A GENERALIZED MARTINGALE BETTING STRATEGY

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1 DAVID K. NEAL AND MICHAEL D. RUSSELL Astract. A generalized martingale etting strategy is analyzed for which ets are increased y a factor of m 1 after each loss, ut return to the initial et amount after each win. The average amount et and the average final fortune are derived for sequences of n ets, for the numer of ets T that results in the first win, and for min(t, n. 1. Introduction In casino gamling, the old strategy is the process of always etting your entire stake or just enough to reach your target goal, whichever is least. It is argued in [2] and [5] that this method is the optimal strategy in terms of minimizing the proaility of ruin. The classical martingale strategy is to doule your et after each loss until you either win or have insufficient funds to continue douling the et. If your goal is simply to get ahead and you receive a 1:1 payoff, then douling an initial et of $1 is equivalent to the old strategy. But with an 8:1 payoff, there would e no need to increase the et until after the eighth loss. By douling the ets after each loss, very large ets may eventually need to e made which often are not allowed due to tale limits. The old strategy that accounts for house limits is discussed in [3]. Other examples of old play with regard to ruin are discussed in [1] and [4]. But is there really any need to doule the ets? Why not choose a smaller factor m so that the et sizes do not increase so quickly and do not run afoul of tale limits? In this paper, we shall analyze a more general scenario for which you multiply your last et y a factor of m 1 after each loss, ut use the initial et amount after each win. Some interesting questions arise using this strategy. Namely, what is the average amount et on the kth wager? What is your average fortune after a sequence of ets? How do we choose m so that we are guaranteed to come out ahead upon any win? Most importantly, to have a high proaility of coming out OCTOBER

2 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES ahead, how many ets would e needed and what initial stake would e needed? If we are wise, then we will quit whenever we get ahead, which should occur after the first win with the right choice of m using this general martingale strategy. So suppose we stop upon the first win. What would e the average amount et on this last wager, and what would e the average fortune after this last et that resulted in the first win? Finally, suppose that we stop upon the first win or after a maximum of n ets, whichever comes first. Now what are the average amount of the last et and the average fortune upon stopping? In this article we shall answer these questions using general proailistic concepts and geometric series. We then shall compare the results of this general martingale strategy with the results otained from some other asic strategies. 2. General Betting Notation Throughout, we shall let X n denote the ettor s fortune after n ets under our general martingale strategy, where X 0 is a random initial stake. There is proaility p of winning each et and proaility q = 1 p of losing. A typical sequence of wins and losses will e denoted y ω = (ω 1,..., ω n, where ω i = +1 to designate a win or 1 to designate a loss. The initial et is 1 = $, which is paid only if a loss occurs, and the initial payoff for winning is a 1 = $a. After the initial wager, the successive amounts et are dependent upon the previous outcome. If there was a win on the (k 1st et, then k returns to. But if there was a loss on the (k 1st et, then k = m k 1, for a fixed multiplicative factor m 1. Thus, for k 2, k (ω = 1 {ωk 1 =1} + m k 1 (ω1 {ωk 1 = 1}. (1 We note that, although the successive amounts et are dependent on the previous outcome, the last actual outcome ω k 1 of a win or loss is independent of the amount k 1 that had een et. If we were to et the same amount each time (i.e., use m = 1, then the average change of fortune after each et would e ap q. The average fortune in this case after n such ets is then E[X n ] = E[X 0 ] + n(ap q. (2 Because casinos operate to make a profit in the long run, they desire to have E[X n ] < E[X 0 ], which occurs if and only if ap q < 0. So the casino must choose a non-advantageous payoff a that satisfies a < (q/p. Under our general martingale strategy, we shall assume this condition as well as a fixed payoff ratio so that the payoff a k on the kth et satisfies a k / k = a/. 184 VOLUME 21, NUMBER 3

