I played this Roulette strategy and getting decent enough results. I just want to know how bad is this as people will analyse and have different views .

It is straight forward Strategy.

Bet on 1-18 and 3rd Dozen in 3:2 ratio and Martingale only incase if your balance is less than your initial bankroll.

This leaves out only 8 numbers and always win one unit on every win..

just aiming for 10% Profit .

If you start with 500 and get to 550 session ends there.

Betting amount 15 on the 1-18 and 10 on 3rd quarter.It can be varied on your Bankroll.

Quote:rajbhah89Hi,

I played this Roulette strategy and getting decent enough results. I just want to know how bad is this as people will analyse and have different views .

It is straight forward Strategy.

A fun way to lose money.

https://wizardofvegas.com/member/oncedear/blog/5/#post1370

Quote:rajbhah89I just want to know how bad is this as people will analyse and have different views .

After some research and analysis, I calculate that this strategy will lose approximately 2.7% (1/37) of all money bet on a single zero game (or 5.26% (2/38) on a double zero game).

Quote:TomGAfter some research and analysis, I calculate that this strategy will lose approximately 2.7% (1/37) of all money bet on a single zero game (or 5.26% (2/38) on a double zero game).

Although true, it ignores the question the OP asks. He wants to know what is the chance he wins $50 before he loses $500 using his system. Figuring it out is above my pay grade, but I’ll make a guess.

Maybe 60% chance for success on double zero table.

Maybe 80% chance for success on single zero.

I don’t think it will be that hard for Wiz or ThatDonGuy or miplet or mathextremist to give exact answer.

I get about a 61% chance of reaching your target before losing the entire bankroll.

On the other hand, if you keep it simple, and just keep betting 50 per spin on 1-18 (or any even-money bet), you have an 85% chance of success.

On yet another hand, if you do a modified Martingale where your initial bet is the amount you need to reach your target and you always bet the smaller of what you need to reach the target and what's left of your bankroll, you have an 89% chance of success:

Bet 1 - bet 50; if you win, you have reached your target, and if you lose, you have 450

Bet 2 - bet 100; if you win, you have reached your target, and if you lose, you have 350

Bet 3 - bet 200; if you win, you have reached your target, and if you lose, you have 150

Bet 4 - bet 150; if you lose, you are bust, and if you win, you have 300 and need to win 250 more, so bet 250

Bet 5 - bet 250; if you win, you have reached your target, and if you lose, you have 50

Bet 6 - bet 50; if you lose, you are bust, and if you win, you have 100 and need to win 450 more, so bet 100

Bet 7 - bet 100; if you lose, you are bust, and if you win, you have 200 and need to win 350 more, so bet 200

Bet 8 - bet 200; if you lose, you are bust, and if you win, you have 400 and need to win 150 more, so bet 150

Bet 9 - bet 150; if you win, you have reached your target, and if you lose, you have 250

Bet 10 - bet 250; if you lose, you are bust, and if you win, you are back to your starting point, so start over with Bet #1

Remember that the more bets you make, the more money you expose to the house edge. If, instead of flat bets of 50, you make flat bets of 10, the chance of success drops to 59%. The modified Martingale does better because if any of your first three bets win, you stop betting.

Quote:ThatDonGuyYour system wins 1 with 28 numbers and loses 5 with the other 8.

I get about a 61% chance of reaching your target before losing the entire bankroll.

Thank you, TDG. I am now quite proud of my 60% guess!

Quote:SOOPOOAlthough true, it ignores the question the OP asks. He wants to know what is the chance he wins $50 before he loses $500 using his system.

Ok, let me try again.

Quote:rajbhah89Bet on 1-18 and 3rd Dozen in 3:2 ratio and Martingale only incase if your balance is less than your initial bankroll.

The way it is written here, and the wikipedia definition of martingale, and the way I understand it, anytime you are under $500, the previous bets get doubled.

On the first bet, there is a 73% chance of winning $5, and a 17% chance of losing $25.

Lose the first bet, and it takes one win with a $50 bet and another with a $100 bet to get back to where you can lower the bet to $25. Lose that $100 bet, and it would require a $200 bet, followed by $400. This happens every time we drop below $500. A very likely scenario is to be down somewhat, but not have enough to "martingale".

It would normally take four escalating losses to wipe out the $500. But the losses do not have to be in a row, and bets can escalate after wins. It only takes five escalating bets to be done with the game.

I estimate the probability of winning $50 to be around 20% (using a very non-rigorous method). But the good news is that a failure usually does not wipe out the $500, just couldn't get back over the initial bankroll before the double-ups reached the limit. This seems to be an interesting spin that actually flips what we normally think of martingale. With this strategy, there is a low chance of winning, but the loss will be small, instead of big.

If I am way off (or even slightly off), someone step in and correct me.

Quote:Ace2“Roulette strategy” is an oxymoron

I’m going to strongly disagree with you. Wouldn’t you agree that different strategies would carry a different risk of ruin? That different strategies would give either a higher or lower chance to achieve a defined goal?

I think what you mean to say is... “there is no roulette strategy that will result in a +EV bet”, or “there is no roulette strategy that will overcome the house edge”.

People are less dismissive of craps ‘strategies’... Iron Cross, continuous come bets, pass line and 10x odds.... all are losing systems... if you solely define a system as being successful if it results in a + EV result.