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16 cops and 8 suspects are sitting in a row, waiting for a train.

It would be desirable that each suspect is flanked on both sides by cops.

If they sit randomly, what is the expected value of the # of suspects so flanked ?

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What about extremities? If $C$ is a cop and $S$ is a suspect, do you consider $$ SCC \dots CS $$ the suspects at both ends flanked or not? – Patrick Da Silva Feb 27 '12 at 1:57
@Patrick: No, they have to be flanked on both sides. – true blue anil Feb 27 '12 at 2:25
up vote 3 down vote accepted

Expectations add, so you can compute an individual suspect's odds of being flanked and then multiply by 8. Each suspect has a $\frac{22}{24}$ chance of sitting somewhere other than the ends (where said suspect cannot be flanked), and then there is a $\frac{16}{23}$ chance the person to his left will be a cop, and then a $\frac{15}{22}$ chance the person to his right will be a cop (given that the person to his left is a cop). $$\frac{22}{24}\cdot\frac{16}{23}\cdot\frac{15}{22}=\frac{10}{23}$$ The expected number of suspects flanked by cops is $\frac{80}{23}$

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+1: Didn't realise it'd be so simple. Thanks a bunch ! – true blue anil Feb 27 '12 at 3:07
I'm sorry if this is basic but how do you obtain 16 out of 23? I tried to repeat the pattern for smaller numbers of cops/suspects in order to understand, yet it didn't work out. – 355durch113 Mar 12 at 18:02

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