3 Thus, a k = (a/ k which from Equation (1 gives, for k 2, ( a a k (ω = a1 {ωk 1 =1} + m k 1 (ω1 {ωk 1 = 1}. (3 3. The Average Amounts Bet Because ω k 1 is independent of k 1 and the average of an independent product is the product of the averages, the average amount et on the kth wager, for k 2, can e written as E[ k ] = E[1 {ωk 1 =1}] + me[ k 1 ]E[1 {ωk 1 = 1}] = p + mqe[ k 1 ]. Thus, E[ 1 ] = and E[ 2 ] = p + mq, while E[ 3 ] = p + mqe[ 2 ] = p(1 + mq + (mq 2 and E[ 4 ] = p + mqe[ 3 ] = p(1 + mq + (mq 2 + (mq 3. By an inductive argument, we have for k 2, k 2 E[ k ] = p (mq i + (mq k 1. (4 i = 0 If mq = 1, then Equation (4 simplifies to (k 1p +. If mq 1, then we can simplify (4 using the geometric series formula n i = 0 xi = (1 x n+1 /(1 x for x 1. We therey otain our first result. Theorem 1. Under the conditions of the general martingale strategy when repeated rounds are played, the average amount et on the kth wager is p(k 1 + if mq = 1 E[ k ] = p(1 (mq k 1 + (mq k 1 if mq 1. Because q = 1 p, E[ k ] simplifies to when m = 1, and E[ k ] is clearly when k = Stopping the Betting Process Suppose that etting stops upon the first win and T (ω denotes the numer of ets needed for string ω, where T (( 1, 1, 1,... = +. Then T is a geometric random variale so that P (T = k = q k 1 p and P (T k = 1 q k for k 1. Then T is the amount of the et that gave the first win. In the case of all losses, we let = if m = 1 and let = + if m > 1. We now shall derive E[ T ], the average of all ets that give the first win. For stopping upon the first win or after a total of n ets, whichever comes first, we shall find E[ T n ], the average of the last amounts et where T (ω n = min(t (ω, n. OCTOBER

4 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES If q = 1, then there is only the string of all losses; hence, T and E[ T ] is either or +, depending on m. For q < 1, the string of all losses has proaility 0, so T can e written almost surely as T = m i 1 1 {T = i}. If we only allow a maximum of n ets, then for all q and all ω, T n = n 1 m i 1 1 {T = i} + m n 1 1 {T n}. Taking the expected value of T (for q < 1, we otain E[ T ] = m i 1 P (T = i = m i 1 q i 1 p = p (mq i, and taking the expected value of T n gives E[ T n ] = n 1 i = 0 m i 1 P (T = i + m n 1 P (T n n 2 = p (mq i + m n 1 q n 1. i = 0 Simplifying the geometric series, we otain the following closed forms for the desired averages. Theorem 2. Let T e the numer of ets needed for the first win. Under the conditions of the general martingale strategy, the average amounts et on the T th et and on the (T nth et are if m = 1 E[ T ] = + if m > 1 and mq 1 p if m > 1 and mq < 1 and p(n 1 + if mq = 1 E[ T n ] = p(1 (mq n 1 + (mq n 1 if mq VOLUME 21, NUMBER 3

5 It can easily e verified that lim n E[ T n ] = E[ T ] y direct evaluation of the limit. However, the result also follows from the Monotone Convergence Theorem. Indeed, T (ω n = n for n < T (ω, ut T (ω n = T (ω for all n T (ω; thus, lim n T (ω n = T (ω for all ω. Moreover, ecause we assume m 1, the sequential amounts et are increasing: 1 2 T. That is, T n increases almost surely to T ; thus, lim n E[ T n ] = E[ T ]. Comparing the results of Theorems 1 and 2, we also can see that E[ T n ] = E[ n ] for all n, even though T n and n are different functions. In fact, T n and n are identically distriuted as we next prove. Theorem 3. For all n 1, T n and n have the same distriution. Proof. If 1 T (ω n 1, then the first win occurred efore n ets were made. Letting T (ω = k, then there were k 1 initial losses followed y the win on the kth et, which occurs with proaility q k 1 p. Because of the k 1 straight losses, the amount of this last et was m k 1. If there were n 1 losses in a row, which occurs with proaility q n 1, then T (ω n = n and the amount of the nth and last et is m n 1. Thus, there are precisely n values in the range of T n which is the set {,..., m n 1 } and q j p for 0 j n 2 P ( T n = m j = q n 1 for j = n 1. For a complete sequence of n ets, the amount placed on the nth et depends on how many losses in a row preceded that et. If there were n 1 losses in a row, which occurs with proaility q n 1, then n = m n 1. If there were fewer losses in a row just efore the nth et, say j losses for 0 j n 2, then a win must have immediately preceded the string of losses. This event occurs with proaility pq j and occurs if and only if n = m j. Thus, n has the same range as T n and P ( n = m j = P ( T n = m j for 0 j n 1. In other words, you may get your first win efore making n ets and stop etting at that point. Or you might make a full sequence of n ets where each win throughout causes the next et to go ack to $. Either way, the possiilities for your last et, T n or n, are precisely the same and E[ T n ] = E[ n ]. However, we shall see that the average fortunes after these last ets, E[X T n ] and E[X n ], are not the same. 5. The Average Fortune After n Bets With our general martingale strategy, the fortune after the nth et has either increased y an amount of a n = (a/ n from the fortune after the OCTOBER

