# Likelihood of bernoulli trials until k successes

Suppose we sample 10 data points where 4 are deemed successes. The likelihood of this given the probability of success $p$ is $p^4(1-p)^6.$

Consider now the problem of continuously sampling until we reach 4 successes. We find that after 10 samples, the $10^{th}$ happened to be the $4^{th}$ success. What is the likelihood of this?

I realize that the second case is negative binomialy distributed i.e the probability of k successes where the $n^{th}$ trial is a success is given by ${{n-1}\choose{k-1}}p^k(1-p)^k$. Maybe I can somehow deduce the likelihood from this?

• What do you mean by "likelihood"? What is meant by "likelihood" in statistics is the probability density (or mass) function evaluated on fixed data, that is maximized in terms of it's parameters. So there is really nothing to deduce in here, you already have the likelihood function...
– Tim
Oct 29, 2017 at 21:15

Answered in comments: In statistics the likelihood function is just the density (or probability mass) function, but evaluated at the observed data, seen as a function of the unknown parameters: $$L(\theta ; x) = f(x ; \theta)$$ So you already have the likelihood function! But look at What does "likelihood is only defined up to a multiplicative constant of proportionality" mean in practice? The likelihood function is only defined up to proportionality. The combinatorial factor in the negative binomial distribution do not depend on unknown parameters, so you can take the likelihood function to be $p^4(1-p)^6$.