# probability around a round table , PMF

A round table has n seats. n people are seated at random around the table. Fred dislikes two of the people. Let X be the number of neighbors of Fred whom he dislikes. Find the p.m.f. of X. (Note that X can only be 0, 1, 2. )

Ans: The total Number of ways of seating arrangements= (N-1)! Fred dislikes two of the people. These two may or may not be his neighbors. so X = 0,1,2 (given)

I am not sure how to proceed further to find the pmf.

Can anyone help with explanation please?

• I'm still wondering what you thought of my solution. – Geoff Pointer Dec 16 '13 at 2:18

Perhaps I jumped too quickly at the solution. If you think so, see Addendum 1 below.

Let disliked people be D and others G. So $L_D$ is a person Fred dislikes on his left, $R_D$ on his right and similarly $L_G$ and $R_G$ for others. With n people in a circle, Fred has one person either side. Wherever Fred sits, there are 2 of $D$ and $n-3$ of $G$ and there are four possible scenarios.

$$L_G \text{ and } R_G \quad L_G \text{ and } R_D \quad L_D \text{ and } R_G \quad L_D \text{ and } R_D.$$

I'll start out with a solution with slightly abused notation. There are conditional probabilities involved as I will explain but I'll keep the notation brief.

\begin{align*} P(0) &= P(L_G)\text{ and } P(R_G) = \frac{n-3}{n-1}\frac{n-4}{n-2} \end{align*}

So the probability is not just a simple independent product. Given there's a choice made on Fred's left the choice on his right is out of $n-2$ people with $n-4$ G people left.

\begin{align*} P(1) &= P(L_D)\text{ and } P(R_G) \text{ or } P(L_G)\text{ and } P(R_D) \\ &= \frac{2}{n-1}\frac{n-3}{n-2} + \frac{n-3}{n-1}\frac{2}{n-2} \end{align*}

Again note how the numbers change according to dependence.

\begin{align*} P(2) &= P(L_D)\text{ and } P(R_D) \\ &= \frac{2}{n-1}\frac{1}{n-2} \end{align*}

If I'm not mistaken $P(0) + P(1) + P(2) = 1$

Addendum 1: The principle I'm applying here is Occams Razor. Look for the simplest solution first. Calculating all the possible permutations and combinations is not often necessary in this kind of problem. Fred has to sit down, it doesn't matter where. Once he's sat down, fill the seats either side of him at random. The remaining seats don't matter. So, the only acts that involve uncertain probabilities are filling the seats either side of Fred. It took me a while to get used to thinking this way about probability problems that involve counting.

And, by the way, they do add up to one, the numerators sum and factorise to $(n-1)(n-2)$.

• @manayay I added some explanation to my answer just in case you wanted to see how it was motivated. – Geoff Pointer Sep 24 '13 at 6:15
• @manayay Does that mean you accept my answer, or do I need to add something? – Geoff Pointer Sep 26 '13 at 4:21
• yes i do accept – manayay Dec 17 '13 at 16:22
• @manayay Until you click the tick this remains listed as an unanswered question, which is why I keep coming back to check it. – Geoff Pointer Dec 17 '13 at 23:16