Odds calculations for d10s

CPA's chart.

Anydice is popping up all over the internet if I look for dice odds distributions etc, but I really have NO idea how to even get started, and their "help page" is not really helping the least. Let alone tweaking it out, I guess it's way over my head unluckily.

What's this ORE you're talking about? If you can direct me to some link I'd really appreciate. I'm italian so googling ORE just leads me to economics newspapers or the meaning of "hours".

It's effectively just the Birthday problem en.wikipedia.org/wiki/Birthday_problem

Your math is wrong.

You have 1 in 10 hitting a called number with 1 die (9).
1 in 100 of hitting a called number with two dice (99).
There are 10 such pairs (11, 22, ... 00) available to hit so you have 10 chances in 100 to get dubs, 10/100 is 1/10 or 10% or 0.1

Each successive die you multiply by 0.1 so trips is 0.01 (1% or 1 in 100) quads as you can guess is one in a thousand, but there is a 100% chance I'll say nice dubs because everybody gets so mad.

In any case it's exactly what said. The math seems simple because it's all d10 in base 10. d6s would be the same except 16.6% for two, 0.166*0.166 for three etc.

Hope that helps.

When I see shit like this, I never understand why people piss all over d20/100 or even 2/3d6 if you need that normal distribution.

Can you please redirect me to the website that screenshot is taken from?

I see Width 1 is the chance of NOT getting a double from 2d10s to 10d10s
Width 2 is the chance of GETTING a double from 2d10s to 10d10s
I don't get what Width 3 to 10 mean

see
Thanks for your answer anyway

> Bro, just use a Taylor polynomial and you can approximate your probablity curves

Fuck, and I thought binomial distributions from Shadowrun were a pain in the ass.

A lot of this is if you roll a d10 and get a 9 what is the likelihood you'll get another 9 rolling again, which is not the same as roll 2d10 get doubles. Some of them seem like if you roll one of each die (d4, d6, etc) and if they all come up with the same value, like 2 on each, 5 and 6 never being able to match a d4.

In short, not at all what you're asking assuming your English is accurate in the OP question.

Dubs, quads, quints is simple for d10, seems like everyone is complicating it with statistics and misunderstanding.

That's wrong tho.

I'm not even sure I can phrase my question correctly in Italian, because we're talking straigth statistic magic spells here to me.
Anyway I hope I rephrased my question better here. With some data as support.

Thanks. Care to explain me why is that wrong?
Please, in laymans terms, because the closest I went to real probability calculus is when I throw the d100.

You've got it, but you're not extrapolating it. If Width 1 is the chance you get no repeating digits from Xd10, and Width 2 is the chance you get dubs from Xd10, then surely Width 3 is the chance of getting trips from Xd10 and on and on.

It's the case where you roll 1, 2, 3, 4, 5, 6, 7, 8, 9, 0.
Which happens 1 times out of the 10^10 combinations you can get.
So technically, it's 1-10^(-10), which isn't really 99.9%, but clearly writing 100% would've been wrong.

Hmmm, I feel I'm getting closer to understanding that.

Any formula I can feed Excel with to get those numbers?

>a double on 2d10s

Think about it. 11, 22, 33, 44, 55, 66, 77, 88, 99, 00

ten possibilities of doubles.

Values of 1 to 100 or 0 to 99 depending on how you read the dice, so 100 numbers.

100 numbers 10 of which are the doubles you're looking for.

10 chances in 100 numbers for doubles.

10 in 100

1 in 10

10%

There is no complicated formula. There is no 98.4% it's all 1s and 0s and a decimal . to me , to you.

It's painfully simple, why are you insisting it's not?

>What's this ORE you're talking about? If you can direct me to some link I'd really appreciate. I'm italian so googling ORE just leads me to economics newspapers or the meaning of "hours".

"ORE" is "One Roll Engine".

Mate, he's not asking for the odds any 3 matches on 3 dice.

