Hello everyone
I wish to know the formula for calculating how much probability do I have of getting same numbers on some d10s.
For example, how much probable is to get a double (1,1 or 2,2 or 3,3 etc) on 3d10?
And a triple (1,1,1 or 2,2,2 etc)?
I just need the formula and some explanation (bonus), I think i can figure out the rest.
Thank you very much in advance
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>Thank you very much in advance
You're welcome.
> You're welcome.
Thanks Idol girl user, care to answer since you already got thanked twice?
Use the binomial distribution for n = 2 (double) or n = 3 (triple) with p = 1/10 (for d10)
This gives you the probability of one set of doubles (1,1) or triples (1,1,1). Multiply by 10 (for d10) to get all possible sets.
Also, >>>/sqt/
double > 1/10 (after the first/second throw, you have 1/10 chance to get the same number of the previous throw)
Triple 1/100 (1/10 to get a dub * 1/10 to get the third number equal to the first two)
You have a 28% chance to get doubles on 3d10, as after the first roll, there are 9 of 10 possible combinations that don't result in doubles, and 8 of ten for the first roll, thus the chance is calculated with
1-((9/10)*(8/10))
if you roll 2 dice there are 100 possible combinations: 1-1, 1-2, 1-3 etc up to 10-10
if you think about the possibility of me getting the dubs that I just got, it's 1 in 10 because there's one possibility in every group of 10 combinations of 100 total
if you add a 3rd die then there are 1000 possible combinations, with 1 appearing in every group of 100, so the probability is then 1/100 and so on
This, it's quite a simple question once you realize that the first die roll is guaranteed to be a number so probability doesn't even begin to enter into it until the second die is cast.
Double on 3 D10:
10*10*10 total possible dice roles,
10 possible double flavors
(3 choose 2) places to put your doubles
9 possible third die roll
10 * (3 choose 2) * 9 / (10 * 10 * 10) = 0.27, 27%
K-ple on X DN:
N^X total possible dice rolls
N possible K-ple flavors
(X choose K) ways to place them
(N-1)^(X-K) possible remaining dice rolls
N * (X choose K) * (N-1)^(X-K) / N^X
It's just the binomial distribution, btw.
OP here, thank you all for your answers
> (3 choose 2) places to put your doubles
Sorry for being dense, but i don't understand how to manage this in a calculation.
I mean,
> 10 * 9 / (10 * 10 * 10)
I can manage, but i don't get how to do this
> 10 * (3 choose 2) * 9 / (10 * 10 * 10)
Look up Combinations and Permutations.
Also, if you type (3 choose 2) into Google, it will know what to do.
(n choose k) is just a way of writing a binomial coefficient: en.wikipedia.org
Its value is n!/k!(n-k)!
In this case (3 choose 2) = 3!/2!(3-2)! = 6/2 = 3
Probability is such a stupid concept, calculating it shouldn't be an academic discipline.
1) It does not make any predictions whatsoever
2) It looks like science because numbers, but it is much closer to astrology
I can understand statistics as a science, but probability? Of an event happening? We either know what's going to happen because we understand it, or it is too complex for us to model.
Nice bait, witnessed.
show me a model without probabilities
Thanks guys, you really cleared it up for me, thanks a lot, really.
One last question, not about probability but rather certainty.
I compiled an excel sheet with all combinations I need, and i noticed that the probability of getting doubles on d10 basically becomes a certainty (98.42%) at 6d10. At 7d10 they go through the roof with 124% and more.
Is this normal? And if it is, can you explain me why?
Thanks again, and sorry for the bother.
>this
I mean, shouldn't the 100% come at higher number of dices? like 10d10?
Think of it as literally being "3 choose 2." You have 3 distinct balls, you have "3 choose 2" different ways to select two of them:
Red, Green, Blue balls:
Red + Blue
Red + Green
Blue + Green
(3 choose 2) = 3
As the others said, mathematically, this corresponds to the binomial coefficient.
My bad, it's double-counting doubles if X is at least twice K: for example the dice roll 1, 1, 2, 3, 2, 2, 5 is being counted twice.
This is why it's generally easier to calculate the probability of something not happening and inverting that then the chance of something happening, since it more intuitively avoids double-counting even if it's less obvious to explain.
You can edit the straight-forward calculation to avoid for this (divide by (floor(X/2) choose flood(x/2 - 1) or something), but it's easier to understand it as a binomial distribution.
>it's double-counting doubles if X is at least twice K
So basically the chance of getting those high probability should be about half of that, if I understood it correctly. Nice.
>divide by (floor(X/2) choose flood(x/2 - 1) or something
Thanks for the advice friend, can you tell me how to edit this formula exactly?
>N * (X choose K) * (N-1)^(X-K) / N^X
>N being the dice faces
>X being the number of dice
>K being the "-ple"
Sorry but I didn't do these things at the university and so I'm like a barren field about this.
Thanks again.
1min in gloogloo.
