chairs

**heathrowjohnny** · February 18th 2008, 03:30 AM

Six individuals, including $A$ and $B$ , take seats around a circular table in a completely random fashion. Suppose the seats are numbered $1, \ldots, 6$ . Let $X = \text{A's seat number}$ and $Y = \text{B's seat number}$ . If $A$ sends a written message around the table to $B$ in the direction in which they are closest, how many individuals (including $A$ and $B$ ) would you expect to handle the message?

**xifentoozlerix** · February 18th 2008, 03:48 AM

Originally Posted by heathrowjohnny

Six individuals, including $A$ and $B$ , take seats around a circular table in a completely random fashion. Suppose the seats are numbered $1, \ldots, 6$ . Let $X = \text{A's seat number}$ and $Y = \text{B's seat number}$ . If $A$ sends a written message around the table to $B$ in the direction in which they are closest, how many individuals (including $A$ and $B$ ) would you expect to handle the message?

Using a purely intuitive argument, the number of people who can handle the message is either less than or equal to 4. The worst case scenario would have A and B directly facing each other, which would leave 2 persons on either side between them.

**heathrowjohnny** · February 18th 2008, 03:50 AM

but you have to use expected values (e.g probability of sitting in a seat is $1/6$ ).

**xifentoozlerix** · February 18th 2008, 05:01 AM

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.

**Plato** · February 18th 2008, 07:55 AM

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 $P(X = 2) = \frac{{2\left( {4!} \right)}}{{5!}} = \frac{2}{5}$ . Or more simply, B can sit in two chairs of the five so that X=2. By the same simple logic we get $P(X = 3) = \frac{2}{5}\,\& \,P(X = 4) = \frac{1}{5}\,$ .
$E(X) = \frac{2}{5}\left( 2 \right) + \frac{2}{5}\left( 3 \right) + \frac{1}{5}\left( 4 \right) = 2.8$ .

**xifentoozlerix** · February 18th 2008, 11:37 AM

Thanks Plato, I knew my method was somehow flawed. Seems I let the constraints posted in the problem constrain my logic.

**heathrowjohnny** · February 18th 2008, 01:14 PM

Yeah it seems in problems where people are sitting in a circle, you always set one person down first to set an order (e.g. circular permutations).

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