Probability that Ken and John set next to each other > A group of ten people sits down, uniformly at random, around a table. Ken and John are part of this group. Determine the probability that Ken and John sit next to each other. There are $10!$ ways to arrange the seating for everyone, there are 10 possible ways for John and Ken to sit together. $$\operatorname{Pr}(J\ \&\ K ) = \frac{10}{10!} = \frac{1}{9!}$$ Am I correct?

Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.