Hi
  It's roulette player again. Ihave a very powerful way to bet on street.
  It requires a $152.30 minimum bankroll with 10 cents minimum bet on streets.
It's a slow bankroll building strategy that requires a lot of patience
  The trigger is that you spin without betting and mark down in which street the number came.
From street 1 to 12 reading the board from left to right. You wait until the number shows up in 9 different streets in the previous 9 spin. It's hard enough to do because there is a lot of repeats before it happens.
  When you meet the condition of your trigger, you are on your way to a profit. You will see why.
You bet 10 cents on each of you 9 different streets and 10 cents on the 0 for a total of $1.00 Most of the time, you will win on the next spin because there is a big chance the number will fall in one of your selected streets. You make 20 cents profit on any street you bet on and $2.60 if the 0 hits.
  The only way to loose is if the number shows up in a new different street. You then go in the second step of your progression. You bet 70 cents on the steet where the ball landed in the last spin and also the same amount on each of the first 9 streets. So you bet on 10 streets and 30 cents on the 0 for a total of $7.30. At this point the chance of a repeat is even bigger and you win. You make 10 cents profit on any street and $2.50 profit if the 0 hits. Very rarely, you will loose as this point. If it's the case, you bet on the 11th different street where the number showed up. You go on the last step of your progression, you bet $12.70 on the last street where the ball landed in the last spin and also the same amount on each of the streets that you bet before and $4.30 on the 0 for a total of $144. I never seen a lost at this point because the odds of having the number falling in all 12 different streets in 12 consecutive spins is  1/18613.92623. I used a random number generator in Excel which generates many sets of spins per second and that never happened. The best I have seen is 11 differents streets in 12 spins and not very often. That's why you need the third step of your progression.
  So try it for fun and make sure you understand this strategy in full before trying it for real money. Good luck with this. I attach also the Excel file for that particular progression. All the formulas are in place for you, I did the hard part. You can change the amounts in each steps for bigger profit but you need a bigger bankrool. Right now my bankroll is $133 just playing the powerful pattern breaker strategy. I am only $20 away to have enough for playing this streets system.

Good luck
Rouletteplayer[smiley]aes/money.png[/smiley]

Hi RoulettePlayer,

You do realize that if playing this system for as little as $1.00 stakes, your final bet is $14,400 with 3 numbers uncovered... 

To reply to green guy
  First of all, you bet on streets, not on single numbers. With a 10 cents minimum starting bet on streets, you never have to bet that high. The highest total bet is $144 on 11 streets and the 0.
The only way to bet $14,400 is by starting at $10 minimum bet on 9 streets.The bet amount of the third step of the progression would be over the table limit. If there is something you don't understand, read my post again and practice for fun.

Good luck
Roulette Player

I have seen all twelve streets within twelve spins!
If you continue long enough, you may see something similar and LOSE.
Losing with progression happens sooner or later.

Hi RoulettePlayer,

I find it instructive to compare bets like these to doubling up on the even chances. So what is the equivalent of your system in terms of a martingale (which it effectively is)?

Your trigger is 9 unique streets. What are the chances?

On the first spin any of the 12 could hit, and in subsequent spins it's required that previous streets are removed, since no repeats are allowed, so on the 2nd spin there are only 11 streets "available", as it were, and on the 3rd spin 10 streets, and so on. So the number of ways (permutations) that 9 unique streets can hit is

12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 = 79,833,600

The number of ways that 9 streets can arrive without any restrictions (i.e. repeats are allowed) is

12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 = 12⁹ = 5,159,780,352

So the probability of 9 unique streets is 79,833,600/5,159,780,352 = 0.01547 or 1 in 64 approximately. In this case, it's easy to see that this is equivalent to a run of 7 on the EC's.

You can work out the probability of getting 12 unique streets in 12 spins in a similar way, which is 1 in 18,614, as you have posted.

In terms of an even chance run length for this probability, you can use trial and error, or logs:

For any even chance, the probability of a run of length n is

p = (1/2)ⁿ

log(p) = n x log(1/2)

n = log(p) / log(1/2) = log(1 / 18,614) / log(1/2) = -4.2698/-0.3010 = 14.184

i.e., a run of length 15 (since we need to round up).

So your system is equivalent to waiting for a trigger of 7 reds (or perhaps Bankers, in terms of Baccarat) in a row, and then doubling up, hoping that the run will not exceed 15.

Here is something which may be of interest.

n 2 = 8.33%.......8.33%
n 3 = 15.28%......23.61%
n 4 = 19.10%......42.71%
n 5 = 19.10%......61.81%
n 6 = 15.91%......77.72%
n 7 = 11.14%......88.86%
n 8 = 6.5%........95.36%
n 9 = 3.1%........98.45%

These are some probabilities showing that it is more likely to hit a repeat street on the 4th and 5th spin.

How about waiting for the 9 original streets and then just betting the last 3 up until a repeat shows (win or lose) and then repeat the process. It would require some calculations working out the progression as you went along.

Quote from: Dane on November 16, 2014, 09:43:01 AM
I have seen all twelve streets within twelve spins!
If you continue long enough, you may see something similar and LOSE.
Losing with progression happens sooner or later.

You must have played a lot to see that. Also the 0 cannot be in that sequence. The 0 ends the progression right away because of the profit.

Roulette player

All push to win progressions lose and they lose so badly, the player never dares to do such thing again. Even a million chip can not ensure you a win of 1 unit with such progressions.