# Avoid the ‘gambler’s ruin’

Oct 17, 2001 1:56 AM

The Gambler’s Ruin Problem was posed ”” and answered ”” by the mathematician Blaise Pascal several hundred years ago. Briefly, it states that a gambler, when faced with a house possessing a large bank roll, and having a limited bank roll of his own, must eventually go broke. Although this is not a certainty, the odds of such a gambler of “breaking the bank” are vanishingly small. It is a principle that all gamblers ”” not just keno players ”” should keep in mind. Somewhat depressingly, the principle applies not just to the usual casino games where the house has an “edge” but even to “fair” games like rolling the highest die or flipping a coin. The larger bank roll will win in the end.

I decided to test the problem using a Monte Carlo simulation on my PC. I arranged for 100 players, each with an initial bank roll of \$10 to flip a coin versus the house. Heads, the player wins a dollar, tails, the house wins a dollar. It’s a 50-50 proposition, a fair game, with the only difference being the house has an unlimited (infinite) bank roll.

Not surprisingly, given knowledge of the Gambler’s Ruin, all 100 players lost their \$10. Although to be fair, this result does seem to fly in the face of common sense. After all, in a fair game, some should win and some should lose, shouldn’t they? And, given the random coin flips, shouldn’t some players lose more than they win (and go broke) while others win more than they lose and eventually break the bank? The problem is TIME. Given enough time, every player will eventually have a long enough losing streak to bust out a limited bank roll. True enough, the house will also have long losing streaks, but given its much larger bank roll, it is far more able to ride them out.

In my simulation, it took over 5,000,000 coin flips to bust out the 100 players, 5,285,132 to be exact. That means that after that many decisions, the house won \$1,000. This was the result of a variation from the expected house win of less than 2/100 of 1%!

The average player lasted 52,851 coin tosses before going broke, but this average includes one lucky player who lasted 3,690,119 tosses (and achieved a net win of \$863 before giving it all back), or about 70% of ALL the coin tosses. The median player lasted only 937 tosses before going broke. The luckiest player played 274,261 tosses before eventually going broke and achieved a net win of \$1,155 before losing it all back again. In this case, the house could have had a bankroll of only \$1,200 and still withstood the efforts of all 100 players!

This result matches fairly well with mathematical theory. In a test where the odds are equal 50-50, the chances of a player going broke with a \$10 bankroll and shooting for a \$1,000 bankroll are:

1- 10/1000 or 99% and in fact only one of the players out of the 100 reached the thousand dollar milestone before going broke.

But this isn’t the end of the story, at least not for keno players, as we’ll find out next week!

If you have a keno question that you would like answered, please write to me care of this paper, or contact me on the web via email at [email protected] Well, that’s it for now. Good luck! I’ll see you in line!