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# Gaming Guru

### Blackjack Ruin Again

19 June 2011

I recently received the following letter from Amy, one of my readers:

Dear Sir,

Alan Krigman says if you go for winning 10 hands at Blackjack (with a bankroll of 50 times your bet) you have a 4:1 chance of winning and a 9:1 chance of winning if you go for winning 4 hands. I take it the math is not linear. So what are your chances of winning 2 hands? Is it in the region of 15:1?

Amy

Well, Amy, this question has come up before. The usual analysis using a one-dimensional random walk is not really appropriate here since the game involves doubles, splits, and 3:2 payoffs. In other words it is a random walk with random wagers and I don't know how to handle that with pencil and paper. What I did in response to an earlier inquiry was to write a simulation and run 150 million trials. You can read the details in my February 1, 2008 article entitled A Blackjack Ruin Problem. Unfortunately, I lost the program in a 2009 computer crash (I know, I now have automatic backups) and have not had the time or energy to rewrite it.

There is another possibility and I describe it in my September 5, 2008 article entitled A Ruin Question. The idea is to take the game's expected value, around -0.5% and use it to determine winning and losing probabilities for a one-dimensional random walk. This is, therefore, an approximation to the actual random walk. For reasons that are not clear to me, this approximation produces ruin probabilities that are smaller than the actual ruin probabilities.

It turns out that using the expected return of -0.5%, the winning probability is 0.4975 and the losing probability is 0.5025. For the classic ruin problem, we need the ratio q/p (which I call r) and it is 1.010050251. See the September article for details on using this ratio. If I use 50 playing for 60 with a unit bet, the ruin probability is approximately 0.21 so the 4:1 figure looks about right since, as mentioned above, this approximation produces smaller ruin probabilities than the actual ones. If I use 50 playing for 52, the ruin probability turns out to be 0.048834445. This means the odds of winning 2 units is about 20.5:1. The actual figure will be smaller than this, so I would estimate it at 18 or 19 to 1.

That's as good an estimate as I can make at this time, Amy. If I reconstruct my simulation program for a future article I'll give you a better answer then. See you 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