Details

Snapshot 1: There are four pairs of shoes. Initially, two of them are at the front door and two at the back door. In the first walk, Mr. A happens to start at the back door (the number of pairs of shoes at the back door drops to 1, as the blue curve shows), and he also happens to stop at the back door (the number of pairs of shoes at the back door goes back to 2). In the second walk, Mr. A happens to start at the front door (the number of pairs of shoes at the front door drops to 1, as the red curve shows), and he happens to stop at the back door (the number of pairs of shoes at the back door grows to 3). In this way the process continues. At the end of the fourth walk, the front and back doors contain 0 and 4 pairs of shoes, respectively. Unfortunately, when going for the fifth walk, Mr. A happens to choose the front door as the starting point. He then finds no shoes. The average number of finished walks when starting with four pairs of shoes is 12 (the number of finished walks was four in Snapshot 1).

Snapshot 2: The initial situation is the same as in Snapshot 1. Now, there are only the minimum number of finished walks, 2. Indeed, at the start of the first, second, and third walks Mr. A happens to choose the front door and at the end of the first and second walk he happens to stop at the back door. At the start of the third walk, he finds no shoes at the front door.

Snapshot 3: Again, the initial situation is the same as in Snapshot 1. Now Mr. A can complete 44 walks.

The average number of finished walks can be shown [1] to be , where is the total number of pairs of shoes. So, if there are initially 1, 2, 3, …, 10 pairs of shoes at both the front and back door (that is, takes the values 2, 4, 6, …, 20), the average number of finished walks is 4, 12, 24, 40, 60, 84, 112, 144, 180, and 220, respectively.

The Demonstration is based on [1, pp. 38–39]. A similar Demonstration is Umbrella Quandary.

Reference

[1] J. Andel, Mathematics of Chance, New York: Wiley, 2001.