Gambler’s Ruin Problem: Some Extensions

For calculating the expected ruin time of the classic three-player symmetric game, Sandell derived a general formula by introducing an appropriate martingale and stopping time. However, the martingale approach is not appropriate to determine the ruin time of asymmetric game. In general, ruin probabilities in both cases, that is, symmetric and asymmetric games as well as the expected ruin time for the asymmetric games are still need to be calculated. The current work is also about three-player gambler’s ruin problem with some extensions. We provide expressions for the ruin time with or without ties when all the players have equal or unequal initial fortunes. Finally, the validity of the asymmetric game is also tested through a Monte Carlo simulation study.


Introduction
In the classical three-player gambler's ruin problem, three players A1, A2 and A3 start a game with arbitrary initial stakes, say 1 a , 2 a and 3 a dollars, respectively. In each round, based on randomness, the winner (with probability i p , where i = 1, 2, 3 and 1   i p ) receives one dollar from both the losing players. The game continues until one of them is left with zero dollars. Of interest is the ruin probability for each player and the ruin time of the game. To date, no corresponding expressions for ruin probabilities have been found for both types of games, that is, symmetric and asymmetric 3 School of Science Volume 5 Issue 3, September 2021 games. Ruin time was claimed as an unsolved problem in the early 1960s, see for example [1]. However, in the late 1980s, Sandell [2] accepted this challenge and proposed the following formula with the help of martingale and the optional stopping theorem: where T denotes the first ruin time. The expression given in equation (1) is valid for both cases, either . The literature on asymmetric game for more than two players is more limited and no general expression is available for the ruin time to date. In this regard, [3,4] provided five different expressions to solve this type of game up to six dollars each as follows: ) , = 5 (2) In equation (2) . For c = 6, their given expressions including matrix and vector are very difficult to solve without a software. Recently, [5] provided a single approximation as an alternative of the five separate expressions of [3,4]. Their proposed structure is stated in the following form: Where, Inquiry and Review  Volume 5 Issue 3, September 2021   3  2  3  2  1  3  1  3  2  1  2  1  3  2  1 p  p  a  a  a  p  p  a  a  a  p  p  a  a In this research, we adapt a direct way to derive the ruin time for the three-player game with unequal initial fortunes. Also, we extend the symmetric game with the involvement of ties in each round of the game. In Section 2, we delineate the mathematical derivations for both the symmetric and asymmetric games with and without ties. Section 3 evaluates the results of the proposed asymmetric game with a Monte Carlo simulation study. At the end, a brief summary with observations is discussed in Section 4.

Mathematical Description
Let us assume that waiting time for the gambler's ruin game is finite. First, we consider the ruin time for two-player game, conditional expectation and random variable (see for example [6,7]), in the following way: where n signifies the remaining steps needed to complete the game, whereas a1 and a2 signify the initial fortunes for players A1 and A2, respectively. Moreover, where p and 1 − p represent the winning probabilities for players A1 and A2, respectively. Then, 2 1 ,a a T can be computed by using the law of total expectation as follows:

Ruin Time: Symmetric Case
For the three-player gambler's ruin problem, waiting time is also finite and expected duration can be written as follows: In this perspective, one out of the following three events has to occur: either the player A1 wins and not the other two players, or the winner player is A2 (whereas players A1 and A3 lose), or player A3 is declared as the winner School of Science Volume 5 Issue 3, September 2021 (players A1 and A2 lose). Similarly, the ruin time of the classic three-player game must satisfy the following difference equation: . We restrict this game up to the stage where at least one of the three players suffers a complete loss with the following boundary conditions: (5) is an inhomogeneous difference equation and should have two solutions, that is, the complementary and the particular solution. Then, the general solution takes the following form: where c a a a  (5) can be written without constant 1 as follows: In this system, there are 9 equations for 9 unknowns. Using the boundary conditions mentioned in equation (6)  . Hence, all other unknowns are also equal to zero and it proves that the homogeneous part (without 1) of system (5) with any arbitrary values of 1 a , 2 a and 3 a is always given a zero value. For the particular solution of system (5), we can take the lowest order of the third-order polynomial with 1 a , 2 a and 3 a . Hence, the value of p a a a T 3 2 1 , , is presented in the following form: where C is an arbitrary constant. We obtained the value of C using system (5) as follows: a a a C and the particular solution can be generated as follows: Hence, the general solution is in the following form:

Ruin Time with Ties: Symmetric Case
By involving the ties in each round of the three-player game, system (5) can be written as follows: It can be written in a more simple form as follows: Hence, the general solution, when ties are involved in the game, takes the following form:

Ruin Time: Asymmetric Case
For the asymmetric game, ruin time of the classic three-player game must satisfy the following difference equation:  where C is an arbitrary constant. We obtain constant C using the system (10) as follows: Now, the particular solution can be generated as follows: Hence, the general solution takes the following form:

Ruin Time with Ties: Asymmetric Case
By involving the ties in each round of the asymmetric three-player game, system (10) can be written as follows: It can be written in a more simple form as follows: . Constant C can be determined by using equation (12) as follows: The particular solution can be written as follows: Hence, the general solution takes the following form: In the next section, we demonstrate our proposition through numerical findings and also compare them with a Monte Carlo simulation study.

Numerical Results with Monte Carlo Simulation Study
In this section, we examine the performance of ruin time for unequal initial fortunes with various settings of probabilities. In order to evaluate the performance of the proposed asymmetric game, a Monte Carlo simulation School of Science Volume 5 Issue 3, September 2021 study was conducted through MATLAB programming language. In Table  1, we present two types of results of the expected duration, that is, the predecided probability for each player and the assigned probabilities with respect to their initial fortunes (probability proportion to size). One may appreciate the vividly close match between the proposed formula and simulation results, especially in the second case, when the probability of each player is based on their own initial fortunes (a real motive to start the game). At the end, two random clicks with different parametric settings are displayed in Figure 1.

Conclusion
For the three-player game, an explicit formula for the symmetric case is available in literature [2]. On the other hand, the asymmetric case is still School of Science Volume 5 Issue 3, September 2021 unsolved (no explicit formula is available). Recently, Hussain et al, 2021 solved this game, approximately [5]. In the current research, we have also solved this game with a different approach. The basic advantage of this approach is that it is conceptually simple for the students familiar with difference equations. Also, for the very first time in literature, tie is being introduced in this game for both cases, that is, symmetric and asymmetric. The legitimacy of the novel derivation of ruin time for the three-player gambler's ruin problem has been established mathematically and verified through a Monte Carlo simulation study.