Another thing I'm curious about, the Wizard's challenge says if your system wins, he will post the system on his website and say he was completely wrong about system betting. MichaelBlueJay promises in writing not to use your system himself in a casino. Personally, if I give you a winning betting system, I definitely wouldn't want you to post it on your website, logical reasons apply, however, I would love for you to play the system in a casino for yourself and put in writing that you'll cut me 15% of the winnings. 20 people cutting you 15% of the profits means you make 300% instead of 100%.

Just curious, how interested you are in a betting system that can beat an HA game.

Quote:JyBrd0403Just curious, how interested you are in a betting system that can beat an HA game.

Is it a method to pick the winning bet, or another

series of progressions to beat the HE?

Quote:JyBrd0403It's a progression that beats the HE. It hasn't been tested, but it wins outright. .

No it doesn't, and its been tested, just not by you.

Don't waste your time.

If you, or anybody else, can run that simulation you'll see the game wins outright. Feel free to post the results here. I think the bankroll would be too high to play in real life. If it's playable in real life, I'd be happy to take a Vegas kick back on the winnings :) But, I think the bankroll would be to high, and the casinos won't have a High Roller table that would handle the biggest bets on that progression.

I've got another game that wins outright, and also is playable in real life casino play.

If you lose betting 1, you bet 2 for a net of 1 as you won 2 but lost 1

If you lose betting 1 and 2, you win at 4 for a net of 1 as you won 4 but lost 3

If you lose betting 1, 2 and 4 and win at 8 for a net of one as you won 8 and lost 7

and so on and so forth, by cutting the bet in half when you win rather than going back down to 1 unit would increase you risk of ruin.

Quote:Gabes22There are many flaws with that system first off, the Martingale is designed you by winning you 1 betting unit

If you lose betting 1, you bet 2 for a net of 1 as you won 2 but lost 1

If you lose betting 1 and 2, you win at 4 for a net of 1 as you won 4 but lost 3

If you lose betting 1, 2 and 4 and win at 8 for a net of one as you won 8 and lost 7

and so on and so forth, by cutting the bet in half when you win rather than going back down to 1 unit would increase you risk of ruin.

No, there are no flaws with the system. First of all you have to understand the martingale does not win on a 50/50 game, it breaks even, like all other systems besides the D'Alembert. The problem is you don't understand the D'Alembert and how it works. I've spent years studying the D'Alembert, I know how it works, and I know why it doesn't work on an HE game. By D'Alemberting the Martingale you produce an outright win even on an HE game. You'll have to do the math yourself to figure out why it works. Good luck. It took me years to figure out how it works. But, hey, if someone has a roulette simulator, we won't have to wait long to find out who's right and who's wrong.

VLSroulette.com and somebody there will gladly

test it for you.

Quote:EvenBobThis isn't a roulette forum, nobody here cares. Go to

VLSroulette.com and somebody there will gladly

test it for you.

mr. charm strikes again

Quote:JyBrd0403I'm curious as to why these betting system challenges have regulations on bankroll amount and maximum units bet for these challenges. Everyone here is 100% certain that no game could win, even with an infinite bankroll, right. So, why the regulations? MichaelBlueJay's challenge requires a real life casino game with a table limit of $5000. In real life, if you're betting a system and you reach the table limit, wouldn't you just move to a higher limit table and place the next bet at that table?

Another thing I'm curious about, the Wizard's challenge says if your system wins, he will post the system on his website and say he was completely wrong about system betting. MichaelBlueJay promises in writing not to use your system himself in a casino. Personally, if I give you a winning betting system, I definitely wouldn't want you to post it on your website, logical reasons apply, however, I would love for you to play the system in a casino for yourself and put in writing that you'll cut me 15% of the winnings. 20 people cutting you 15% of the profits means you make 300% instead of 100%.

Just curious, how interested you are in a betting system that can beat an HA game.

The goal of the challenge is to develop a system which can be used by an actual player in improving the Edge into the player's favor. The regulations are designed in support of that purpose.

1.) Bankroll

-There must be a limited bankroll because, technically, every single person to ever walk into the casino would have a finite bankroll. If an individual were the richest person on the face of the planet, with a net value (liquidated) that exceeds the value of the entire casino, then the casino's, "Bankroll," could not cover the player's bankroll and the casino could not risk more money than it actually has.

-It is with this that, regardless of how much theoretical money you want to give to the player, some bankroll will always come into play because the player's cash cannot exceed the casino's cash or the casino cannot necessarily make the bet.

-The other reason is that I don't need your system to know a system that would work with infinite bankroll. I already know one: The Martingale. Given infinite money and infinite time, the Martingale MUST win.

