Sampling Without Replacement : Draw x Red before Drawing RGB

Question

I am struggling to get my head around how to solve the following problem and can only find solutions to simpler versions

Consider a bag containing x Red, y Blue and z Green marbles What is the probability that one will draw n Red marbles before drawing at least one marble of each colour?

So say we wanted 2 red from a jar containing RRGB then RRGB, RRBG, RGRB and RBRG would be successful results for n = 2 but RGBR RBGR BGRR and GBRR would be failures. I'm assuming n, x, y, z are > 0

I have so far worked out my unique combinations but am struggling to formularise my success combinations in a general form

I would like to extend this to cover more colours.

Any help in this matter would be greatly appreciated

Answer 1

OK, I think I may have cracked it for 3 colours in a general form

Assuming r red marbles, b blue and g green and wanting n red before blue and green have shown i get

Using notation 0! = 1 x! = x(x-1)(x-2).....(1) xCy = x! / ( y! (x - y)! )

Total Unique Combinations T = (r+g+b)!/(r!b!g!)

k = 1 to g i = 1 to b

for n = N Tn = sum over k [ ((k+n)Cn)((r-n)+(g-k)+(b-1))C(r-n) ] + sum over i [ ((i+n)Cn)((r-n)+(b-i)+(g-1))C(r-n) ]

Gives the combinations

So probability = Tn / T

I suspect this can be simplified with a double summation.

Anyone fancy validating this ;)