A Hard Combination Probability Equation From A Reality T.V. Show I was watching the T.V. show Survivor tonight and miserably failed at wrapping my mathematically out-of-practice post-college brain around a probability equation. 
Background: 
At the start of the show, 18 people are randomly divided into 2 tribes (red & blue) of equal size, so 9 each. Then tonight the 15 remaining people (6 red & 9 blue) found out that they would be randomly splitting again...but this time into 3 tribes of 5 (red, blue, & green). When they finished the random tribe swap, they were shocked to find that 5 of the 9 old blue team members remained together on the new blue team, 5 of the 6 old red team members all stayed together (but moved to green...not sure if it's relevant), and finally the remaining 4 of the 9 old blue team members and the 1 remaining old red team member ended up on the red team together. 
Long story short to the non-mathematically inclined viewer...wow nothing changed (rigged reality tv). But what is the probability of that actually happening?
More specifically what is the probability that after all the randomization of teams, 2 of the 3 newly created teams would have 5 people that had previously been on a team together (when there were only 2 teams).
Thanks ahead of time!
 A: So the question is, if you have 6 red objects and 9 blue objects , and you generate three random groups of five, what is the probability that two of these groups contain only a single colour of object (one blue group and one red group)?
The total number of ways of choosing three groups of five from the 15 objects is: 
$$\text{Total number of ways} = \frac{{15 \choose 5} \cdot {10 \choose 5} \cdot {5 \choose 5}}{3!} = \frac{3003 \cdot 252 \cdot 1}{6} = 126126.$$
To yield two groups containing only a single colour of objects (one blue and one red) we can first choose an all-blue group and then an all-red group, and then the remaining objects become the third (mixed) group.  By the multiplication principle, the number of ways of doing this is:
$$\text{Number of ways with two single-colour groups} = {9 \choose 5} \cdot {6 \choose 5} = 126 \cdot 6 = 756.$$
Thus, assuming simple random sampling, the probability of obtaining two single-colour groups at random is:
$$\mathbb{P}(\text{Two single-colour groups}) = \frac{756}{126126} = 0.005994006.$$
So this would happen coincidentally about one in every 167 times you did the randomisation.
A: Calculation of probabilities would closely depend on the way of drawing team members and the way information (on interim team assignments) is made available. 
For example, randomly choose five out of 15 people and assign to team (1) another five to team (2) and the remaining go to team (3).
Another way may be to randomly choose one person out of 15 and assigned to team (1), then another is chosen from the remaining 14 and assigned to team (2), and the next to team (3). Then the steps are repeated.
Not knowing the setup, let's assume that the drawing was according to the first case above.
Probability of randomly selecting five blue team members from 15 is $ {9 \choose 5} / {15 \choose 5} = 0.04195804$.
Given that all five were blue, the probability of selecting (in the next step) five red from remaining six red and four blue is ${ 6 \choose 5} / { 11 \choose 5} = 0.01298701$.
The probability of both events happening is $0.04195804 * 0.01298701 =  0.0005492342$. 