6 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES previous et, or has decreased y an amount of n. So for n 1, X n can e written as ( a X n (ω = X n 1 (ω + n (ω1 {ωn=1} n (ω1 {ωn= 1}. Because the amount et n is independent of whether or not this et is won ω n, we have ( a E[X n ] = E[X n 1 ] + E[ n ]E[1 {ωn=1}] E[ n ]E[1 {ωn= 1}] ( a = E[X n 1 ] + E[ n ] p q. (5 Applying recursion, we otain ( a E[X n ] = E[X n 2 ] + (E[ n ] + E[ n 1 ] p q ( a = E[X n 3 ] + (E[ n ] + E[ n 1 ] + E[ n 2 ] p q = = ( a n E[X 0 ] + p q E[ k ]. (6 k = 1 If m = 1, then E[ k ] =, which gives E[X n ] = E[X 0 ] + n(ap q. Otherwise, we use the result of Theorem 1 to otain E[X 0 ] + ( a p q n (p(k 1 + if mq = 1 k = 1 E[X n ] = E[X 0 ] + ( a p q n ( p(1 (mq k 1 1 mq + (mq k 1 if mq 1. k = 1 Simplifying this result gives us our next desired average. Theorem 4. Under the conditions of the general martingale strategy when repeated rounds are played, the average fortune after a sequence of n ets is ( (n 1p E[X 0 ] + n(ap q + 1 if mq = 1 2 E[X n ] = ( ( ap q E[X 0 ] + np + q(1 m(1 (mqn if mq 1. We note that for m = 1, the expressions simplify to E[X 0 ] + n(ap q in oth resulting cases of q = 1 and q 1. Moreover, ecause of our assumption that ap q < 0, we see from Equations (5 and (6 that E[X n ] < E[X n 1 ] and E[X n ] < E[X 0 ] for all n VOLUME 21, NUMBER 3

7 6. The Average Fortune Upon Winning We again let T denote the numer of ets needed to attain the first win. Then X T denotes the final fortune when stopping after the first win no matter when it occurs. But if we stop upon winning or making a total of n ets, then X T n denotes the final fortune. We first shall derive E[X T ]. This average depends on several conditions regarding the terms m and mq. But in all cases, we will assume that q < 1 so that T < with proaility 1. Suppose we win for the first time on the ith et. Then the first i 1 wagers all resulted in losses which means that j = m j 1 for 1 j i. The payoff for winning when T = i is a i = (a/ i = m i 1 a. We now let D i denote the total deficit accrued with the initial i 1 losses. Then (i 1 if m = 1 i 1 D i = m j 1 = (7 j = 1 mi 1 1 if m > 1. Hence, the fortune after the first win can e written (almost surely as ( X T = X 0 + m i 1 a D i 1{T = i} (8 = X 0 + a (i 11 {T = i} if m = 1 X (( a m i 1 1 {T = i} if m > 1. Taking the expected value when m = 1, we otain E[X T ] = E[X 0 ] + a i P (T = i + P (T = i = E[X 0 ] + a E[T ] + ( 1 = E[X 0 ] + a p 1 = E[X 0 ] + ap q p. (9 Several cases arise if m > 1. The first case is when a /( = 0, which is equivalent to m = 1 + /a. This situation occurs for instance with 1:1 payoffs and douling the ets so that a = and m = 2. In this case, the infinite series term in X T in (8 is identically 0, and /( = a; hence, X T X 0 + a and E[X T ] = E[X 0 ] + a. OCTOBER