He's asking for the odds of any 2 matches on 3 dice. And then also any 3 matches on 4 dice. And etc., etc.

That's not doubles though, that's the chance of rolling another die that matches the previous dice. They are two different things mathematically, and that's not what OP asked for.

You're basically making it the three doors problem in game show where statistically you should always switch blindly after picking the first door. It's not at all what OP asked.

Thank you for your thorough explanation, but as I said that before, that I already got it down.
And it's in the 1st row of the table you're quoting from: it's the chance of having doubles with 2d10.
I already know that the chance of getting triples on 3d10s is 1%. And the chance of getting 4ples in 4d10s is 0,1%.
This is simple. It's a division.

I need to know the changes for the other cases, that is, the chance of having doubles with 3d10s.
With 4d10s. 5d10s. And then the chance of having triples with 4,5,6d10s And so on.
And here it's not that painfully simple. I wouldnt be asking here if it was.

>I'm trying to figure out the chance of doubles, triples, quads, quints, etc using d10s. I'm plotting the whole range from 2d10s to 10d10s.
>I need to know how to adapt this formula for triples, quads, etc. AND for calculating odds of doubles on 3d10s, doubles and triples on 4d10s, etc.
OP explicitly states trips, quads, and more, at all numbers of dice rolled. That chart should be exactly what he's after. He wants to know the chance of 5d10 having trips? Width 3, # of dice 5. .081% chance. I don't know if the chart's accurate, and I don't know the math behind it, I'm not the guy that posted it. But in theory, that's exactly what he's after.

I literally do not see where OP asked for doubles in rolling 10d10. I see where he mistook that chart for what he was looking for, but I'm going from the OP which is possibly unclear.

But if he looking for trips in 10d10 do you count the possibility for two sets? Do quads count as trips for purposes of calculation? If it's not as simple as I read the op then there is insufficient information about what op is doing to answer.

OP here.
Just for clarification, is that the difference between getting 1,1,2 and getting 1,2,1?
Meaning, the first example is getting a match after the 1st number rolled (1,1,2)
The second one is getting a double, because there are two 1s (1,2,1) albeit not one after the other.
Right?

Because I'm asking to calculate the second one.
I don't care what position they are in, I just care that they come up.

>Do quads count as trips for purposes of calculation?

OP here. No, they don't. They count as quads, because they have the same number "up to" 4 times. They neither count as trips nor as doubles. Just quadruples.

When you say dubs, trips, etc. are you saying they have to be sequential?
For dubs in 3d10, is 1,5,1 good or does it need to be 1,1,5 with the 1s adjacent in roll order?

In that case you will need a little bit of combinatorics. If you have Xd10 and want Y matching numbers, you'll want to be doing X choose Y ( X!/(Y!(X-Y)!) ) on top of your (.1)^(Y-1).

As i said here, I don't care about the order they show up, I just care about the fact that they show up.

So yeah, 1,5,1 is a dub.
And 1,1,5 is still a dub.

That's because we throw all the dices together, but it honestly wouldn't change if we threw them one by one, mechanic wise. The mechanic triggers when there are exact same numbers after the d10s dice roll.

And actually the (.1)^(Y-1) won't hold beyond the 3d10 case. I'm interested enough now that I actually want to make a formula for you.

Is this formula any good? I asked Veeky Forums before coming here and they gave me this.

Also, this seems to lead me to a problem when the number of d10s increase too much with respect to the number of matches I want.
That is, with that formula the chance of getting doubles in 7d10s brings me to 124%, which is obviously not correct. I suppose it's not, at least.
I corrected this for doubles with the formula in OP, and this is why I was asking for analogues for the other matching numbers.

I would say that it's not a good formula if it ever loses accuracy, that makes me lose trust in it everywhere else.

