>>>/vg/146322514
What is the probability of getting only one 6 in three six-sided dice?
>>8158675
Binomial
>>8158675
Just a single 6? not at least one 6? If its just a single 6 its equal to the probability of not getting a 6 at all which is 5/6 subtracted from the probability of every combination of numbers which is just 1 or what ever whole fraction you want to represent 1 as.
so a potential of 216 different outcomes means that the probability of not having a single 6 is (5^3)/(6^3) which is 125/216.
So 216/216 - 125/216 = 91/216. Therefore you have a probability of 91/216 to get just a single 6. Which I my calculator says is 0.42129
>>8158675
50% as you either get it or not.
Dank memes aside. You want to get the probability of getting one '1' which is 1/6 and then not getting a 1, twice.
(1/6)(5/6)(5/6) = 25/216
This neatly rounds down to a 10% chance.
>>8158684
Why would rolling multiple times reduce your chances of getting a 6 if 1 dice yields 1/6 which is 13%
>>8158685
you have a conditional that out of three rolls, you get one and only one 6. So you have to exclude the chances of getting a 6 twice or three times. This lowers the probability.
>>8158685
>Why would rolling multiple times reduce your chances of getting a 6 if 1 dice yields 1/6 which is 13%
But you want only one and only one six so every time you roll you are risking getting a second six, reducing your chances.
>>8158681
this
let X be the # of sixes in 3 rolls, then X ~ Binomial(3, 1/6) and P(X = 1) = 25/72 = ~0.347
>>8158683
>Which I my calculator says is 0.42129
>42%
>There is an almost 50% of this to be true
You must have made a mistake. The chance of rolling 3 dices and getting exclusively one six is definitely not fucking 42%
>>8158698
>34%
Maybe.
>>8158684
>10%
Maybe?
Well, 3 mathematicians with 3 different results already. Looks like once again we will have to rely on the physicists to try this out experimentally.
Can anyone ask for an NSF grant to get 3 6-sided dices?
>>8158698
Check your math.
>>8158713
>Well, 3 mathematicians with 3 different results already. Looks like once again we will have to rely on the physicists to try this out experimentally.
It's beautiful, to be honest. Really shows the quality of posters we have here.
>>8158718
its definitely not a 1% chance to get a single 6 in 3 rolls
>>8158684
the problem here is that there's more than one way to get 1 six and 2 non-sixes specifically there's 3 choose 1 = 3 ways so 3(1/6)(5/6)(5/6) = 0.347
>>8158718
>Three of these outcomes have "only one 6".
Demonstrably wrong.
Lets find your first 3 possible outcomes
(6,1,1) and (6,1,2) and (6,2,1)
BUT WAIT. You can also roll (1,2,6) or (5,6,2), etc.
You are really fucking retarded. How did you reach that conclusion?
>>8158721
>its definitely not a 1% chance to get a single 6 in 3 rolls
because....?
If I wanted to go somewhere where people vote on the answers to math questions, I'd hang out in Yahoo Answers.
>>8158723
Nigga you are right.
Well, I was pretty close when it came to modelling the problem.
>>8158725
because your 1/74 metric doesn't make fuck all sense, you're just removing 5/6 because you somehow think it includes all the 6's you already aren't getting
>>8158715
okay it's checked i even did a simulation
/dog/ is sorry about our autism
>>8158736
This is really interesting.
When you asked the question analytically it gave you 25/72
When you asked it to perform a simulation with its own random number generator is gave 0.35, which is 0.01 above the actual probability.
From that you can see how bad (or good, depending on who you care) our pseudorandom number generators are at approximating actual randomness.
Pretty neat. +0.01 is not bad at all, good job programmers.
>>8158675
1/2
it either happens, or it doesn't
>>8158743
Probability has finally been solved.
>>8158717 again here. Drop-out CS major. Do I have to help you guys out?
First branch: 6 is the first die roll. There are 6 second die rolls, 5 are valid. There are 6 third dice rolls, 5 are valid. The second and third rolls are independent, thus (5/6)^2=25/36 valid combinations and 11/36 invalid.
Second branch: First die roll is anything but 6. Second die roll is 6. Same shit. (5/6)^2=25/36
Third branch. First two die rolls are anything but 6. Third roll is 6. (5/6)^2 = 25/36.
Fourth branch. No sixes. 5^3=125 ways for this to happen.
Sums: (25*3)/(36*3 + 125) = 125/233 = 0.5364... = 53.64%
Let me know if I fucked up, but I'll be proud I didn't fuck up as hard as some of you.
>>8158750
>53.64%
That is definitely not it.
WHERE ARE THE PHYSICISTS FOR FUCKS SAKE. NSF GRANT FOR 10 BUCKS TO BUY A SIX SIDED DIE !!!!
>>8158750
On second thought there are still 216 possible outcomes. 125/216 =57.87%? I dunno mayonnaise.
>>8158758
well theres 5 outcomes on a die plus 6 is the one you want
so 6/11 is 54.5%
>>8158755
I mean I could make a spreadsheet of the 216 outcomes and figure out analytically who is wrong and why, but I am le tired and organizing 216 rows doesn't sound particularly fun. $1000 grant?
>>8158750
You are thinking it all wrong.
Here is an intuitive way of doing this:
First, consider the set X = {1,2,3,4,5}. Really simple set.
Now you want all the row vectors of the form (6,a,b) and (a,6,b) and (a,b,6) where a and b are elements of X.
How many of such row vectors exist?. Well, as 6 is a fixed, we want all of the possible configurations of (a,b)
We can easily get this by calculating the number elements in the cartesian product of X, which is 25 (5 times 5)
But as we have 3 different configurations of these we multiply 25 times 3 = 75.
And to get out final probability we divide by total possible configurations of rolling 3 dices (216)
75/216
which is what other people have been getting using the traditional methods.
it always astounds me how sure people who have no fucking clue can be about themselves.
>>8158742
>Pretty neat. +0.01 is not bad at all, good job programmers.
You know what variance is right? You need to consider the number of trials (10000 in this image) for the the error to mean anything at all.
>>8158675
assume the dice are numbered
number of total possible dice rolls: 6^3
number of total possible dice rolls with one 6: 3C1* 5^2
probability is 75/216, near 33%
>>8158789
>You know what variance is right? You need to consider the number of trials (10000 in this image) for the the error to mean anything at all.
GOOD JOB PROGRAMMERS
Anyways, do it with a billion then.
>>8158791
Fuck my ass, I feel dumb for how simple that is. Thanks for the best condensed explanation.
