Closed form expression for this?

[aaaa202]

Say we have NA blue balls and NB red balls mixed together in an urn. When we pick a ball it leaves the urn. I want to find the probability of picking n red balls in a row.
The probability of picking a red ball is:
NA/(NA+NB)
But each time a red ball leaves the urn the probability of picking the next will be:
(NA-1)/(NA+NB-1)
So in general for n successive red ball picks the probability must be:

p(n) = ∏_i=0ⁿ(NA-1)/(NA+NB-i)

First of all I want to ask if this is correctly deduced. Secondly, I want to know if you can evaluate a product like the above in a closed form.

 

[Ray Vickson]

Say we have NA blue balls and NB red balls mixed together in an urn. When we pick a ball it leaves the urn. I want to find the probability of picking n red balls in a row.
The probability of picking a red ball is:
NA/(NA+NB)
But each time a red ball leaves the urn the probability of picking the next will be:
(NA-1)/(NA+NB-1)
So in general for n successive red ball picks the probability must be:

p(n) = ∏_i=0ⁿ(NA-1)/(NA+NB-i)

First of all I want to ask if this is correctly deduced. Secondly, I want to know if you can evaluate a product like the above in a closed form.

I'll change to a nicer notation: we start with N balls altogether, R red and B blue, with R+B=N. If you draw n <= R balls, what is the probability all n are red? Note: this is very different from asking that if we pick all balls, one at a time, we ever get n red balls in a row; that would be a much harder problem.

As you said, the answer is
p(n) = \frac{R}{N} \frac{R-1}{N-1} \cdots \frac{R-n+1}{N-n+1}.
This can be written in terms of factorials and binomial coefficients (if you think that is nicer and fits your definition of "closed form"):
p(n) = \frac{R!}{(R-n)!} \frac{(N-n)!}{N!}= \frac{C(R,n)}{C(N,n)},
where C(a,b) = \frac{a!}{b!(a-b)!} is the binomial coefficient.

RGV