I know how to use gloogloo, but gloogloo is useless if I cannot interpretate the results. And since i didn't have a proper education in statistics and probability, this is just pretty shapes for me.
you can't read a graph?
please close your Veeky Forums tab and google "statistics 101"
don't come back until you know how to read graphs
Yes you're so smart.
This graph isn't helping me understanding how many doubles I can get on 3d10. Neither has an answer to this question, which at this moment is much more important.
Do you have one, smart user?
3 dice
call them a b and c
you have 3 doubles and 1 triple
ab, ac, bd, abc
since there are 10 sides you multiply by 10
30 unique doubles and 10 triples
the probability is (amount of positive results that exist)/(total amount of results)
for 3d10 the denominator is 10^3, or 1000
so that leaves us with 3% chance of any specific double and 1% chance for any specific triple
since you probably don't care about which number the double/triple is, we can multiply by 10
30% double, 10% triple
btw there is actual math developed to make this shit ezpz to calculate. this is just an indepth explanation
Thanks for the in depth explanation. My problem is I had one exam on basic statistics like 10 years ago, never used it again, never had any kind of education in probability.
So that ezpz math is lost to me, and I probably would have a hard time understanding it thoroughly.
Anyway, I set up an excel sheet trying to get all combinations running (from 2d10 to 10d10), using the
>N * (X choose K) * (N-1)^(X-K) / N^X
formula.
I have problems with the chance of doubles, it reaches 100% at 6d10, and goes to 124% at 7d10, which I am not understanding why.
Do you have a fix to the formula I can use?
can you pls define what each letter means
I'm a man, not a compiler
It's as I said here.
>N * (X choose K) * (N-1)^(X-K) / N^X
>N being the dice faces
>X being the number of dice
>K being the "multiple": double triple quadruple etc
what's the "choose" operand?
>X choose K
>X!/K!(N-K)!
The binomial coefficient, as it was explained to me.
if the formula is good, maybe you just need to clean up your brackets
ie (N * X!/K!(N-K)!) * ((N-1)^(X-K)) / N^X
btw the pattern is:
1-(1*0.9*0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1*0)
The dice count is equal to how far down the bracket chain you go
idk how to do any of the math though
feeling pretty brainlet rn
the problem is simple when you try to solve for the inverse
imagine you want to keep every die rolled unique from all the other dice
first one is free, the second one has to avoid a single number, the third one has to avoid 2 numbers... etc etc etc
It gives me the same results even cleaning the brackets as you did.
What i want to know, is if it's normal to reach the almost certainty (98.42%) of getting doubles on 6d10s and then break the roof (124%) of the chance of getting doubles at 7d10s.
I find it weird that i have more than 100% chance of an event occurring. Does it even make sense?
something's definitely broken
skip the hard math and just use my 1 line of ezpz math
>something
Can you tell me what exactly? Because that would definitely help.
I can post screenshots of the excel sheet if it's helpful.
>my 1 line of ezpz math
Which multiple is that calculating the percentage of? Is it double? Triple? How can I see that?
Thanks in advance.
so the series 1*0.9*0.8... is calculating the chance of having every die be unique. That means no double, no triples etc
The inverse, which is 1-(1*0.9*0.8...) is the chance of getting at least a double.
you reach 100% at 11 because if the first 10 are all unique then every number is already taken.
For at least a triple, the series would be:
1-(1*1*0.99*0.98*0.97...) reaching 0 at the 102nd
... actually I'm not sure about this one
Thank you, this makes much more sense than the progression I had with the other formula, at least regarding doubles. It's also easier to understand why it's the way it is, thanks for your explanation.
Can you explain me better how to adapt this formula to calculate the progression for trips, quads and so on? What should I add/change to the original formula?
If it doesn't count just for doubles I mean.
Oh... do I just go with hundredths for triples, thousandths for quads and so on?
my intuition tells me that it's just a question of the order of magnitude like I have here it kinda makes sense if you treat it like the doubles one but you can only have double 5s (for example). ((because the first dice is the hypothetical 5)).
also don't forget to reach the minimum amount of required dice to hit the target.
eventually one of the 20D10 IQ people will show up to help you out better
I'm asking because I just checked for the triples progression, and increasing the number of d10s it gives me the same problem as before, that is they break the 100% roof of chance to get triples.
That is, it seems the function doesn't "plateau" at 100% probability, so I need to clearly understand how to get the right series and how to build the same series for quads, quints and so on.
It is wrong.
P(at least a double)=1-P(all different)
P(at least a triple)=1-P(all different)-P(exactly a double)
Can you please answer me about how to translate that in numbers for the d10s?
I am looking to plot the chance of doubles, triples, quads, quints etc on a range from 2d10 to 10d10.
Just to formulate better what my question is:
I need to know what's the chance of getting doubles, triples, quadruples etc on ALL the range from 2d10s to 10d10s.
For example, pic related is the probability of getting doubles from 2d10s to 11d10s. I need a solid formula to do the same for the other multiples
(bump)