Variations on the Gambler s Ruin Problem

Transcription

1 Variations on the Gambler s Ruin Problem Mat Willmott December 6, 2002 Abstract. This aer covers the history and solution to the Gambler s Ruin Problem, and then exlores the odds for each layer to win in three variations on the roblem: the attrition variation, a variation in which one layer has an infinite number of oints, and a three layer variation, in which two layers lay against a house layer. 1. Introduction. Throughout the years, the Gambler s Ruin Problem has been rominent in alied mathematics. With differing levels of comlexity, variations on the roblem arise in all tyes of games, from a child s board games to comlicated casino games such as cras and blackjack. The theory behind the roblem is also used in horse racing and dog racing, generally by the track to be sure that it makes a rofit. There are many variations on the roblem, and as fast as older ones are being solved, newer ones are being formulated. Four-layer and five-layer variations, for examle, have already been roosed, and some have even been solved. The Gambler s Ruin Problem is clearly an imortant and growing toic in discrete alied mathematics; it was relevant in the 1600s and is still relevant today. Section 2 covers the history of the Gambler s Ruin Problem. Then Section 3 discusses and gives a solution to the most common form of the roblem, closely following the solution resented by DeGroot [2, ]. Sections 4 through 6 discuss some of the more common variations on the roblem. Section 4 resents and analyzes a game where the layers do not win oints on successful trials, but only lose them on unsuccessful ones. Section 5 discusses a variation in which one layer has an infinite money suly. Section 6 concludes the aer with a discussion of a three layer variation on the roblem. 2. History. The Gambler s Ruin Problem is one of the oldest roblems in robability theory. According to Edwards [3,. 73], Pascal osed a roblem similar to the Gambler s Ruin Problem in 1656 in a letter to Fermat. Caravaci later summarizes the letter: Let two men lay with three dice, the first layer scoring a oint whenever 11 is thrown, and the second whenever 14 is thrown. But instead of the oints accumulating in the ordinary way, let a oint 1

2 be added to a layer s score only if his oonent s score is nil, but otherwise let it be subtracted from his oonent s score. It is as if oosing oints form airs, and annihilate each other, so that the trailing layer always has zero oints. The winner is the first to reach twelve oints; what are the relative chances of each layer winning? This roblem, however, is not the Gambler s Ruin Problem in its common form. Edwards goes on to exlain that the common form comes from Huygens, who read and restated the roblem. Originally, Huygens looked at the above roblem as if oints were accumulated normally, and the winner were the first layer to lead by 12 oints. Later, he again restated the roblem as follows: Problem 2-1 Each layer starts with 12 oints, and a successful roll of the three dice for a layer getting an 11 for the first layer or a 14 for the second adds one to that layer s score and subtracts one from the other layer s score; the loser of the game is the first to reach zero oints. What is the robability of victory for each layer? The above roblem is known as the Gambler s Ruin Problem because one of the layers will run out of money be ruined at the end of the game. This form is the form most commonly used today. 3. The Basic Form of the Gambler s Ruin Problem. To solve Problem 2-1, we look at the game from the oint of view of one of the layers, say Player A. We name Player A s oonent Player B. Say Player A starts with i oints, and Player B starts with k i oints, so that the total number of oints in the game is k. For examle, in Section 2, we see that i = 12 and k = 24. Player A wins when he has k oints, and loses when he has zero oints. Also, say that the robability of A winning the next oint is, so the robability of B winning the next oint is q = 1. Finally, say that a i is the the robability of A reaching k oints before reaching 0 oints when starting with i oints; so a 0 = 0 and a k = 1. Let us refer to the event that A reaches k oints before B as W, the event of A winning the first oint as A 1, and the event of A losing the first oint B winning the first oint as B 1. Then P W = P A 1 P W A 1 + P B 1 P W B Substituting in the robability definitions made above, we get the following system of equations: 2

