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Coscott comments on Open Thread for February 11 - 17 - Less Wrong
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Comments (325)
My Answer
Coscott's solution seems incorrect for N=3. label 3 cars 1 is fastest, 2 is 2nd fastest 3 is slowest. There are 6 possible orderings for the cars on the road. These are shown with the cars appropriately clumped and the number of clumps associated with each ordering:
1 2 3 .. 3 clumps
1 32 .. 2 clumps
21 3 .. 2 clumps
2 31 .. 2 clumps
312 .. 1 clump
321 .. 1 clump
Find the mean number of clumps and it is 11/6 mean number of clumps. Coscott's solution gives 10/6.
Fix?
My solution gives 11/6
Dang you are right.
Coscott's solution also wrong for N=4, actual solution is a mean of 2, Coscott's gives 25/12.
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.
You got it.
I am not sure what the distribution is.
The distribution; see e.g. here.
Ah, yes, thank you.