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99.92%

Started by Let Me Win, Jan 20, 05:57 PM 2019

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Let Me Win

There is a system by Turbo Genius that has a 99.92% probability of success.

You will either win at least one unit (sometimes more) or bust out with a loss of 1,783 units.

The system is called 50 max and can be read about on this forum from a few years ago here:
link:s://:.rouletteforum.cc/index.php?topic=6920.0

My question is can anyone provide a "system" with a probability of success higher than 99.92%

Or can anyone match the same probability of success but by risking less than 1,783 units?



Steve

Exactly what are the calculations that gave you 99.92%?
"The only way to beat roulette is by increasing the accuracy of predictions"
Roulettephysics.com ← Professional roulette tips
Roulette-computers.com ← Hidden electronics that predicts the winning number
Roulettephysics.com/roulette-strategy ← Why most systems lose

Let Me Win

I stole them from member Bayes.

But I remember him as always honest, helpful and a good coder and mathmatican.
I wonder where he is these days?

The answer given was:

1 - (19/37)6 x (25/37)2 x (31/37)4 x (33/37)3 x (34/37)3 x (35/37)7 x (36/37)24 = 99.92%

The General

Quote from: Let Me Win on Jan 20, 06:19 PM 2019
I stole them from member Bayes.

But I remember him as always honest, helpful and a good coder and mathmatican.
I wonder where he is these days?

The answer given was:

1 - (19/37)6 x (25/37)2 x (31/37)4 x (33/37)3 x (34/37)3 x (35/37)7 x (36/37)24 = 99.92%

Which means that you're not going to win frequently enough to overcome the short payout in the long run.
Basic probability and The General are your friend.
(Now hiring minions, apply within.)

Steve

Ok looks like it's a progression. If you play short term and use progression, the chances are you'll profit. But if you lose, you lose big.

So a system that has a high probability of winning in this case is not unique.

And you cant just play short term for a small profit, then leave... and repeat this daily.

What happens if 100 players all tried this? Most would win, and maybe 1 would lose. But when you look at the overall result from all players combined, the result is a loss. What determines who wins and who loses? Consider the bigger picture.
"The only way to beat roulette is by increasing the accuracy of predictions"
Roulettephysics.com ← Professional roulette tips
Roulette-computers.com ← Hidden electronics that predicts the winning number
Roulettephysics.com/roulette-strategy ← Why most systems lose

Let Me Win

Thanks, I've been playing roulette for around 30 years all over the world and online with the RX bot.

My question was could anyone post a progression with a higher or equal probability of success using the same or less units.

Steve

If you cover almost the whole table with mostly outside bets, and use a progression, you'll get around the same probability of profit. But what I said still applies. It isnt a good strategy.
"The only way to beat roulette is by increasing the accuracy of predictions"
Roulettephysics.com ← Professional roulette tips
Roulette-computers.com ← Hidden electronics that predicts the winning number
Roulettephysics.com/roulette-strategy ← Why most systems lose

Firefox

I think Victor's parachute was better.

For the sole reason that you bet less. Both are a series of independent trials but the final expected loss of either is always equal to the total amount bet multipled by the house edge.

So with Victor's parachute, you make a flat bet of 1 for 36 spins and risk far less capital at a chance of winning back more than 1 unit in some cases. It still loses on average 36/37 units every 36 spins though.

I don't know when Victor named it, but it was in Norman Squires book, How to win at Roulette which was published around 1970 or maybe earlier.

I know that doesn't answer the question, but systems similar to Max50 will lose more, simply because they bet more.

Taotie

Let Me Win, you use the term "bust out". So the game is set in stone, we either take the 1st profit or continue play until the bust out. Does the system show any profit after 10000 games with 9992 wins x 8 busts? > 8 busts equalling 14264 units.

My own system uses a short progression of 1.2.3.4 over and over, and therefore it "busts out" often. To calculate the success of this type of system requires use of another factor, session length.

Given a long enough session length it will exceed the 99.92% win rate. However it could not always guarantee a drawdown of less than 1783 units.

So the jury is still out on the risk side.

Let Me Win

Thanks for the replies.

Yeah probably Victor got it from Norman's book and brought it to the attention of the Internet.

I agree the parachute with a maximum loss of 36 units is more attractive but its quite easy to bust it at only 95% so Turbo's version feels safer at 99%.

I think the Turbo version could be used online only with a bot.

Firefox

It seems that subjecting this to a large number of trials via a bot would expose the frailty of the system.

Eg 10000 trials 9992 wins 8 losses

Outcome: Win 9992 lose 8x1783=14264
Total  -4272 units

If you get lucky 9996 wins 4 losses

Outcome: Win 9996 lose 4x1783=7132
Total  +2864 units

If you get unlucky Win 9988 lose 12

Outcome: Win 9988 lose 12x1783=21396
Total -11,408 units

In short, likely large losses for potential small gains, and any tweaks one makes via different progressions will have similar results.


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