3 a 1 = a 2 + qa 0 = a 2 a 2 = a 3 + qa a k 2 = a k 1 + qa k 3 a k 1 = a k + qa k 2 = + qa k 2. Since + q = 1, we can relace each a i with a i + qa i. Then Equations 3-2 become a 2 a 1 = q a 1 a 3 a 2 = q q 2a1 a 2 a 1 = a 4 a 3 = q q 3a1 a 3 a 2 = 3-3 a k 1 a k 2 =. q k 2a1 a k 2 a k 3 = 1 a k 1 = q q k 1a1 a k 1 a k 2 =. And finally, we can sum Equations 3-3 to get k 1 q i. 1 a 1 = a For a fair game, = q = 1/2, and so Equation 3-4 becomes 1 a 1 = k 1a 1, that is, a 1 = 1/k. From Equations 3-3, we can see that for = q, we have a 2 = 2a 1 and a 3 = 3a 1 and so on, so that i=1 a i = i/k. 3-5 For an unfair game, q, and so Equation 3-4 becomes This exression can be simlified to 1 a 1 = a 1 q k q q 1. a 1 = q 1 q k 1. 3

4 Using Equations 3-2, this formula can be generalized to a i = q i 1 q k We can use Equation 3-6 to solve Problem 2-1. Assume Player A is rolling for an 11. Since the chances of rolling a 14 on three dice is 15/216 and the chances of rolling an 11 is 27/216, we have q/ = 5/9. Plugging this ratio into Equation 3-6 and using i = 12 and k = 24, we get a i = , which is about 1156/1157, the same solution that Pascal came u with. The Gambler s Ruin Problem can be modified and generalized to aly to many different tyes of games with different numbers of layers, different tyes of layers, and different rules. 4. The Attrition Variation. One of the most common variations on the Gambler s Ruin Problem is called attrition. In attrition, one layer does not win a oint from the other layer on a successful lay, such as a roll of an 11 for Player A in Problem 2-1. Instead, the losing layer simly discards a oint. As W. D. Kaigh describes it in [4,.22], this variation alies to many situations from oular board games like Risk to best-of-seven sorts series like the World Series or the Stanley Cu finals. The robability that Player A wins the game, and B is ruined, can be found by examining A s total score at the end of the game. We denote the number of oints that A has lost at the end of the game by L A. We define the event W, the robabilities a i,, and q, and the variables i and k as in Section 3. We now see that 0 L A i 1. As a result, we see that i 1 P W = P W L A = x. 4-1 x=0 For simlicity s sake, let us refer to Player B s starting score as b, where b = k i. Then for A to win while losing x oints in the rocess, we want Player A to win b oints in the same amount of time that it takes for Player B to win x oints. Using basic binomial robability [2,.84 85], we rewrite this condition as b + x 1 P W L A = x = b q x. 4-2 x Combining Equation 4-1 with Equation 4-2, we obtain the robability that A wins the game, or that B is ruined, as i 1 b + x 1 a i = b q x. 4-3 x x=0 4

5 5. One Player with Limited Points vs. One Player with Infinite Points. In another common variation on the Gambler s Ruin Problem, one layer, B say, has an infinite oint suly. Obviously, this hyothesis eliminates the game art of the roblem because B can never lose. However, we can still examine the numerical consequences of having a layer with an infinite oint suly. For < 1/2, we have q/ > 1, and as k goes to infinity in Equation 3-6, we see that a i always aroaches zero. For > 1/2, we have q/ < 1, and as k goes to infinity in Equation 3-6, we see that the chance that A is not ruined is q i. a i = 1 Player A can never win because B has an infinite number of oints and therefore can not be ruined, so this equation only gives the robability that A will continue to lay forever. For = 1/2, we use Equation 3-5 to find the chance that A is not ruined. Clearly, as k aroaches infinity, i/k aroaches zero, so Player A loses all of his oints. We can conclude that A is always ruined eventually for 1/2, and has a robability of a i = 1 q/ i of laying forever for > 1/2. 6. A Generalization of the Problem to Three Players. We can generalize the Gambler s Ruin Problem to three layers. Player A and Player B lay games against each other, but they also lay a combined game against a searate layer, called C. In this variation, however, A and B lace two half-oint bets on every lay, rather than one full-oint bet. In the game between A and B, say A wins the half-oint bet with robability 1, and B wins with robability q 1 = 1 1. Also, A and B both contribute a half-oint to a combined full-oint bet against C. Say A and B win this bet with robability 2, and C wins with robability q 2 = 1 2. Now suose that A, B, and C start with i, j, and l oints resectively. Set k = i + j + l. Then k is the total number of oints in the game; it is the number of oints the winning layer has at the end of the game when the other two layers are ruined. We denote the event that Player X gains a oint by G X, and the event that Player X loses a oint by L X, where X = A, B, or C. Then we define the four ossibilities on each lay as := P G A and L C = 1 2, q := P L A and G C = q 1 q 2, 6-1 r := P G B and L C = q 1 2, s := P L B and G C = 1 q 2. 5

