round table seating probability There are $6$ people, let's call them - (a,b,c,d,e,f), to sit at a round table. The number of ways they can arrange themselves is $(6-1)! = 5! = 120$ ways. 
What is the probability that person 'a' will have person 'b' sat to his immediate left, and person 'c' sat to his immediate right? I'm confused on how to go about this.
 A: You have already counted the number of arrangements (with rotations being equivalent) correctly as $$\frac{6!}{6} = 120$$
Now you need to count the number of arrangements (with rotations being equivalent) in which $a$ has $b$ to his left and $c$ to his right. 
To do this, treat the group of $a$, $b$, and $c$ as one person, and count the number of arrangements of the four people as $$\frac{4!}{4}=6$$

Your final probability is the number of "successes" ($a$ has $b$ to the left and $c$ to the right) divided by the total number of possibilities, or $$\frac{6}{120}=\frac{1}{20}=\boxed{0.05}$$
A: Given that $a$, $b$ and $c$ are placed in the order $bac$, with $a$ in a given position, there are $3!=6$ ways to arrange the remaining three people relative to these. The probability of such a placement ocuring is then just $ \frac{6}{120}=\frac{1}{20}$.
A: Actually, the way you had the question worded originally, there is no way to solve this question without more information and/or making assumptions.  For example, there could be several seated positions that are to 'a's left but we don't know for sure because we don't know the spacings of how the people are seated (such as are they equally spaced around the round table?).
