Re: Fibonacci and the roullette wheel
- From: digikey@xxxxxxxxx (D. Akers)
- Date: Tue, 25 Apr 2006 18:10:33 -0400
Alexander Donis wrote:
"You are of course right about the numbers. However with regards to the
odds - a block of 12 does pay 2 to 1 but you also get your stake back
which makes it the equivaqlent of 3 to 1.
Using the fibonacci sequence 1,1,2,3,5,8,13,21,34......n, at any given
point in the sequence the total bets placed for that sequence is equal
to 2 interations along the sequence minus 1.
Therefore in the example above if you had got to betting 8 units when
the sequence finally ends you would have bet 20 units in total. If the
winning bet pays 2 to 1 plus return of stake then the return is 24
units.
In fact the further along the sequence you go the higher is the total
profit when the sequence ends.
Most people though seem to think that the casino always wins because of
the 0 or double 0. However these are irrelevant to the sequence since
they merely represent a losing bet and you move further along the
sequence.
Since the bets would be placed on each of the block bets, thus covering
every number other then the 0 or double 0, you would require a long
sequence of results landing on 0 or double 0 to prevent a profit. Or
alternatively for one of the block bets to run for a very long time
before finishing.
As stated in my original question, other than a long losing sequence on
one of the block bets, are there any other reasons why this would not
work."
______________________________________
Re:
For starters 2:1 is not equivalent to 3:1; that's not proper use of the
terminology. For a bet on a group of 12 the house pays 2 to 1; that is
2 units paid for every unit wagered. On an American wheel, there are 38
slots in the wheel, 1 through 36, 0, and 00. The probability of winning
on any given spin of the wheel, betting a group of 12, is simply 12/38
or 31.58%. Conversely, the probability of losing is 68.42%. Your
fibonacci method smacks of the Martingale system which also attempts to
recover previous loses and profiting with each successive bet; the
difference being wagering the even money (1:1) bets (odd/even or
black/red) and the doubling of the bet with each loss.
One of the problems with your proposed system is the diminishing returns
with escalating wagers. At 5 units the potential return is 25% while at
the next iteration, namely 8, it is only 20%. By the time you are
betting 34 units (something you will do 4.8% of the time you run your
system), your return has been reduced to 16%. At that point, you're
facing a 68.42% probability of losing for a gain of only 16% of the
total risked capital. In other words, your risk is constant while your
expectation decreases with each wager.
You are correct; your system will always profit provided you have enough
capital to sweat out every win and you don't hit the wheel limit before
winning.
For other approaches take a look at the "Reverse Labouchere System"
which attempts to risk less capital for greater gains.
Dan Akers
.
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- Re: Fibonacci and the roullette wheel
- From: Alexander Donis
- Re: Fibonacci and the roullette wheel
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