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Gaming Guru  A ruin question

5 September 2008

Dear Don,

My name is Padme, living in India and playing Blackjack in Nepal and Cambodia. Thanks for your great columns and advice. I am writing a very small, and very simple, book on betting at Blackjack for my friends only, mainly because I believe that an average to good player, with good to excellent betting decisions, will make more money at Blackjack than a good to excellent player with poor betting decisions. Can you tell me if the examples below are mathematically correct?

If you bet, say, \$10 with a bankroll of \$500 and want to win \$100, then quit, the odds are 4 to 1 you'll succeed before going broke, with only 5% risk of ruin for 500 hands (about six hours of play).

For a \$40 win goal, with \$10 bets, and a \$500 bankroll, the odds go up to 9 to 1 in your favor or for a \$30 win goal the odds are 12 to 1 and for a \$20 win goal the odds are 19 to 1.

Thanks,

Well, Padme you certainly don't ask easy questions. Let me first address an issue in your first question. You can't have both a 4 to 1 chance of success and a 5% risk of ruin. I suspect that you are mixing the answers to two different questions here, so let's ignore the 5% figure for the 500 hands.

The classic ruin problem is a one-dimensional random walk with a unit step in either direction. What makes the ruin question in Blackjack so difficult is that because of splits, doubles, and 3-to-2 payoffs for Blackjack the size of the step becomes random. I don't know how to handle a random walk with a random step size. If I were going to try to seriously tackle this question I would probably create a simulation of some sort and crunch the numbers on a computer. That said, there is an approximation that one can make that will give you "ball park" figures.

Blackjack, when played using basic strategy, has a house edge of around a half percent. So, we can create a simple random walk with a house edge of 0.005 and see what the ruin probability looks like for such a game. If p represents the probability of success and q the probability of a loss then p - q represents the player's expected return. By our assumption

p - q = -0.005

and since q = 1 - p

2p - 1 = -0.005

Solving for p we see that p = 0.4975. Thus q = 0.5025. For the classic ruin problem we need the ratio q/p (which I'll call r) and it is 1.10050251. For your first question, rather than use 500 playing for 600, I'll use 50 playing for 60 with a unit step. The classic ruin calculation is then r raised to the 60th power minus r raised to the 50th power all divided by the quantity r raised to the 60th power minus 1. This number, representing the ruin probability, turns out to be approximately 0.21. Hence your 4 to 1 figure looks about right.

For the 40 question the ruin probability is approximately 0.094, which represents a chance of success of about 11 to 1. For the 30 question the ruin probability is approximately 0.072 so the success rate is about 14 to 1. I'll leave the 20 question for you Padme. See you all next month.

Don Catlin can be reached at 711cat@comcast.net

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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