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Some Gambler's Ruin Questions - Part 2

4 June 2005

By Donald Catlin

Last month I addressed a Gambler's Ruin problem that was raised by one of my readers, Ralph Scalitzia. You can refer to that article by scrolling back through recent articles on this site. The question raised there was a standard Gambler's Ruin problem and I gave the solution without any explanation of how one arrives at such. This month I want to continue with a more difficult question. The solution I give here uses standard techniques and these would apply to last month's article.

The question posed is this. If one makes a pair of dozen wagers on a single zero Roulette game, what is the probability of ruin? Notice that at any step in the process the player can at most win one unit (each dozen pays 2 to 1) but can lose two units. Specifically, if one's initial stake is 50 units and the goal is to obtain 51 units, then what is the probability of ruin? I'll solve the general case first and put the numbers in later.

Let p represent the probability of a win at any step and let q be the probability of a loss. Then as usual

p + q = 1 (1)

If uz represents the probability of ruin when the player's stake is z then at the very next step (wager) the player's ruin probability becomes uz + 1 with probability p or uz - 2 with probability q. Thus we have the difference equation

uz = puz + 1 + quz - 2 (2)

If the player reaches his goal, call it a, then we also have the condition

ua = 0 (3)

since at this point there is no chance of ruin. Actual ruin is a matter of definition. One might consider ruin to be at one unit since the player can no longer make a wager. On the other hand, one could consider a scenario wherein the player reaches into his pocket and pulls out an additional unit in order to make the wager. If he wins and puts the additional unit back in his pocket then he has two units. If he loses, his stake is -1. I am going to use this scenario. In other words I'll define ruin as being a stake of either 0 or -1. Thus

u0 = 1 (4)


u-1 = 1 (5)

conditions (4) and (5) representing certain ruin.

Very well, our problem is to find a function that satisfies the equation (2) as well as the conditions (3), (4), and (5). The usual technique, known as the characteristic equation method, assumes that a function of the form sz satisfies (2). If we substitute sz into (2) we obtain

sz = psz + 1 + qsz - 2 (6)

Multiplying relation (6) through by s2 - z and rearranging the terms one gets

ps3 - s2 + q = 0 (7)

This is a cubic equation. Note, however, that if we substitute s = 1 into (7) we just get the expression p - 1 + q = 0, which by (1) is obviously true. Thus (7) has 1 as a root and hence s - 1 as a factor. Dividing s - 1 into the expression on the left of (7), we end up with the equation

ps2 - qs - q = 0 (8)

If you don't trust my long division you can multiply the left side of (8) by s - 1 and see if you get the left side of (7) (note: you may have to use (1) ).

Equation (8) is our old friend the quadratic equation and (remembering your ninth grade mathematics class) can be solved using the quadratic formula. This done (I'll let you check my arithmetic), the three roots of (7) are

s1 = q/2p[ 1 + (1 + 4p/q)1/2]

s2 = q/2p[ 1 - (1 + 4p/q)1/2] (9)

s3 = 1

Thus s1z and s2z are solutions of (2). It is a trivial matter to show that a constant times any solution of (2) is again a solution of (2) and that the sum of any two solutions of (2) is also a solution of (2). The condition s3 = 1 therefore implies that any constant is a solution of (2). It follows at once that if I define

f (z) = (s1s2 - s2)s2z - (s1s2 - s1)s1z + (s2 - s1) (10)

then f(z) is a solution of (2). The strange constants in (10) are designed to obtain the following calculations:

f(0) = s1s2 - s2- s1s2 - s1 + s2 - s1 = 0 (11)


f(-1) = (s1s2 - s2)s2-1 - (s1s2 - s1)s1-1 + (s2 - s1) = 0 (12)

Finally, we define

uz = [f(a) - f(z)] / f(a) (13)

Clearly uz is a solution to (2) that obviously satisfies (3). It also satisfies (4) and (5) because of the calculations in (11) and (12). That's it!

Okay, now let's answer Ralph's question from last month's article. The probability of hitting one number in a pair of dozen wagers is 24/37 and the probability of not hitting any of the two dozen is 13/37. Putting these numbers into the expressions in (9) we obtain

s1 = 1.0055063759

s2 = -0.513397092

and so from (10)

f(50) = 21.72139892


f(51) = 23.00382611

Substituting these values into (13) we get the final result:

u50 = 0.05575

There you have it, Ralph. If you bet two of the dozen numbers in single zero Roulette, start with $50 and quit at $51, there is still a 5.575% that you will go broke. From last month's article recall that the ruin probability for playing an even money bet with the same stake and goal was 5.6197%. So the two dozen wager is slightly better in terms of ruin than the even money bet although both have the same house edge (2.7%).

If any of you have other gambling questions or need a bit of help with the algebra in this article, I can be reached at 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