Aye.
They told me to correct using
>divide by (floor(X/2) choose flood(x/2 - 1) or something

Not that i really got what that meant. And then I started experimenting with the formula in OP. And then I came here for advice on that.
But it seems this formula is better than the one in OP, since it does not consider only sequential numbers but all dubs coming up.

Hell I don't know...

I posted this exact question three months ago and it led Veeky Forums to an autistic breakdown. Math Stackexchange's explanations led me to believe that it's easier to just code a diceroller that rolls a few thousand dice and counts the results. It's not completely exact, but who fucking gives a shit about being 0.0004 off.

I've let the thing run from 2d10 to 10d10 and copied the results in here. This gives you the probabilities for any given combination of dubs, triples, quints, double dubs, quadruple heptuples or any other possible combination with your dicepool.
The number in front of the equals sign is the "height" of the set, the number after that is the number of sets.

>Result: {1=2}
Means two Singles

>Result: {2=1}
Means one doubles

>Result: {1=4, 2=1, 3=7}
Means four singles, one dubs and three heptuples in that roll

Wow...

OP here, thanks for this gem. Let me have a look

>three heptuples in that roll
That's seven triples
Fuck

Just roll the dice faggot.

I was thinking about homebrewing up a system in the vein of LotW or ORE, but I wanted to know the math beforehand.
These numbers told me that the LotW guy never fucking bothered to figure out the math of his system and the whole pokerdice shtick is really more trouble than it's worth it.

Does it show anywhere the number of dice rolls to get results?
Because I don't get the "two singles" thing.

I've found dice to be surprisingly difficult to model. A while ago I looked into finding something to calculate the density of 4d6 drop and it turns out it only exists as a derivative of a polynomial.
Also, as a minor critique, I would say you should collapse those outputs to just whatever the highest-achieved repeat was (pents, quads, etc.) Another gripe is you never seem to make it past octs. But I like it.

Veeky Forums had their heads up their asses, that's just telling you to divide by X/2 rounded down to an integer.

Forgive me I think I got what that means. It means that on a 2 dice roll it has got two numbers which are not matching, therefore single.

Sorry for dumbness

Man has been bested by machine once more. If I ever figure out a formula (if it's even possible), I'll post it anyway and maybe it will make its way back to you. Simulation tends to be the way to go for dice, though.

>that's just telling you to divide by X/2 rounded down to an integer

Thanks for that friend. Should I divide the whole formula for X/2 rounded down, or what else?
Because I think this pretty much may solve it.

What do you mean? How many dicerolls the thing did to get these results? About ten thousand.
Or what the results mean?

>Result: {1=1, 2=1, 3=1} ------- 4.32948%

Means: The probability of rolling EXACTLY one single (1=1), one doubles (2=1) and one triples (3=1) is 4.3...%

>Also, as a minor critique, I would say you should collapse those outputs to just whatever the highest-achieved repeat was
I hear you, but I wanted the exact results for my homebrew idea. For that, I needed to know how likely getting, say, a doubles and a triples with seven dice is.

>Another gripe is you never seem to make it past octs.
Yeah, by nature of literally rolling ten dice ten thousand times, incredibly unlikely results just don't show up. I played around with going higher than ten thousand rolls, but it slowed down the program exponentially because I'm still a shitty coder and barely changed the numbers. Since the thing was made for me anyway, I just figured "well, the likelyhood of eightuples on ten dice is 3.9999999999999996E-5%, don't worry about beyond that"

Yes thank you I understood it only a couple minutes after posting that, it's very clear after you get the key to read it.

Please do it.
After all it seems it's kind of a recurring theme around here, and I looked for an answer around the internet and just found Anydice.
Which is a fucking bitch to wrap one's head around.

>that's just telling you to divide by X/2 rounded down to an integer

Any info on What should I be dividing by X/2?

>Which happens 1 times out of the 10^10 combinations you can get.
Only if you have to roll them in order, which is not the case here. The actual number of combinations is 10! which means the probability is 0.00036288