…maybe. Last week, I had an interview with a quantitative trading firm. I was asked to solve a probability problem that’s directly related to this course. The question was:

Five friends are seated at a dinner table randomly. If no one shares a birthday, what is the probability that they will be seated in order of age (either decreasing or increasing) around the table?

We came across a similar problem in the beginning of the course when discussing cycle structures. You’ll probably remember that the number of distinct cycle structures for a set of 5 elements is 4! = 24. Now how many of these correspond to decreasing order of ages? 1. Increasing order of ages? 1. So 2 out of 24 arrangements work. Therefore the probablility we are looking for is 1/12. In general, for n friends, the probability of correct age ordering is

That’s one way to solve the problem, but we can also apply another math 152 concept to solve this problem – permutations. Assume without loss of generality that the 5 friends sit down in order of age and then perform a random permutation by switching seats. How many permutations are possible. This is simply

How many permutations will lead to another correct age ordering of the friends? We know that any permutation group of a polygon will preserve a cycle structure (if we are indifferent between forward and backward ordering). Thus the permutations of a pentagon’s symmetry group will give us correct orderings. Remember that for a pentagon, we have one identity, four rotations, and five reflections. Thus, the following number of permutations will work:

Since 10 out of 120 permutations will lead to the correct age ordering, the probability of 5 friends sitting in order of ages is 10/120 = 1/12. We get the same result either way!

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Harvard’s Math 152 « Metroplex Math Circleon November 26, 2008at 11:39 am