An always winning casino roulette system- PART 3

3. Choosing a betting type.

As you can see from the bets descriptions above, only the five-number bet efficien value is different from the others (with 2,6% smaller). So one of the eight bet possibilities fades out from the choice of "wanna be a winner" player. But the other seven bet types have the same efficient value. So, the next thing, in which we compare these seven types is the chance to win the betting. The bigger chance means that you'll need less start money to fulfil your strategy. And the 8) betting type has the biggest chance to win the betting. But there are six possible bets from this type. You can place money on red, black, even, odd, low or high. The "winning order" comes here…

4. Introduction to the strategy.

Do you know, what is a geometric progression? Here is one example for those of you who don’t know: The numbers 1, 2, 4, 8, 16, 32 are given. Can you guess which is the next number? It's 64. The relation between the numbers is that every number from the sequence is twice bigger from the previous. A sequence from numbers with any relation valid for all numbers from the sequence is called progression. If you haven't understand that good, look here -http://en.wikipedia.org/wiki/Geometric_progression. The betting strategy rely on a simple geometric progression. Check the following table:

Bet Number	Stake ( $ )	Gain ( $ )
1	1	+ 1
2	2	+ 1
3	4	+ 1
4	8	+ 1
5	16	+ 1
6	32	+ 1
7	64	+ 1
8	128	+ 1
9	256	+ 1

This table describes the following thing:

You bet on "Red" 1$. If you win you get 2$, you're ahead with 1$. If you lose: You bet on "Red" 2$. If you win you get 4$, you're ahead with 1$. If you lose: You bet on "Red" 4$. If you win you get 8$, you're ahead with 1$. If you lose: You bet on "Red" 8$. If you win you get 16$, you're ahead with 1$. If you lose: You bet on "Red" 16$. If you win you get 32$, you're ahead with 1$. If you lose: You bet on "Red" 32$. If you win you get 64$, you're ahead with 1$. If you lose: You bet on "Red" 64$. If you win you get 128$, you're ahead with 1$. If you lose: You bet on "Red" 128$. If you win you get 256$, you're ahead with 1$. If you lose: You bet on "Red" 256$. If you win you get 512$, you're ahead with 1$.

So, now you ask yourself – what is the chance to have 9 times no red number?
And here comes the probability theory in help. The chance the ball to stop on a black number, 0 or 00 is: 100% - (your chance to win the betting on "Red") = 100% - 47,37% = 52,63%. The first time you place a bet on "Red" the chance to lose is 52,63%. But the second time this chance changes. Why? The "Independent Events" is one chapter of the probability theory. Two events are said to be independent, if the result of the second event is not affected by the result of the first event. If A and B are independent events, the probability of both events occurring is the product of their individual probabilities. If you can't understand this, here you'll find one very nice explanation - http://regentsprep.org/Regents/Math/mutual/Lindep.htm. The following table is based upon the probability theory and shows how the chance to lose changes with every next bet you place on "Red":

Bet Number	Chance to lose
1	52,63 %
2	27,69 %
3	14,57 %
4	7,67 %
5	4,03 %
6	2,12 %
7	1,11 %
8	0,58 %
9	0,30 %
10	0,16 %
11	0,08 %

This table is the answer to your question – the chance to have nine times in a row no red number is 0,3%. This means that from 1000 times you've played with this system, 3 times you will lose and the other 997 you will win. 997 times you win 1$, this makes (+997$) for you, and 3 times you lose 511$, this makes (-1533$) for you. Overall (-536$). So, this shows that the system isn't a winning one… yet. But look what is the chance to lose 10 times in a row – 0,16%. This means that from 1000 times, you'll lose 1,6 times. This is the same like: from 2000 times to lose 3 times. This gives you 1997$ - 3 x (511$) = 464$. The system is now a winning one. And 464$ / 2000 = 0,25$ per one play time. And this money will be more, if you can afford more bets. If you lose 0,08% of the time you play (this means 11 same bets in a row), you'll get 0,6$ per one play time. But why you need to afford paying the bets, when you can simply make spins without making a bet. And here comes the system, which gives you the incredible 0,9$ per play time.