Re: Fibonacci and the roullette wheel

Alexander wrote:
<snip>"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:
If you run your system on just one block of 12, then the probabilistic
breakdown of the winning spin breaks down like this for a double-0
wheel:
31.6% chance of winning on spin #1 (F=1), 21.6% on spin #2 (F=1), 14.8%
on spin #3 (F=2), 10.1% on spin #4 (F=3), 6.9% on spin #5 (F=5), 4.7% on
spin #6 (F=8), 3.2% on spin #7 (F=13), 2.2% on spin #8 (F=21), and 1.5%
on spin #9 (F=34). Thus, there is a 96.6% probability of winning before
spin #10. This leaves a 3.4% probability of having to bank-roll more
than 9 spins. So, if you can bank-roll 9 spins of the wheel in your
system, then about 96.6% of the time you will walk away from the table
with a relatively small profit. However, it is the other 3.4% of the
time, when you take a loss, a huge loss relative to the unit value, that
previous profits will be wiped out. Example: walking away from the
table after a loss on spin #9 results in a loss of 88 units. That is
the primary reason your system, as stated, will not "work" in the long
run. In fact, it is the "long run" that makes the 0 and 00, along with
the implicit 5.2% house advantage, very relavent and less than a
zero-sum game for your system.

Dan Akers

.

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