# How state 1 has a 0.5 chance of terminating on the left, and state 950 has a 0.25 chance of terminating on the right?

Sutton-Barto's RL book (page 203)

Example 9.1: State Aggregation on the 1000-state Random Walk: Consider a 1000-state version of the random walk task (Examples 6.2 and 7.1 on pages 125 and 144). The states are numbered from 1 to 1000, left to right, and all episodes begin near the center, in state 500. State transitions are from the current state to one of the 100 neighboring states to its left, or to one of the 100 neighboring states to its right, all with equal probability. Of course, if the current state is near an edge, then there may be fewer than 100 neighbors on that side of it. In this case, all the probability that would have gone into those missing neighbors goes into the probability of terminating on that side (thus, state 1 has a 0.5 chance of terminating on the left, and state 950 has a 0.25 chance of terminating on the right). As usual, termination on the left produces a reward of −1, and termination on the right produces a reward of +1. All other transitions have a reward of zero. We use this task as a running example throughout this section.

Question: I did not understand how state 1 has a 0.5 chance of terminating on the left, and state 950 has a 0.25 chance of terminating on the right. Could someone explain this calculation in detail?