Two rivals kept apart

Question

Eight people, including two rivals, are seated uniformly at random. What is the probability that the two rivals do NOT sit next to each other (a) in a row of 8 seats, (b) around a circular table of 8?

Accepted Answer

"Not adjacent" is easiest as one minus "adjacent," so first find the adjacency probability by conditioning on one rival's seat. (a) Row. Fix one rival; the other is equally likely in any of the 7 remaining seats. Counting adjacent seats: if the first rival is at an end (2 of 8 seats) only 1 seat is adjacent, otherwise 2 of 7 are: P(	ext{adjacent}) = \frac{2}{8}\cdot\frac{1}{7} + \frac{6}{8}\cdot\frac{2}{7} = \frac{2 + 12}{56} = \frac{1}{4}. So P(	ext{not adjacent}) = 1 - 	frac14 = 	frac34. (b) Circle. By rotational symmetry, fix one rival anywhere. Every seat at a round table has exactly …

Two rivals kept apart

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