Fibonacci Betting Strategy
Here is a Fibonacci-based betting strategy that I have been developing. It is the simplest and the safest of my strategies for beating a standard house at craps. It seems effective but I wonder whether its reliability can be decided more precisely
The basic game of craps is to match your point on the dice before you "seven-out". The probability that you will seven-out (and fail to match your point) depends on the point, but on average is .66.
Let us define a loss as sevening-out six times in a row. The probability that you will lose is .08. The probability of a win is then .92. And the probability of winning 5 times in a row is .65 and of winning 8 times in a row is .51. The strategy outlined below yields a 25% profit with roughly 5 to 8 wins. So it is likely on this betting algorithm to come out ahead by 25%.
Here's the strategy.
1.
Begin with $400. Remember to quit when you have earned $100 (or go bankrupt, which is about 6 losses in a row).
2.
Bet only Fibonacci numbers (in order) on the "pass line"
1, 2, 3, 5, 8, 13 ...
So, always begin with $1 on the pass line. If you fail to match your point, then bet $2 on the passline. If you fail to match your next point, then bet $3, and so on, traversing the Fibonacci series. If at any stage you do match your point, then begin the sequence again.
Most importantly, always take 10x odds behind the line. Completely ignore wins and losses on the come out roll, since the gains and losses there will be negligible (i.e., the game play that matters begins only after the point is established and 10X odds are placed behind the line ). The game is over when you're up $100 (and walk with $500 total) or you go bankrupt (and walk with $0 total).
Give the method a try here. Click on "options" to set your virtual bankroll and to set the game to 10X odds.
What is the precise reliability of the strategy described above? If the problem is undecidable (as I expect it is), then might one nonetheless be able to design a computer program that would run the strategy a sufficient number of times for statistical probability?
6 comments:
Why did you make it so complicated? More importantly, why Fibonacci? It seems arbitrarily chosen.
Your method, or at least its premise, seems derivative of the "double your bet until you get 1 dollar" strategy. Let's say you have about 50% chance of getting double your money and about 50% of losing all. Like in craps.
Then you bet 1 dollar. If you lose, bet 2 dollars. If you lose, bet 4 dollars. Eventually you will probably win and end up with 1 dollar above what you started with. Repeat.
This strategy works but can spark some very interesting philosophical/mathematical debates. Assuming a "fair game" for simplicity, the expected value of this strategy is 0. Yet you are almost guaranteed to make 1 dollar every time.
How is the expected value 0? Well it comes from the fact that every time you play, unless you have infinite money, you have a tiny chance of losing so many times in a row that you go bankrupt before you replenish yourself. So even though you probably make 1 dollar, you might lose all the money you ever owned. It's like reverse Lotto. And still not in your favor.
I used this when I was a student in the 80s. Works fine but it's a little boring (use the behind the line chips for the fun).
Trouble is pit bosses spot you in 20 minutes so it is not a road to success unless you ramp up the stakes (or use the behind the line chips to add randomness).
Cardcounting is sooo much better.
The big problem is if you are at the 8 0r 13 bet and win on come out before you place odds
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