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Hey guys!

As you may know, I recently got a Casino Bonus that I failed to release. Now, this bonus got me thinking a little bit about the roulette. Where I play, PartyPoker, the minimum bet is $0.10 and the maximum bet is (if I remember correctly) $500.

Now, this opens up quite a big possibility. I mean, you need a BIG amount of money to start with but the benefits could be quite good.

As we all know, the house has an edge on every single game, but in roulette, I think it could be minimized enough to make it profitable for the player. I know I know, in the long run the casinos always wins, but bare with me for a moment.

Now, a bet on any given number gives you the odds of 1/37. Not that good, right? Now, consider putting a bet on 30 numbers instead, that's 30/37, a lot better, although not good. But here's where things get better.

Say that you happen to come over a big amount of cash (maybe you won a high payout MTT?), say that you would want to risk $588 of that on the roulette. Now, if you place a minimum bet of $0.10 on 30 numbers, it's a total bet of $3. Now, the casino has 7 out of 37 chances to win and if they do, you lose your $3 bet, but if you win, you only get a profit of $0.60.

Still not worth the risk, right? Now, keep placing that bet until you lose (odds are that you'll lose about 1 out of 5 times), and when you do, you place a continuation bet on the very same numbers but you bet 6 times what you did before. This would give you a total bet of $18. A win on any of your numbers would give you $21.6. If you now take those $18 that you just bet and add the $3 that you bet before, you get $21. You made up for your loss and still made $0.60.

It still seems risky to you, I get it. But hang on just a little bit longer. The odds of you losing this second time in a row is less then 380 out of 10,000 (about 3,8%). Still not good enough odds in our favor.

If we now say that you would lose that second time also, then you place another continuation bet of 6 times of your last bet. This gives you a bet of $108. A win would give you $129,6. Your previous bets of $3, $18 and $108 sums up to $129. You'll make a profit of $0.6. The odds of you losing this third time in a row are less then 68 out of 10,000 (about 0,68%). The odds are now starting to look a lot better.

But let's take it one step further and say that you would lose this third time as well. Place another bet but this time you can't just go 6 times your previous bet since that would give you a $648 bet and the limit is $500. So now what you should do is place $16.6 on your chosen 30 numbers to give you a $498 bet (you could just do $16.7 on 20 of the numbers to get it up to an even $500 but I'm not going to just for the simplicity of the calculation). A win would give you $597,6. Add together all your bets and you'll get: 3+18+108+498=627. This means that a win would give you a total loss of $29.4, but hey, it's better then it would have been if you had left it after the last bet and lost $129.

Now, the odds of losing this fourth bet in a row is less then 13 out of 10,000 (about 0,13%). Now it looks a lot better, right?

And here's where it gets interesting. If we say that you use this betting system 10,000 times, 13 times you would lose $627, 68 times you would lose $29.4 and 9919 times you would come out with a $0.60 profit. Now, let's get started on these calculations:
13*627=8,151
68*29.4=1,999.2
9919*0.6=$5,951.4

This means that for every 10,000 times this system is used, the casino would win 4198.8.
But, think about it, the whole idea was that you had some money that you actually was going to gamble with. This means that heading into this, you still have a 99,19% chance of leaving your gamble with a profit.

I think that this is probably the safest thing if you're going to gamble on the roulette with some money that you don't even care about. It is, theoretically, possible to turn those $588 that you wanted to risk into $5,951.4. This is, of course, for the most extreme case and the biggest amount of luck possible. But I would say that it is probably possible to make it into $1,800, about 3 times what you started with, and with the lowest risk possible in a casino game.

You could make your odds even better by adding numbers that you bet on, but then your losses would be even greater if you did lose and you would also need to adapt the 6xBI rule to fit your own system.

So, after reading this, what are your thoughts? Is it even possible to do this and what are the odds of actually getting a decent profit out of this before you lose?

