Some Gambler's Ruin Questions

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