Binomial probability distribution (sort of)

03032016, 03:57 PM
Post: #1




Binomial probability distribution (sort of)
I know that you can use a binomial probability distribution to determine the odds of x successes within n independent trials, each with chance of individual success P  the usual B(n,P,x) = COMB(n,x) * P^x * (1  P)^(n  x).
Suppose I'm concerned with a target number of successes  say 2  at which point I stop carrying out trials. I should be able to calculate the odds of needing to carry out n trials in order to reach that many successes, but I'm not quite getting there, and I'm not enough of a statistician to know what to search for. Naively, one might think you would just calculate B(n,P,x) for a given n, but if those x successes came before the nth trial, you wouldn't have actually performed n trials. It almost sounds like a Bayesian sort of thing, e.g. P^x chance of x successes in x trials, 1  P^x chance of needing to carry out more than x trials, etc. but I imagine there's a more elegant way to compute it than tree traversal. Note that these would still be fully independent trials, just that we'd be stopping after the target number of successes, and want to calculate the odds of having to carry out a given number of trials. 

03032016, 05:14 PM
Post: #2




RE: Binomial probability distribution (sort of)
Geometric distribution allows you to set probability of success on a given trial (assume each trial has that given probability), set number of intended successes, and set probability of having intended # of successes (perhaps you want p=.7 of obtaining your target # of successes).


03032016, 05:53 PM
(This post was last modified: 03032016 06:00 PM by Dave Britten.)
Post: #3




RE: Binomial probability distribution (sort of)
(03032016 05:14 PM)lrdheat Wrote: Geometric distribution allows you to set probability of success on a given trial (assume each trial has that given probability), set number of intended successes, and set probability of having intended # of successes (perhaps you want p=.7 of obtaining your target # of successes). Geometric distribution sounds fairly close, but everything I'm reading describes it as being based on the first success, rather than a specific, fixed number of successes. e.g. Suppose I want to know the odds of a particular number of trials being required to roll a die and get 6 twice. A few sample experiments might look like this: 6, 1, 2, 5, 6 1, 3, 3, 4, 3, 5, 4, 4, 6, 4, 4, 3, 5, 2, 4, 6 4, 3, 4, 5, 5, 3, 1, 4, 4, 5, 1, 6, 3, 3, 2, 1, 2, 3, 4, 3, 5, 4, 1, 3, 5, 3, 2, 5, 1, 4, 3, 3, 4, 2, 2, 1, 4, 1, 1, 1, 6 In other words, what are the odds of having to carry out that many trials before hitting the desired number of successes? Is there a more generalized case for the geometric distribution that allows for that? EDIT: Negative binomial distribution sounds like it might be what I'm after. I'll do some more reading/number crunching. 

03032016, 08:42 PM
Post: #4




RE: Binomial probability distribution (sort of)
You look to be on the right track...I think I meant hypergeometric distributions, but this applies if you are not replacing, say, balls from an ern as your experiment goes on. Binomial assumes replacement.


03032016, 08:49 PM
Post: #5




RE: Binomial probability distribution (sort of)
(03032016 08:42 PM)lrdheat Wrote: You look to be on the right track...I think I meant hypergeometric distributions, but this applies if you are not replacing, say, balls from an ern as your experiment goes on. Binomial assumes replacement. Yup, negative binomial distribution was what I needed. Geometric distribution is just a special case of negative binomial where r = 1, so that actually got me steered in the right direction, funny enough. 

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