Edited by erru9107 (19 November 2014 @ 13:23 GMT)


     
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Reading this % sounds nice, but was trying roulete with red/black martingale x2 and saw how fast u lose 7+ times in row with that system althou the chances are very low (cant count random like this imo, each game starts anew with same chances as previous, it sounds logical to expect some certain win but in the end it comes out different)

Try it would like to see if you have luck Smile

You are assuming you wont lose the first 833 games via this method, is how much u need to win to get +500$ and if you lose that 500 just once out of this 833 games u are already done. (assuming u have 500 only)

     
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Posted by jsikonja:
Reading this % sounds nice, but was trying roulete with red/black martingale x2 and saw how fast u lose 7+ times in row with that system althou the chances are very low (cant count random like this imo, each game starts anew with same chances as previous, it sounds logical to expect some certain win but in the end it comes out different)

Try it would like to see if you have luck Smile

You are assuming you wont lose the first 833 games via this method, is how much u need to win to get +500$ and if you lose that 500 just once out of this 833 games u are already done. (assuming u have 500 only)

The big difference between the red/black strategy and this one is that with red/black, every bet only gives you 18/37 to win. This gives you 30/37, that's 32,4% higher chance of winning. Granted, the win isn't as big for each time, but give it a couple of goes and you'll see that you have more by the end of it.

I'm not saying this will work every time, in the long run, the house ALWAYS wins! But for a one time only shot in the dark, it might give you some extra cash in your pocket or you'll lose what you were already prepared to lose. And I would say that 99,19% chance of winning is probably the best one you're gonna get with a one time only gamble.

I don't have the cash to try this myself, but I might go with play money later tonight and give it a shot to see how it works out!

     
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The main flaw with your reasoning is that on the 4th time, you say the odds of losing are 0.13%... but this is not true - The odds of losing 4 in a row may very well be that amount, but still the odds of losing that specific bet are 7/30 or around 20%

The house edge is not beatable - The more you stake, the more you'll lose.

     
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Posted by yout85:
The main flaw with your reasoning is that on the 4th time, you say the odds of losing are 0.13%... but this is not true - The odds of losing 4 in a row may very well be that amount, but still the odds of losing that specific bet are 7/30 or around 20%

The house edge is not beatable - The more you stake, the more you'll lose.

It should 7/37, there are 37 possible scenarios.

And yes, I do see what you mean, but according to probability, the probability for that specific game in it self is indeed 7/37, but the probability for that event to have occurred 3 times in a row before this game and to do it again is much much smaller, in fact it's only 7/37*7/37*7/37*7/37=2401/1874161=0.00128=0.128%

     
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It looks like you have put a lot of thought into this but as with all roulette systems you can make the maths look attractive but the fact is after the first few losses you are gambling a huge amount of money to win 60c.Also,how much profit is your objective?Win one bet and leave?
You say "The odds of you losing this third time in a row are less then 68 out of 10,000 (about 0,68%).".....in theory it maybe true but the truth is the odds are always the same as your very first bet,the roulette table has no memory and ultimately you are always betting a large amount to win a very small amount.In that respect I am sure all casinos would welcome you with open arms. Smile
On the flip side you could get lucky and win the first few bets and you have won a few dollars for little risk. Smile Why don't you try a budget of $1000 on play roulette and see how it goes and let us know,try it 10 times and then use your own money. Smile

     
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Hey, hey, hey - lil' baby's first meet with the Martingale system. How adorable!

     
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Posted by erru9107:
It should 7/37, there are 37 possible scenarios.


Nope... 7:37 would mean 7 wins and 37 losses expected for every 44 bets.

In the same way that 2:1 odds equal 66%, not 50%

     
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Posted by yout85:
Posted by erru9107:
It should 7/37, there are 37 possible scenarios.


Nope... 7:37 would mean 7 wins and 37 losses expected for every 44 bets.