8 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES Secondly, ecause a < (q/p, then m 1+/a implies mq q+(/aq > q +p = 1. So in the case of m > 1 ut mq < 1, we cannot have m = 1+/a. In this case we have E[X T ] = E[X 0 ] + + = E[X 0 ] + + = E[X 0 ] + + ( a ( a ( a p m i 1 P (T = i (mq i i = 0 ( p = E[X 0 ] + ap ( p + (( = E[X 0 ] + ap (1 p mq + (( = E[X 0 ] + ap + q(1 m (( = ap q E[X 0 ] +, (10 which gives the same result as in Equation (9 if we were to let m = 1. Thirdly, if m > 1 + /a (which implies mq > 1, then E[X T ] is as in the eginning of Equation (10, ut i=0 (mqi diverges to +. Thus, E[X T ] = + ecause a /( > 0. Finally, if 1 < m < 1 + /a and mq 1, then i=0 (mqi = +, ut E[X T ] = ecause a /(m 1 < 0. Collecting the results in (9 and (10 and considering all cases, we otain our formulization for E[X T ]. Theorem 5. Let T e the numer of ets needed for the first win. Under the conditions of the general martingale strategy with q < 1 and a < (q/p, the average fortune after the T th et is ap q E[X 0 ] + if m = 1, or if m > 1 with mq < 1 E[X 0 ] + a E[X T ] = if m = 1 + /a + if m > 1 + /a (so that mq > 1 if 1 < m < 1 + /a and mq VOLUME 21, NUMBER 3

9 We therey otain two cases for which E[X T ] > E[X 0 ]. For instance, if a = and m = 2, then you will always have $(X 0 + a if you quit after the first win. However, for a = with ap q < 0, we have p < q so that q > Thus, mq 1 when m = 2. Therein lies the paradox of the classic martingale strategy. Although, E[X T ] > E[X 0 ] in this case, y Theorem 2, the average amount of your last et will e The Maximum Guaranteed Numer of Bets Given an initial stake $X 0, how many ets n can we e guaranteed to make with the general martingale strategy? Of course, any win along the way will allow more ets to e made; ut we must e ale to cover the deficit D n+1 accrued from n initial losses in a row. So from Equation (7, we must have n if m = 1 n X 0 m i 1 = D n+1 = ( m n (11 1 if m > 1. Solving for n gives us the maximum numer of ets that we can e sure to cover: X 0 / if m = 1 n = (12 ln (1 + X 0 (/ / ln m if m > 1. For example, with a $3000 stake, an initial et of $20, and tripling the et after each loss, then ln (301 / ln 3 = 5 ets definitely can e made, which exhaust $2420 if all ets result in losses. 8. Choosing the Multiplier m If we are always making the same et of $, then y Equation (2 the est to hope for is to come out at least even with our initial stake of $X 0. We then can sustain a/ losses in a row and still reak even with a win on the next et. For instance, in American roulette, a square et is a et on a lock of four numers for which p = 4/38 and q = 34/38, with an 8:1 payoff ratio. Thus, we will still reak even with 8 losses in a row followed y a win. And the chance of 9 losses in a row is only (34/ , so there is 63.25% chance of winning within 9 tries. But how many wins W n are necessary to reak even after any sequence of n ets of $? The final fortune is now X n = X 0 + aw n (n W n. In order to have X n X 0, we need aw n (n W n, which means that the numer of wins W n must satisfy W n n/(a +. OCTOBER

10 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES Here, W n is a inomial distriution where P (W n = k = ( n k p k q n k for 0 k n. For example, in a sequence of n = 12 roulette square ets, with a = 8, then at least 2 wins are needed to reak even. But this event only has proaility P (W 12 2 = 1 P (W 12 = 0 P (W 12 = 1 = 1 (34/ (4/38(34/ So with a square et, we are much etter off trying to win once within 9 wagers than trying to get 2 wins out of 12. So with 9 or more losses on a square et, we can no longer reak even with just one win when using constant ets of $. The purpose of increasing the ets y a factor of m > 1 is to guarantee that we come out ahead after any win, no matter how many losses have occurred. But will any m > 1 guarantee that X T > X 0? From Equation (8, we see that we should choose m 1 + /a, which is equivalent to a /( 0. If m = 1 + /a, then the fortune after the first win will always e X T = X 0 + a. And if m > 1 + /a, then the overall gains are an increasing function of the numer of plays needed for the first win. Indeed, ecause now a /( > 0 and m > 1, we have X T 1 {T = i} = X 0 + ( + a m i 1 < X 0 + ( + a m i = X T 1 {T = i+1}. Equation (8 also shows that the overall losses are an increasing function of the numer of plays needed for the first win when 1 m < 1 + /a. 9. The Average Fortune Upon Stopping With a finite initial stake X 0, we may not e ale to keep etting until a win occurs. So suppose we decide in advance to make at most n ets, ut quit if we ever win. Then with m 1 + /a, we have X T n > X 0 except if we lose all n ets. We now shall derive the average final fortune E[X T n ]. Theorem 6. Let T e the numer of ets needed for the first win. Under the conditions of the general martingale strategy, the average fortune after the (T nth et is E[X 0 ] + n(ap q if mq = 1 E[X T n ] = E[X 0 ] + (ap q(1 (mqn if mq 1. Proof. We must adjust Equation (8 to account for a maximum of n ets, and then sutract D n+1, the total deficit accrued through n initial losses, 192 VOLUME 21, NUMBER 3

