Probability question

Question

If a group of 54 students are randomly broken into 3 groups of 18 students each every school year. What are the odds that any two students will be in the same group for three years in a row? Thanks

Answer 1

OK, this question is worded somewhat ambiguously, so first I'll answer the more likely interpretation, and then cover an alternate one.

Let's rephrase the question in the following way: Jesse and Kris are 2 of 54 students. For each of the next 3 years, these 54 students will randomly be broken into 3 groups of 18 students. What is the probability they will be in the same group the next three years in a row?

To begin, we first must calculate what is the probability that Jesse and Kris will be together in year 1. Let's give names to the 3 groups of year 1, call them A, B, and C. Jessie is randomly in some group, let's assume it's group A. There are now 17 other spots in group A left, and 18 spots left in each of group B and group C. Kris is equally likely to get any of the remaining spots, so the probability that Kris is also in group A is 17/(17+18+18) = 17/53. Notice that this is slightly less than 1/3. This might seem unintuitive at first, so you have to convince yourself of this to understand it.

So the probability for year 1 that Jesse and Kris are in the same group is 17/53. For years two and three, again it is the same probability. So the probability they will be together all three years is (17/53)^3.

As mentioned before, the use of the phrase "any two students" is ambiguous: it could mean "pick two particular students - what is the probability they are together all three years". This is what we just answered.

However, it could also be interpreted as "what is the probability that there exist two students who are together all three years" (i.e. does this happen to *any* pair of two students). This is the less likely interpretation, because it leads to a less interesting problem. Still, I'll solve that version now.

What is the probability that there end up being two students who are together all 3 years. Well, let's count the number of different class sequences someone could have over the next three years. Someone could be in group A all three years (AAA), or A, then B, then A again (AAA), or any of 27 sequences (there are 27 3 letter "words" using A,B,C). Each of the 54 students ends up with one of these sequences. Since there are 54 students, and 27 sequences, there must be two students that end up having the same sequence (this is known as the "pigeonhole principle"). So the probability for this interpretation of the question would be 1. (which is why I think this is not what was meant).

Cheers,

Brian