6 Let x and y be the total scores for A and B resectively. Player A is ruined when x = 0; Player B is ruined when y = 0; Player C is ruined when x + y = k. Therefore, the first ortion of the game, that is, the three layer ortion before one layer is ruined, can be described as a two dimensional random walk within the triangle bounded by x = 0, y = 0, and x+y = k. Once one of the boundaries of the triangle is reached, the three layer ortion of the game ends and the game becomes a standard two layer Gambler s Ruin Problem. This two dimensional walk is illustrated in Figure ,k B wins C ruined A ruined r q * s 0,0 C wins B ruined k,0 A wins Figure 6-1. Reresentation of the two dimensional random walk created by the three layer game. The oint may move u, down, left, or right according to the robabilities given. As Barnett describes in [1, ], there are six ossible ways for the game to lay out: any one of the three layers may be ruined first, and then either of the two remaining layers may be ruined, leaving a single victorious layer. Thus, each layer has two ways of winning. For examle, A may win by first ruining B in the three layer game, and then ruining C in the two layer game or by ruining C in the three layer game, and then ruining B in the two layer game. Informally, this organization of the game is described as: 6

7 P A wins = P B is ruined first P A goes on to ruin C + P C is ruined first P A goes on to ruin B, P B wins = P A is ruined first P B goes on to ruin C P C is ruined first P B goes on to ruin A, P C wins = P A is ruined first P C goes on to ruin B + P B is ruined first P C goes on to ruin A. Converting these equations into more useful ones, however, is more difficult than it first sounds. The robability of a layer winning the two layer ortion of the game deends on the starting number of oints of the two layer ortion, that is, the ending number of oints of the three layer ortion. Therefore, we must sum over all the ossible scores for the start of the two layer ortion of the game. Define the function ux, y to be the robability that the three layer ortion of the game ends at the oint x, y. Thus, we combine Equation 3-6 with Equations 6-2 and simlify to get the robabilites that A, B, or C resectively win the game: P A = P B = P C = k 2 q 2 2 n 1 k 1 q 1 q 2 n=1 2 k un, n 1 1 q 1 n=1 1 k un, k n, 1 k 2 q 2 2 n 1 k 1 1 q q 2 n=1 2 k u0, n + 1 k n n=1 q 1 k un, k n, k 2 2 q 2 k n 1 un, u0, n. q 2 k 1 n=1 Barnett continues to show that we can break down the robability of ultimate victory for each layer by defining a better ux, y. Define E i,j x, y to be the average number of times the two dimensional walk starting at i, j leaves the oint x, y. Then we see that the ux, y that we used in defining the robabilities is written as follows: un, 0 = se i,j n, 1 for n = 1, 2,..., k 2; u0, n = qe i,j 1, n for n = 1, 2,..., k 2; 6-4 un, k n = re i,j n, k n 1 + E i,j n 1, k n for n = 1, 2,..., k 1. Finally, we can ut Equations 6-4 into Equations 6-3 to obtain formulas for P A, P B, and P C. 7

8 References [1] Barnett, V. D., A three-layer extension of the gambler s ruin roblem, Journal of Alied Probability , [2] DeGroot, Morris H., Probability and Statistics, Addison-Wesley Publishing Co., [3] Edwards, A. W. F., Pascal s roblem: the gambler s ruin, International Statistical Review , [4] Kaigh, W. D., An attrition roblem of gambler s ruin, Mathematics Magazine ,