Hello Veeky Forums
Could you tell me what the chances are of me rolling the following results:
With 4d6 = 3x6 and 1x3
With 3d6 = 2x6 and 1x3
With 2d6 = 1x6 and 1x3
Do the same percentages also apply to rolling fours and twos instead of sixes and threes?
If not what would the percentages for those be?
Sorry for this noob question, I'm really bad at maths.
>With 4d6 = 3x6 and 1x3
>With 3d6 = 2x6 and 1x3
>With 2d6 = 1x6 and 1x3
Your notation is unclear. Explain your problem more clearly and thoroughly so we can help.
Do you mean is it the same probability to roll a four as it is to roll a 2 or 3?
The percentages would be the same for fours and twos as sixes and threes if it wasn't for the indents in the dice. You're most likely to roll a six and least likely to roll a one for this reason.
If we assume that each dice has a 1/6 chance of rolling all numbers then the math becomes much easier.
Without the differing weights taken into account you would have a 1/6 chance (0.166666667) of rolling any single number on a die. Moving the decimal over twice and you get your percentage of 16.666667% or round up to 16.7%
When you want to roll 2 specific numbers you need to multiply the chance by itself.
2d6 would be (1/6)*(1/6)=02.77777779%
This could also be written as (1/6)^2
3d6 would be (1/6)^3= 0.462962966%
4d6 would be (1/6)^4=0.0771604944%
When you apply rounding you get:
2d6=2.8%
3d6=0.463%
4d6=0.077%
No I meant is it the same probability to roll 3x4 and 1x2 instead of 3x6 and 1x3
Thank you for this very detailed explanation, I am very grateful!
D6 refers to a fair 6 sided die
Additionally your probabilities are wrong
OP asked for the odds of n-1 6s and 1 3 for n D6
The odds of this are 6^-n * n
Thus 2/(6*6) ; 3/(6*6*6) ; 4/(6*6*6*6)
I'm confused now.
If you would be so kind, how would this transfer to
>2d6=2.8%
>3d6=0.463%
>4d6=0.077%
then?
>With 2d6 = 1x6 and 1x3
Given first rolls:
100000 (0/6)
010000 (0/6)
001000 (1/6)
000100 (0/6)
000010 (0/6)
000001 (1/6)
So there are 36 possible configurations for two dice rolls and only 2 out of 36 yield the correct outcome so the probability is 2/36
>With 3d6 = 2x6 and 1x3
100000 (0/36)
010000 (0/36)
001000 (1/36)
000100 (0/36)
000010 (0/36)
000001 (2/36)
So out of 216 possible outcomes, only 3 yield the correct one so the odds are 3/216
I am too lazy to compute the probability matrix for the one with 4 throws so do it yourself. I already learned you some probability.
What you are saying is retarded because you are not taking into account what OP is actually asking.
The idea of multiplying odds together only works when the events are mutually exclusive.
I dont think thats what he meant
Assuming that you mean by XdY to roll with X amounts of Y sided dices and getting the outcome to be a N-tipple of one number and a single of another number it should be [math]X choose (X-1) * Y^X[/math] assuming i didnt fck up.
I thought OP was asking for the results of his rolls being:
2d6 6 3
3d6 6 6 3
4d6 6 6 6 3
Without weight factored into it, there is a 1/6 chance of rolling a 6 as well as a 3.
Based on the way you're calculating it, what are the percentages of rolling those specific numbers? (Or two of the same number, as without holes in the dice, the calculation would be exactly the same.)
>2d6 6 3
>3d6 6 6 3
>4d6 6 6 6 3
I also interpreted it this way, but I don't think he is asking to take order into account, which is why is wrong.
>No I meant is it the same probability to roll 3x4 and 1x2 instead of 3x6 and 1x3
Again no one seems entirely clear on what you mean. Literally spell it out, stop using random notation.
I can interpret this as rolling a 3 four times, or rolling a three on one dice and a four on a second dice. Even similarly, but less likely, rolling a 4 three times.
Please describe completely what the problem is. Regardless, it isnt super tough, but there is no consensus on the original problem and what exactly you're asking.
And repeating exactly what you have typed before does not qualify as describing more completely or specifically.
I see my error now, but I'm not sure on how to correct it.
Say's that 2d6 would be 2/36 or 05.556%
3d6=3/216 or 1.3889%
Is this correct?
Well already found one error [math]Y^X[/math] should be [math]\frac{1}{Y}^X[/math]
My question is, how likely are these results when rolling (from top to bottom) four 6-sided dice, three 6-sided dice, two 6-sided dice
Also, what would the answer to 4d6 be?
And no, order does not matter.
As individual rolls not taking the other two rolls into account?
>Is this correct?
Yes. The behind is that you take the number of different outcomes that result in a "win" and divide it by the total number of outcomes.
So if you roll two dice, the chances of getting a 6 and 3 are simple.
You either roll a 3 and then 6, or you roll a 6 and then 3. Those are 2 outcomes that "win" out of 36 total possible outcomes.
Now, there are obviously formulas and theorems but the notation is heavy so you can't really access those immediately, so it is better to have the intuition on what probabilities really are.
Like look at this fag Pretentious motherfucker. Like if someone asking a probability questions knows that X choose Y even means.
Gotcha.
1) (1/6)^4
2) (1/6)^3
3) (1/6)^2
This is only the case in which order does matter.
If you're looking for combinations instead, order doesn't matter, ill just leave that for someone else.
Thank you! This means that 4d6 would have 4 winning outcomes out of a possible 1296.
1d6=1/6=0.166666667=16.667%
2d6=2/36=0.0555555556=5.556%
3d6=3/216=0.0138888889=1.3889%
46d=4/1296=0.00308641975=0.3086%
>that 4d6 would have 4 winning outcomes out of a possible 1296.
That is correct.
That might be a mess to read.
1d6=1/6
2d6=2/36
3d6=3/216
4d6=4/1296
1d6=16.667%
2d6=5.556%
3d6=1.3889%
4d6=0.3086%
your sample space is every possible outcome
take 4d6 = [1,1,1,1]
this can only happen in a single way, that is, die 1-4 are all one.
on the other hand, consider 4d6 = 3x6 and 1x3
since this outcome encompasses multiple points in the sample space, it's more likely than the previous outcome, which was just a single point in the sample space. how many different permutations can you make with [6,6,6,3]?
>And no, order does not matter.
To elaborate, there are 3 permutations for 3d6 and 4 for 4d6, thus the respective probabilities without order is
1) 2/26 = .0555555555...
2) 3/216 = .0138888888...
3) 4/1296 = 0.030864...
Yup
Thank you for your time lads. Sorry for all the confusion.
np, come back any time mate
Is anyone here good with sequences?
I could use some help over at .