11 to account for T > n. If q = 1, then X T n = X n = X 0 D n+1, where D n+1 is as in Equation (11, and the result follows. For q < 1 and m = 1, we have n X T n = X 0 + a1 {T n} (i 11 {T = i} n 1 {T > n}. Using the fact that n k=1 kxk 1 = (1 x n nx n + nx n+1 /(1 x 2, for x 1, we otain the result in this case (mq < 1, p = y E[X T n ] = E[X 0 ] + (a + P (T n = E[X 0 ] + (a + (1 q n p n i P (T = i np (T > n n i q i 1 nq n ( 1 q = E[X 0 ] + (a + (1 q n n nq n + nq n+1 p (1 q 2 nq n ( ap q = E[X 0 ] + (1 q n. p For q < 1 ut m > 1, we have X T n = ( X 0 + ( m n 1 1 {T n} + 1 {T > n}, n (( a m i 1 1 {T = i} which gives E[X T n ] = E[X 0 ] + (1 ( qn + a n 1 p (mq i i = 0 ( m n 1 q n ( E[X 0 ] + n ap p 1/q 1 if mq = 1 = ( ( E[X 0 ] + (1 (mqn m 1 + ap p 1 (mq n m 1 1 mq if mq 1 E[X 0 ] + n(ap q if mq = 1 = E[X 0 ] + (ap q(1 (mqn if mq 1. OCTOBER

12 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES For q < 1 and a < (q/p as in Theorem 5, we have lim n E[X T n ] = E[X T ] except when m 1 + /a (which implies mq > 1. In that case, lim n E[X T n ] =. We also see that E[X T n ] for mq = 1, and more importantly, E[X T n ] < E[X 0 ] when a < (q/p. 10. The Desired Numer of Bets Finally, suppose we want to have a high proaility r of coming out ahead within n ets. With m 1 + /a, we will come out ahead after the first win. So we want the proaility of a win within n ets to e at least r. That is, we want P (T n = 1 q n r. Solving for n, for q < 1, the minimum numer of ets that we must e ale to make is ln(1 r n =. (13 ln q Then to e ale to make these n ets with an initial et of $, we must have an initial stake X 0 that satisfies Equation (11. Using m 1 + /a, we will then have at least proaility r of coming out ahead when quitting after the first win or after a total of these n ets. 11. Comparison of Results We shall illustrate our main results using a column et in American roulette, for which p = 12/38, q = 26/38, with a 2:1 payoff ratio. Our initial et will e = $10 (with a = $20, and we wish to have r = 0.95 proaility of coming out ahead. How many ets n must we e ale to make; what multiplier m should we use; what initial stake X 0 do we need? When quitting after the first win or after n ets, what is the average final fortune? We then shall compare the results to the scenarios of (i making n ets of $, and (ii making ets of $, ut quitting if we ever get ahead or after at most n ets, which is the strategy discussed in [6]. From (13, we need to e ale to make n = ln(0.05/ ln(26/38 = 8 ets, which actually gives us a 1 (26/ proaility of winning within these 8 ets. With a 2:1 payoff, we need m 1+1/2 = 1.5. We shall use m = 1.6 (and hence mq > 1. We note that y choosing m < 2, our ets will not increase dramatically and should stay within the tale limits. Now from (11, we need an initial stake of at least 10( /0.6 $ (With m = 2, we would need X 0 $2550. Theorem 2 then gives the average amount of the last et as E[ T 8 ] $ From Theorem 6, the average final fortune, with X 0 = $700, is E[X T 8 ] $ In actual practice though, we would proaly round the ets.6 i 1 10 up to the nearest dollar, which requires X 0 = $701. These results, with adjusted averages, are given in the following tale. 194 VOLUME 21, NUMBER 3

