I am totally new at this forum and I hope somebody can help me. I have been studying the roulette for many years (developing and testing systems only as I do not like to gamble or to play at all, just interested in the odds and statistics). In all cases so far I have been able to determine the chances of something happening or not happening but now I have developed a strategy to play for which I have difficulty to determine the chances and the odds. Below I describe the strategy and I add what I think (!) the odds are but I am very uncertain about this so please correct me when I making a mistake in my thinking process:
The strategy is as follows (and for easy of explaining I skip the zero!)
Red and Black have each a 50 % of falling
Clearly if we have a row of 8 spins than the chance of 8 R or 8 B (or any other sequence of R-B) is 1/(2^8)= 0,39%
A perfect spread of chances would be: R B R B R B R B R B R B R B R B etc.
In reality this will not happen for a long time so I thought of betting on a max number of single appearances of Red or Black
If one uses a progression (such as Martingale) the player can bet that a single B or R will not occur too often:
We want to play on Black:
We see: R R R B R R B and now we start betting that B will not be another single but a second B will fall. In principal we have a 50% chance of winning this time. If we win we start again, if we loose we wait for the next B and make another bet to win (double the bet to recoup the loss and make a small profit).
Now I am trying to find a method to calculate the odds that there are 10, or 11 or 12 subsequent singles of one colour.
My thinking is (similar to the first example of the 8 red or 8 black) that if we loose say 12 Blacks than the MINIMUM row of spins needed will be: 24 spins.
R B R B R B R B R B R B R B R B R B R B R B R B
If R has one or more doubles or longer rows than the total number of spins before we loose must be even greater than 24 spins.
Well, the occurrence of this specific row of 24 spins is 1/2^24 = 0.000000059604644775390625 so that is a chance of loosing of 0.0000059604644775390625% or in other words a chance of loosing of 1 in 16,777,216 ( If we only play for 12 R or B not showing than the chance of loosing would be 1/(2^12) = 0.024% or 1 in 4096 ! )
Am I right or am I making a mistake in this calculation. If I am right this method would reduce the chance of loosing with roulette to less than the chance of being killed in traffic.........
So far my problem/question. I hope somebody is out there who can help me with this. Thanks upfront.