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/sci/ - Science & Math

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4141703 No.4141703 [Reply] [Original]

Hey /sci/ how do you feel about helping out a fa/tg/uy?

So lets say you had some 12 sided dice. When rolled, getting a result of 10,11, or 12 is considered a success. Lets say you want to calculate probability of getting certain amounts of successes...

For example, lets say you need 2 successes and have a dice pool of 5. Each die has a success chance of 25%. You dont need exactly 2 successes, just 2 at minimum.

Can /sci/ produce a formula with the variables r=Successes needed and n=dice pool?

>> No.4141708

http://en.wikipedia.org/wiki/Binomial_distribution

>> No.4141709

http://en.wikipedia.org/wiki/Bernoulli_trial

>> No.4141802

>>4141709
>>4141708

Yay for reading material!

Sure this is what i need to learn though? it's going to take me a while to figure out.

>> No.4141814

>>4141802
Yes.

>> No.4141895
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4141895

I was solving this but my girlfriend came over, and she's too good at giving head for me to concentrate. It's going to be a bitch to manually compute, but if you can use a graphing calculator, or an equivalent, then you can use the "binomcdf" function, which is the solution to exactly what you're asking for.

I just came; pic related.

>> No.4141982

>>4141814
>>4141708
>>4141709

Hmm, I think I understand how to calculate any single outcome now (such as the odds of getting EXACTLY 2 successes) but I still don't see how to calculate the odds of getting 2 or more.

>>4141895

Ya I have a method of getting the probability using a program on my computer, I just dislike not understanding the system on which I'm relying. Thanks for trying, at least nothing's wrong with your priorities.

>> No.4143437

p=probability of success in one trial=1/4
q=probability of failure in one trial=3/4
r=number of required successes
n=number of trials

P(X=r)=(nCr)(p)^r(q)^(n-r)=(nCr)(.25)^r(.75)^(n-r)

Sorry it's not in latex, couldn't work out how to actually use it on /sci/

>> No.4143452

>>4141982
>such as the odds of getting EXACTLY 2 successes
As in getting two successes in so many rolls?