Six individuals, including and , take seats around a circular table in a completely random fashion. Suppose the seats are numbered . Let and . If sends a written message around the table to in the direction in which they are closest, how many individuals (including and ) would you expect to handle the message?
We can assume that X = 1.
Then E(X) = 1 and
E(Y) = (1/5)(2+3+4+5+6) = 20/5 = 4.
You can then define a random variable Z = 4-abs(abs(X-Y)-3). This is the number of people who handled the letter.
E(Z) = 4-abs(abs(E(X)-E(Y))-3) = 4-abs(abs(1-4)-3) = 4.
I could be wrong though, since I don't know if you can just assume that wherever A is sitting will be where you start numbering the chairs. The method I used to find Z is kind of complicated to explain. A general solution is beyond my grasp of Prob/Stats.
I disagree with the reasoning in the replies to this question.
Also I see no need to confuse things by numbering the chairs.
Just seat person A at table. Now the table is ordered.
There are (5!) ways to seat the other people.
If X is the number of people handling the note, the X=2,3, or 4.
Where are two ways to seat B next to A and then (4!) ways to seat the others. So . Or more simply, B can sit in two chairs of the five so that X=2. By the same simple logic we get .
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