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Coscott comments on Open Thread for February 11 - 17 - Less Wrong Discussion
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4 with prob 1/24, 3 with prob 6/24, 2 with prob 11/24, 1 with prob 6/24
Mean of 25/12
How did you get 2?
Must have counted wrong. Counted again and you are right.
Great problems though. I cannot figure out how to conclude it is the solution you got. Do you do it by induction? I think I could probably get the answer by induction, but haven't bothered trying.
Take the kth car. It is at the start of a cluster if it is the slowest of the first k cars. The kth car is therefore at the start of a cluster with probability 1/k. The expected number of clusters is the sum over all cars of the probability that that car is in the front of a cluster.