If anyone is curious about the image in OP here's a video about it: https://www.youtube.com/watch?v=i4EFkspO5p4

That's a redraw of one of the earliest digital artworks.

>>1557829
>This means you can disregard those 2 dice and just rephrase the questions as such:
The last part of the question is why I mention that, if it makes a difference.
>0.00000108
What's that as a percentage? Because hearing it like that sounds WAY lower than I was hoping.

Getting four or maybe even five on a very good day was my main hope. Currently I'm seeing something like maybe 10% for a three result, my end goal is to get a variety of four 4x31 result rolls or maybe even a five or two.
I've the possibility of guaranteeing three of the rolls as being 31 instead of just two, but it would be so much more of a pain in the ass to do that I deemed it not worth doing.

Still seeking help with this, Would love to know my % odds of getting a four or even a five.

>>1557941
By following the math logic of this anon and a bit of adjustment >>1557831.
It is said 2/6 is guaranteed to get the number between 0 to 31(0 is included so that makes 32).
So by that logic I only need this for:
3 dice; 1/32=0.03125×100=3.125%
4 dice; (1/32)^2= 0.0009765625×100=0.0977%
5 dice; (1/32)^3=0.000030517578×100= 0.00305%
6 dice; (1/32)^4=0.000000953674×100=0.0000954%

>specific combo
Let me check with ai, since you need specific face, ah yes.
Total outcome = 32^4 = 1,048,576
Arrangement = 4! = 24
24/1048576= 0.00228%

In conclusion no chance, well... except if you add more rolls the chances are increase, a bit like 8 rolls to get 4 specific number = 24.98%

I hate math, and my calculation could be wrong.

Wait you did say 2/6 is guaranteed to get the number you want so...
Total Outcomes=32^2=1,024
2/1024= 0.195%

By adding more rolls the percentage increases.
Adding 2 more rolls = 2.15%
2 more = 6.24%
Oh well I'm tired.

>>1557951
>>1557952
Really? Only 3.125% for 3 to hit 31? That feels a little low. I'm only about 100 or so tests into my own practical as it takes me several minutes for one, and I'm working in batches of 30, but so far I've had 9 results of 3x31 and 0 higher out of the first 90.
Wouldn't 3.125% be the odds of rolling a single 31 from a single dice, as opposed to the odds of rolling it on any one of four different dice, thus making the odds four times more likely or thereabouts?

>>1557954
>3.125% be the odds of rolling a single 31 from a single dice
Actually yes, but I'm confused isn't 0 included too, wouldn't that make it 32 face? Otherwise the percentage would be 3.226%
>thus making the odds four times more likely or thereabouts?
Let me recalculate with ai here, 32 face and I only 1 desired face and I get to roll 4 times.
31^4= 0.8756
1(1 specific face)−0.8756≈0.1244×100= 12.44%
So ye... your desired outcome checks out.

For other calculation.... man I'm so tired.

File: Thumbs up.gif (1.2 MB)

1.2 MB GIF

>>1557956
>Actually yes, but I'm confused isn't 0 included too,
Yes it is, 32 possible faces effectively but they're labelled 0 through 31.
>So ye... your desired outcome checks out.
Cheers.
>For other calculation.... man I'm so tired.
Thanks for the partial solve. I'm aiming to get and end result of either eight results of 4x31, two results of 5x31 or a single god roll of all six, and I'd like to estimate how long it's going to take me.
The catch is that I need a variety. For example with the two results of 5x31 one of the dice is inevitably not 31, let's call it a roll of X. If dice 1 2 3 4 and 5 are 31 and 6 is X, then on the second 5x31 then dice 6 *has* to be 31 and X is one of the other numbers, or it's a miss for me and I'll need a third one.
There's one other thing I've not bothered mentioning as it probably complicates things way too much for calculations, but more numbers than just 31 are valuable to me:
>If dice two is 0 or 30, that's not quite as good as 31 but it's worth about 3/4th of a 31 roll to me
>Dice three, four or five can be 30. and that's still a pretty desirable result.
>Dice six can be 30 and that's good like with the others, 0 has some low amount of value but more than zero value.
>The first dice pretty much always needs to be 31 or bust though, 30's not worth anything to me.
This is why I've repeatedly insisted that this is six dice rolls with two of them being guaranteed to be exactly 31, rather than discounting those and rounding down to just calling it four dice.

Batch 91 to 120 had my best result yet: 31/X/31/31/X/30 and a 31/X/30/30/X/31
Other than that two more 3x31 and nothing else. The journey for a 4x31 or better continues.

File: 32-side-dice-3Y2IavlU_200.jpg (4.6 KB)

4.6 KB JPG

Disregard my comment, but I think you should have made it more clear.

>>1557964
Beyond including the Wikipedia description of what dice are or something, I'm unsure how I could have made it clearer.

