Snapshot 1: There are 10 cards and initially player A has five cards. The game lasted 313 steps and player A won the game. In this example, the expected duration is approximately 584 steps. The lower-left plot shows that the duration does not depend much on the initial number of cards; this is the result of the random walk having the tendency to move away from the absorbing barriers. Even with one or nine cards, the expected duration is only somewhat smaller, 511. Similarly, the probability of ruin does not depend much on the initial number of cards and is quite near to 0.5 for all initial numbers of cards. Only the case of having initially one or nine cards has a somewhat larger effect to the probability of ruin.

Snapshot 2: The initial situation is as in the first snapshot. However, now the game lasted the minimum number of steps, five.

Snapshot 3: Again this starts as in snapshot 1, but the game lasted 4765 steps!

Snapshot 4: With only six cards, the expected duration of the game is 37.

Snapshot 5: With 20 cards, the expected duration of the game is over a half-million steps!

Here are the expected durations of the game (rounded to the nearest integer) for 2, 4, 6, …, 20 cards where player A initially has half of the cards:

The discrete-time Markov chain is as follows. The walker A moves among the points 0, 1, 2, …,

. The chain is said to be currently in state

if player A has

cards. The walker moves from state

to state

with probability

and to state

with probability

. States 0 and

are absorbing, so that the chain remains in state 0 or

once it is achieved. For example, the walker moves from state 1 to state 0 or from state

to state

only with probability

. This implies that the walker hesitates to stop.

Consider the expected duration of the game. Let

be the expected duration if player A initially has

cards. After the first step, A has either

cards with probability

or

cards with probability

. Thus, by the theorem of total expectation,

. This recurrence equation can be solved for

in terms of

and

. Clearly, the initial condition is

. It can be shown [1] that

.

Consider then the probability of ruin. Let

be the probability of ruin if player A initially has

cards. By the theorem of total probability,

. This equation can be solved for

in terms of

and

. As to initial conditions, clearly

. It can be shown [1] that

.

The reluctant random walk can be compared with the so-called gambler's ruin process; see

The Gambler's Ruin. Assume, for example, that player A initially has five dollars and that the casino also has five dollars. At each step, player A wins or loses one dollar with probability 0.5. The game stops once A has 0 or 10 dollars. The expected duration of this game is only 25. Recall that in the reluctant random walk the corresponding expected duration was 584.

This Demonstration is based on [1, pp. 34–37].

[1] J. Andel,

*Mathematics of Chance*, New York: Wiley, 2001.