13 Outcome Proaility Last Bet Final Fortune W p $ = $721 LW qp $ = $723 LLW q 2 p $ = $727 LLLW q 3 p $ = $731 LLLLW q 4 p $ = $740 LLLLLW q 5 p $ = $752 LLLLLLW q 6 p $ = $773 LLLLLLLW q 7 p $ = $807 LLLLLLLL q $ = $0 X 0 = $701, m 1.6, p = 12/38, q = 26/38, n = 8, E[ T 8 ] $48.47, P(X T 8 > , E[X T 8 ] $ Following are the results of two other strategies: (i For sequences of 8 ets of $10 each with X 0 = $701, Equation (2 gives the average final fortune as E[X 8 ] $696.79, which is higher than E[X T 8 ] using m 1.6. However now we require at least n/(a + = 3 wins to come out ahead, which occurs only with proaility The following tale shows the possile outcomes. Outcome Proaility ( of k Wins Final Fortune (any order 8 k p k q 8 k k 10(8 k 8W $861 7W, 1L $831 6W, 2L $801 5W, 3L $771 4W, 4L $741 3W, 5L $711 2W, 6L $681 1W, 7L $651 8L $621 P(X 8 > , E[X 8 ] $ (ii Suppose now that we make at most 8 ets of $10 each, ut quit if the fortune X i ever surpasses X 0 = $701. By scaling down the values, we can compute E[X T 8 ] and P (X T 8 > 701 using matrix products. We first divide all values y gcd(a,, which is 10 in this case. Now X 0 = $70.1, = 1, and a = 2. After 8 ets, the minimum possile fortune is $62.1; thus we sutract 62.1 from each fortune value. We now have a Markov chain that egins at height X 0 = 8, moves up 2 units at a time with proaility p = 12/38 or down 1 unit at a time with proaility q = 1 p, and stops after at most 8 steps or if ever reaching an upper oundary of 9 or 10. We then let A e the 1 11 matrix of possile position heights, let B e the 1 11 initial proaility state matrix, and let C e the OCTOBER

14 MISSOURI JOURNAL OF MATHEMATICAL SCIENCES matrix of transition proailities, where c i j = P (X k+1 = j X k = i, for 0 i, j 10 and all k 0. A = ( B = ( q 0 0 p q 0 0 p q 0 0 p q 0 0 p C = q 0 0 p q 0 0 p q 0 0 p q 0 0 p Then B C 8 gives the proaility state of eing at height i upon stopping or at most 8 steps. These proailities coincide with the actual fortune eing 10 (i E[X T 8 ] = 10 (B C 8 A T $ The following tale shows the possile outcomes. Final Fortune Proaility $ $ $ $ $ P(X T 8 > , E[X T 8 ] $ Because of the one case of losing all ets and thus losing the whole stake, the martingale strategy yields the smallest average fortune. However, it provides the highest proaility of coming out ahead, which also can e chosen in advance. In our example, you will come out ahead aout 19 out of 20 times, with manageale ets, provided you are willing to risk it all. References [1] D. Connolly, Casino Gamling, the Ultimate Strategy, College Mathematics Journal, 30 (1999, [2] L. E. Duins and L. J. Savage, How to Gamle if You Must, McGraw-Hill, New York, [3] W. J. Ernest, The Bold Strategy in Presence of House Limit, Proc. Am. Math. Soc., 32 (1972, VOLUME 21, NUMBER 3

15 [4] J. D. Harper and K. J. Ross, Stopping Strategies and Gamler s Ruin, Mathematics Magazine, 78 (2005, [5] R. Isaac, Bold Play is Best: A Simple Proof, Mathematics Magazine, 72 (1999, [6] D. Neal, Quit When You Are Ahead, The Mathematical Scientist, 23 (1998, MSC2000: 60G40, 60D05 Department of Mathematics, Western Kentucky University, Bowling Green, KY address: david.neal@wku.edu Department of Mathematics, Western Kentucky University, Bowling Green, KY address: michael.russell@wku.edu OCTOBER