>>1557831
>Doing the easiest (rephrased) one first:
>Chance of all four getting 31 is: (1/31)^4 = 0.00000108
nevermind, this is wrong since the dice has 32 sides, not 31.

Why don't you just say that you're playing pokemon and trying to get good IVs? You could get more specialized advice.

File: Mass Ditto farming shenanigans.jpg (561.8 KB)

561.8 KB JPG

>>1558006
Firstly because there's no specialized advice I could get that I'm not already aware of, I'm extremely sure of that.
Secondly because I've been curious as to how long it'd take for somebody to put the clues together and point it out.

Redoing this whole mess, once again I hate math XD.
This is dependent event, 32 faces to roll for specific face, things are rounded to 4, percentage rounded to 2.
Any 2 faces are guaranteed to get, so instead of needing to roll for 3 face I only need 1 face.
If 2 faces that are guaranteed to roll included in the rolling, instead of rolling 4 it is 2(will include the exclude).

3 face(only need 1 face)(first is excluded roll, second included roll);
(31/32)^4 = 0.8807, 1 - 0.8807 = 0.1193 × 100 = 11.93%
(31/32)^2 = 0.9385, 1 - 0.9385 = 0.615 × 100 = 6.15%

4F(2F) using binomial probability...; P(x) = (n/x)p^x q^n-x
n = 4 or 2 being rolled, x = 2 is same face, p = 1/32 successfull, q = 31/32 unsuccessfull
4/2 = 4C2 = 4!/(4-2)!4! = 6, 6(1/32)^2 (31/32)^4-2 = 6×961/1024^2=5766/1,048,576 = 0.55%
2/2 = 2C2 = 2!/(2-2)!2! = 1, 1(1/32)^2 (31/32)^2-2 = 0.098%
With how low the percentage is to achieve 10% you need to roll more successively(8/12/16/20)
8C2 = 28(1/32)^2 (31/32)^8-2 = 2.28%
12C2 = 66(1/32)^2 (31/32)^12-2 = 4.62%
16C2 = 120(1/32)^2 (31/32)^16-2 = 7.71%,
20C2 = 190((1/32)^2 (31/32)^20-2 = 11.13%, so you need to roll 20 times to achieve >10% to obtain 4 same faces, 18 rolls around 9.82%, 19 = somewhere 10%, probably this number....

5F(3F); 4C3 = 4, 4(1/32)^3 (31/32)^4-3 = 0.0118% (because even if 2 are guaranteed the chances are abysmal)
20C3 = 1140(1/32)^3 (31/32)^20-3 = 2.31%
25C3 = 2300(1/32)^3 (31/32)^25-3 = 3.61%
50C3 = 19600(1/32)^3 (31/32)^50-3 = 12.8%
45C3 = 14190(1/32)^3 (31/32)^45-3 = 10.8%, roll 45 times successively to achieve >10% of same 5 face,
[Chatgpt says about 28[35 if guaranteed] rolls to achieve 10% of 5 same face].

6F(4F); 4C4 = 1(1/32)^4 (31/32)^4-4 = 0.0000954%
30C4 = 27405(1/32)^4 (31/32)^30-4 = 1.20%.....
60C4 = 487635(1/32)^4 (31/32)^60-4 = 8.18%...
70C4 = 916895(1/32)^4 (31/32)^70-4 = 11.1%
65C4 = 677040(1/32)^4 (31/32)^65-4 = 9.82%, so around 67-68 idk anymore....
[Chatgpt: 50-51[57 if guaranteed] rolls to achieve 10% of 6 same face].

Okay finally I get to rest XDXDXD, anon buy me hamborger with chips[not really I don't need it]
Also my calculation once again could be wrong, feel free to ask other places.

>>1557970
"I have six 32-sided dice"

File: Pokemon full odds shiny.jpg (610.8 KB)

610.8 KB JPG

>>1558011
Turns out it's quicker and easier to get a full odds shiny than a 4iv Ditto. The picture doesn't show it well but it's violet.

>>1558048
>>1558049
>>1558050
I'm struggling to parse this, but thanks for the help.

>>1558051
I guess. I have a bit of 'tism so occasionally I word things a little more awkwardly than I should.

File: Pokemon shiny 3.jpg (478.8 KB)

478.8 KB JPG

>>1558065
Make that two shiny Pokemon at full odds because what the fuck is reality even?

>>1558049
Okay so does this just mean that 0 and -0 exist and the law defining the meaning of 0

>>1558191
Idk, my advanced math was ungraded, go ask someone. Math is nothing special its just a knowledge. I'm leaving this thread as I fullfilled OPs request to the best of my ability.

>>1558194
Math has to be converted into law, for it to hold meaning... Every equation has to prove that 0 exists or it means nothing, transient from a=100 what does 0 =

>>1558195
Idc, I'm not going to argue with you.
Have a great day/night.