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Best of Donald Catlin

Gaming Guru

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Some Gambler's Ruin Questions

1 May 2005

One of my readers, Ralph Scalitzia, sent me the following email after an earlier response to one of his questions:

Dear Dr. Catlin,

Thank you very much for your help and informative replies. What I am interested in knowing is which of the following betting scenarios has the lowest Gambler's Ruin probability:

Case (1):

Initial capital 50, goal 51, betting one dollar at a time. Probability of winning (.493). Example: like betting on either black or red in single zero Roulette.

Case (2):

Initial capital 50, goal 51, betting two dollars at a time. Probability of win (.648) but if you lose you would need three consecutive wins to reach your goal and if you lose again you will need five consecutive wins to reach your goal. Example: like betting on two dozen [numbers] in single-zero Roulette. How do we calculate Gambler's Ruin here?

Finally, when they say that the house edge is 1.35%, does this mean that the chances of winning on even bets are (.493) instead of being (.5)?

Once again, thank you for your HELP!!!

Respectfully yours,

Ralph Scalitzia

Wow, Ralph, you have lots of questions here. Let me start with the last one first. The short answer is no! Let me explain. The 1.35% house edge figure you state is for European Roulette (single zero) with the added feature of the en prison rule. So I'll first explain the en prison rule and then we'll determine the correct probabilities.

If one bets on (say) Red and the ball lands on a red number, the player wins even money. If the ball lands on a black number, the player loses his wager. On the other hand, if the ball lands on the zero then the player neither wins nor loses and the player's bet is carried over to the next spin - the bet is said to be "in prison." On the next spin if the ball lands on a red number then the player's wager is returned. If the ball lands on a black number, the player's wager is collected. If the zero occurs again the bet remains en prison and the wager is carried over to the next spin. This procedure continues until the bet is settled.

First of all, note that there are three outcomes here: a win, a loss or a tie. The player can only win on the first spin and that probability is 18/37 or approximately 0.4864865 (hence my no response to your question). Likewise there is an 18/37 chance of losing on the first spin. The probability of the zero occurring is 1/37. Of course, it could occur again, and again, ad infinitum. So as you might expect there is an infinite geometric series lurking in the background here. Although one can analyze the problem using the series, there is a simple observation one can make that avoids that analysis. Simply put, once the first zero occurs, there is an even chance of subsequently either losing or obtaining a tie. Thus the probability of tying is one half of 1/37 or 1/74. Likewise, the probability of losing after the initial spin is also 1/74. Thus the probability of losing is 36/74 + 1/74 or 37/74 (exactly ½). In summary, we have the following table

Event

Probability

Payoff

Product

Win

36/74

+1

+36/74

Loss

37/74

-1

-37/74

Tie

1/74

0

0

Totals

1

---

-1/74

Figure 1
Expected Return for European Roulette with En Prison

1/74 expressed as a percentage is approximately 1.35135%.

Now I'll address your Case (1) question. I'll assume that you are playing a single zero Roulette game with no en prison feature. Although some Atlantic City games have a Surrender feature (different from en prison), I know of no en prison games in the U.S. Thus the correct win probability is not 0.493 but rather 18/37 or approximately 0.4865 (the same as it is for the en prison game by the way); losing probability is 19/37 or approximately 0.5135 (in the en prison game it is 1/2).

The standard formula for Gambler's Ruin in such a game is determined as follows. If p is the probability of winning one unit and q is the probability of losing one unit we define

f(x) = (q/p)x (1)

Then the ruin probability is given by the formula

r(z) = [f(a) - f(z)]/[ f(a) - 1] (2)

where z represents the player's current stake and a represents the player's goal. Note that r(a) = 0 (no chance of ruin if the goal is reached) and r(0) = 1 (ruin is certain). For the problem at hand (q/p)50 =19/18. Hence (q/p)50 = 14.92983 and (q/p)51 = 15.75926. The difference here is 0.82943 and the ruin probability according to the formula in (2) is 0.82943/14.75926 or 0.056197. By the way, the formulas in (1) and (2) apply to the en prison game as well, but the probabilities used must be those in Figure 1, that is q/p = 37/36.

You may wonder how one arrives at the formulas in (1) and (2). Next month I'll address Ralph's Case (2) question in detail and you'll then see the techniques used in deriving these ruin formulas. Not all of them are tractable but fortunately the question Ralph raised in Case (2) can be answered (though not without some struggle). See you next month.

Donald Catlin

Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers