Bernoulli doesn't work..

Tippe auf eine hypergeometrische Verteilung..?

I would try to answer this question but I only speak English.
The translation is poor and I don't understand what the question is.

Ich würde versuchen, diese Frage zu beantworten, aber ich spreche nur Englisch.
Die Übersetzung ist schlecht und ich verstehe nicht, was die Frage ist.     

Hi Melody..here's a short translation:

Imagine a bowl with three types of colored balls, eg. red/green/blue. In this example let's assume a total of 30 balls, 10 of each color. I now draw always two balls (not necessarily at the same time but they remain as a pair) without putting them back. The first pair of balls may now be either of two different colors or it may be of the same color, e.g. a pair of green balls. I now do continue to draw a second pair, and so on, until the bowl is empty. 

The question is:

What is the overall probability that I will draw at least three identical pairs, all of the same color? (e.g. 3 pairs of red balls).

What I meanwhile have figured out is that with a hypergeometric distribution I would be able to set up the probabilities of the first draw. Then, according to the results of the first draw - reducing the balls inside the bowl - the next draw can be calculated. The first 2-3 draws is ok to calculate somehow, however it ends up in approximately 60,000 different branches on 10 different levels, all backed with hypergeometric distributions. That is an incredible amount of potential ways to find a minimum of 3 pairs of identically colored balls..thus binomial tree is no solution, hypergeometric distribution only delivers the result for one draw but the overall probability after 10 draws is still not found..

I thought about a hypergeometric permutation (variation?) or something? Thanks for your interest..if you find any solution, please give an example, too, as I am not very skilled in reading mathematical symbols. One main problem is with n..I could use it for both, n=2 for collecting the pair as the random grab sample, others use it for the draws - if they grab one only. Here we've got both..Nobel prize?

Christof

I am stating very clearly that I do not know how to do this problem.  I am just thinking out loud.

I am also goint to simplify the question to 18 items instead of 30

How many ways can 18 different objects be paired.

This is probably too simplistic but I am thinking that I could chose one item randomly and pair it with any of 17 others.

Then I could chose another item randomly and pair it with any of 15 others.  etc

so that would be

17*15*13*11*9*7*5*3*1 = 34 459 425 ways

Now say there are 6 red, 6 blue and 6 green but they are labelled blue1, blue2  etc  so they are all distinct.

How many ways can I pair all the red together, all the blue together and all the green together?

5*3*1    *   5*3*1    *    5*3*1   =   3*5*3*1 = 45 ways

How many ways can I have the red altogether?

5*3*1*11*9*7*5*3*1  = 155 925

How many ways can I have the green altogether?

5*3*1*11*9*7*5*3*1  = 155 925

How many ways can I have the blue altogether?

5*3*1*11*9*7*5*3*1  = 155 925

So how many ways can I have 3 pairs of the same colour

3*155 925-2*45 = 467685 

So the likelihood that there are at least 3 pairs the same colour is  

467685/34 459 425 = 0.0135720488661665

Now the original question also wanted to included  triples of  RG.RB, and GB

This is getting much more complicated and I am not at all confident that what I have already presented is correct anyway.

Anyway, Chrisof does that give you any ideas.   I am not saying that what I have presented is correct.  It might be total rubbish.....

If you would like to discuss what you think of my simplified question and answer (good or bad) then please feel free to do so.    

Hm..I like the approach, however still have to check if its correct..could be..thanks for your thoughts.

Christof