dynamic partitioning scheme :OS

Question

Consider a dynamic partitioning scheme.
Show that on an avg. memory contains half as many holes as segments??

i am not able to understand the concept of relation bw holes and segments in context to this question..

Answer 1

Let s and h denote the average number of segments and holes, respectively. The probability that a given segment is followed by a hole in memory (and not by another segment) is 0.5, because deletions and creations are equally probable in equilibrium. So with s segments in memory, the average number of holes must be s/2.