The Fibonacci system, can actually hold its own

May 27, 2008 7:00 PM


Crapshooter by Larry Edell | Progressive betting systems, such as the Martingale, often draw mixed reviews because they hold out the possibility of sustaining lengthy losing streaks that are difficult to overcome.

One system, The Fibonacci system, can actually hold its own when applied to shooting craps.

This is a somewhat advanced system, but once you get used to it you'll see that it almost always wins if you stick to the rules and don't make any other bets.

Fibonacci is a mathematical term and means that you have a number series containing two consecutive numbers, and their total is larger than all of the preceding numbers.

For example, a Fibonacci series would be 1, 1, 2, 3, 5, 8, 13. As you can see, 13 + 8 (21) is more than 1 +1 + 2 + 3 + 5 (12). Also 5 + 8 (13) is more than 1 + 1 + 2 + 3 (7). Another attribute to this series is that the next number would be the sum of the two previous numbers. For example, 13 is the sum of 8 + 5; and the next number after 13 would be 13 + 8 or 21.

Let's see how craps professionals use the Fibonacci to increase their profits.

The Fibonacci can be very profitable to a crapshooter. You can bet flat bets on the pass, don't pass, come or don't come lines. If you win any two consecutive bets, no matter where you are on the progression, you will make money. And, if you win any two out of three bets you will also have a profit if you stick to the progression.

The progression you must bet is, of course, the Fibonacci series. You will bet multiples of 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 and so on. With the exception of the first two bets, your new wager is always the sum of your prior two bets, so you don't have to memorize anything.

Let's look at some examples assuming $1 unit bets.

You start betting, first $1 on the pass line, lose, and bet another $1. Now once the first two bets are established, the next bet is always the total of the two previous bets, so you bet $2, which you lose also. The next bet is $3, which you lose. Now, 3 + 2 is 5 so your fifth bet is $5, easy to remember, but you lose that one also. Now let's see what happened.

You lost the first five bets: $1 + $1 + $2 +3 + $5 for a total of $12. You win the next two bets for $8 and $13, so you've won $21 and lost $12 for a net profit of $9.

Let's say the dice are really cold and you lost the first 11 bets so you lose: $1 + $1 + $2 + $3 + $5 + $8 + $13 + $21 + $34 + $55 + $89 for a total of $232. If you win only the next two bets at $144 and $233, you would win $377 and be ahead by $144. You will also profit if you win any two out of the three bets. If you won the 11th bet of $89, lost the 12th bet of $144 but won the 13th bet of $233, you would still be ahead by $178.

Your goal in this system is to make a profit. Once you win any two-out-of-three bets you pocket the profit and start over again at bet No. 1. You should quit when you win (or lose) 50 percent of your buy-in, so if you buy in for $500, you should stop when you win or lose $250. You can find $1 tables on weekday mornings (the limits are usually raised on weekends in Las Vegas) or try budget casinos such as those downtown or ones dedicated to locals.

Now that you're familiar with Fibonacci, give it a try and rake in some chips!

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