In the same way that 2:1 odds equal 66%, not 50%

Well yeah, 7:37 does, 7/37 doesn't. And I was counting the losses, not the wins.

I know all about the no memory thing on the tables and whatever, but it's still done by an RNG meaning that the numbers are just that, random. So probability for every game in itself yes, the odds remain the same. And yes, theoretically, you could lose every single time.

Probability is always a factor, even here.

An no bowie, it's not my first meet with it, it's just the first time that I've actually put any thought into it to see what I can come up with.

As I've said, in the end, the house always wins, I'm just trying to figure out if there's a way to just even the odds out a little bit.

I did find something wrong in my calculations though. It's when I added all the numbers together by the end. 13 out of those 68 times I'd lose in the 3rd bet shouldn't count as they are already covered in the 4th bet after losing those 3 times. So the actual numbers are a little bit lower on the casinos take after 10,000 games, more precisely $382.2 less.

And as I've said, I'm not even sure that it would actually work, I just got to thinking about if there is a way to even out the odds just a little bit so it might be worth taking a shot at some point. And since it's all run by an RNG (or at least should be) the rules of probability DO have something to do with it no matter if the table has any memory or not.

It's basically the same with dies. The odds of throwing a 6 is 1/6. The very next throw, the odds remain the same, 1/6. But, combine both those and the odds for hitting a 6 two times in a row suddenly changes to 1/36.

I will try this with play money a bunch of times and I'll update you on it.

     
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This system is a Martingale on steroids. If you win, you win peanuts, but if you lose, you lose big. Yes, one can stear the variance a bit at roulette, as there are bets with very different chances available, but the house edge remains the house edge. If you have spare money to blow on roulette, bet it all on red/black or high/low or odd/even. Far higher chance to pocket a good win than with tricky betting systems.

     
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It's quite similar to a thing I did a few years ago with football betting...

Basically, I took every English Premier League team, and bet on them until I was in profit on them, then stopped.

So, first weekend I bet $1.00 on every home team... the one's that won, I would not bet on again - the one's that lost, I would calculate how much I needed to bet on them to be at least $1 in profit, then continued until all teams were winners. The theory is the same - no team will lose forever... but if just one team loses 5 or 6 times in a row, the profit from the 19 winners doesn't even equal out the loss of the losing team... It's called chasing your losses - and can only result in tiny profits, or huge spiralling bets to win back what you've already lost. It is not sustainable, and for anything to be deemed "a system" it has to work long term...

     
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Yeah, and as I've said, the house always wins in the long term. What I'm talking about here is basically just a way to try and double your money (that you are prepared to lose either way) with as small of a risk as possible. It's basically just a "system" for a one time only shot in the dark, not one that will last forever and make you rich or anything.

And since roulette is the only casino game (that I know of at least) where you can actually improve on your odds of winning a certain bet, it seemed to be the logical game of choice to try something like this.

Again, not meant as a long term thing, just increasing your odds for a one time only gamble.

     
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Too much to read but just skimmed over what you wrote and saw that it was a bout casino and party poker and stuff and you guys know already what i will say, but let me do it casino games are mug an you will lose money off it if you play them as the house have the edge and so on! Also roulette is a mug game, so have a strategy and hopefully you can make some money off those in the short term cheers good luck once again!!

     
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As i posted before: If you want to double up, bet it all on red (or a comperable bet). Chances are much better to actually make a double up, than with any martingale system. And much quicker too.

     
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Posted by shokaku:
As i posted before: If you want to double up, bet it all on red (or a comperable bet). Chances are much better to actually make a double up, than with any martingale system. And much quicker too.

And as I've said before: I'm not after that, I'm looking for the safest way to do it by minimizing the houses edge as much as possible.
Your way is:
1 - Easier? Yes
2- Quicker? Yes
3. Higher risk of losing? Yes

The "system" I wrote about here isn't meant to last forever, just enough to get you some money in the safest way possible (knowing that a casino bet is never safe of course).

     
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