Odds of anyone in a group getting picked 3 times in a row I asked this question earlier today, but now need to extend it and can't seem to generalize it out to a third event:
My original question was: You have a list of 10,000 people. Every week, you randomly select 2% (200) of those people. What are the odds of one or more of those 2% getting picked the following week?
Now, I need to know, what are the odds of anyone picked in that first week getting picked the two following weeks?
 A: A straightforward Monte Carlo method can be used to find a decent approximation. In Python2, we can do the following.
import random
def run_trials(num_people_total, num_people_selected, num_rounds, num_trials):
    matched = 0
    people = set(range(num_people_total))
    for _ in xrange(num_trials):
        resultant = set(random.sample(people, num_people_selected))
        for _ in xrange(num_rounds-1):
            resultant.intersection_update(random.sample(people, num_people_selected))
        if resultant:
            matched += 1
    return matched

running this simulation with 10000 people, 200 selected, 3 rounds, and 50000, I got 3909 hits, or about 7.8%.
Alternatively, since you said 5600 employees with 112 in your comment to Batman, I ran the code with 5600 people, 112 selected, and 100000 trials, obtaining 4315 hits, yielding 4.3%
A: Hint: Look at the complement event. For picking none of those initial two hundred the following week, you have (10000-200) people to choose from.
A: Sir, I have been looking this over and I looked at the answer to your previous question. You asked the probability of a person being selected in week 2 given that he was selected in week 1. Now you are wanting to know the probability of this person getting selected again in week three. This is mathematically equivalent to asking the probability of selection in week 3 given that he was selected in week two. 
Then why not treat this in the same manner as the probability of rolling heads n times in a row on a fair coin. (so .5^n in that case). Here using the same method of solution from your earlier question and your new numbers of 5600 and 112, the probability of one of the people selected in week 1 being selected in week two is approximately 89.8%.
Here is some supporting math:
P(person from 1 being selected in week 2)=1- P(none from week 1 selected in week 2)
P(person from 1 being selected in week 2)= 1 - $\frac{5488 \choose 112}{5600 \choose 112}$=.898 
Then the probability of one of those people being selected in week 3 is .898^2. This is .8064 or 80.64%. I hope I did not make a gross conceptual error here.  
So there is a high probability that this can happen. If I made a mistake, please let me know. My other thought is to use Bayes' theorem.
