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.