2.) Table Limit

-There will always be table limits, but it's not the Table Limit itself that is important to the casino, at least not when it comes to systems, it is the spread. For example, if you have a Table Minimum of $5.00, Max $200.00, a Martingale line fails at six consecutive losses (5, 10, 20, 40 80, 160-can't bet 320)

If you increase the Maximum to $500, then it is seven consecutive losses (5, 10, 20, 40, 80, 160, 320-can't bet $640)

If you make it $1,000, then it is eight consecutive losses (5, 10, 20, 40, 80, 160, 320, 640-can't bet 1280)

-I understand your point about moving to higher limit tables, but eventually you will reach the point where a real casino would actually say, "I'm sorry, Mr. so-and-so, but we can't cover a bet of $5,000,000/hand. Casuarina may be able to cover that kind of action, give them a call."

(The last part was a joke)

3.) Further, and to stay with Martingale, even if you were to reach a point where the next bet would be $1,310,720 (assuming $5.00 base bet) then he has lost eighteen hands in a row and $1,310,715. Most players cannot afford to lose $1,310,715, so the system would be patently useless to them, even in the event of a win on that nineteenth hand.

Quote:JyBrd0403Buzzpaff, could you run a simulation of this system for me. It's for single zero roulette. Only problem is that it has to be a roulette simulator. The system requires that you rebet the last bet whenever a zero comes up. That way the progression itself is 50/50 18 blacks vs. 18 reds. What you do is basically D'Alembert the Martingale, but instead of dropping back down to 1 unit after a win, you just bet half the last bet. So you double the bet every time you lose, and you bet half of the last bet when you win. Like this 1-2-4-8-16-32-64-32-16-8-4-2-1. And, if a zero comes up you bet the same amount again. So, if you bet 32 and a zero come up, you don't double the bet, you just bet 32 again.

If you, or anybody else, can run that simulation you'll see the game wins outright. Feel free to post the results here. I think the bankroll would be too high to play in real life. If it's playable in real life, I'd be happy to take a Vegas kick back on the winnings :) But, I think the bankroll would be to high, and the casinos won't have a High Roller table that would handle the biggest bets on that progression.

I've got another game that wins outright, and also is playable in real life casino play.

This system would fail, and quickly, for a variety of reasons.

1.) You are basically playing the Martingale, just changing the base bet.

-The purpose of the Martingale System is that you double the bet after each loss and restart the system with the base bet after a win. By doing so, you will ensure that a completed Martingale line wins back all of the previous losses plus one unit.

-Your system does absolutely nothing, except continuously plays the Martingale, but changes the base bet in the event of a win. I will illustrate:

If you lose five bets in a row starting with 5 and win one:

-5 -10 -20 -40 -80 +160 = 5

You have profitted $5.00, which we are calling one unit.

If you accomplish this, but instead of returning to the normal Martingale, you cut the winning bet in half and lose five then win one (Table Max. $500-for example)

-80, -160, -320...You Can't Make The Bet!!!

The advantage, to the extent that it can be said the Martingale offers one (it doesn't) is that it takes a substantial number of losses in a row to wipe you out. Your system will wipe you out faster because, instead of returning to the base bet and starting a new Martingale line, you are simply dictating a higher initial minimum bet for yourself, in this case, $75 higher.

For example, let's just say you have a pattern where you lose three spins in a row followed by one win:

-5 -10 -20 +40 (+5 total) -20, -40, -80, +160 (+25 total) -80, -160, -320 (The next bet cannot be made)

In comparison, if you were guaranteed to have a pattern whereby you would lose three spins consecutively followed by winning one, the Martingale could never fail. That is the purpose of the Martingale. In the sequence above, you have lost a total of $535.00 due to the table limit.

In short, the Martingale system relies upon ONE WIN to break a losing streak and still result in a player profit. Your system not only counts on a win to break the losing streak, it also relies upon winning streaks to keep you away from the betting limit. Furthermore, where a losing streak that brings you to the Max Bet will demoralize your bankroll, you are not taking advantage of your winning streaks because you are decreasing your bets everytime you win.

2.) Rebet on Zero

-This is probably the single worst part of your betting system because it deliberately ignores a loss.

Let's say you lose two in a row, then lose on a zero, lose one more, then win:

-5, -10, -20, -20 +40 (-$15 total)

3.) I understand where you think the advantage in your system lies. With the Martingale System, you come ahead the +5 units and then re-start. Your system does, to a certain extent, take advantage of winning streaks. For instance, if you lose five in a row then win three:

-5, -10, -20, -40, -80, +160 (+5 total) +80 (+85 total) +40 (+125 total)

Then you have taken advantage of three consecutive wins in a fashion that the Martingale fails to do.

I have also taken note of the fact that if you went, win-loss-win-loss, you would also come out ahead.

-5, +10 (+5 total) -5 (Even) +10 (+10 total) -5 (+5 total) +10 (+15 total)---This is the same thing Martingale would do.

So, don't think I don't understand the purpose of your system.

CONCLUSION:

The main difference between your system and the Martingale is that you system chooses to conveniently ignore a specific type of loss (on zero) and the Martingale would never ignore any loss. If the Martingale did ever ignore a loss, then it could not succeed, even theoretically.

The other problem with your system is the Martingale seeks nothing more haughty than breaking a losing streak. In other words, the Martingale can ultimately succeed while only winning a fraction of the time. With a $5-$500 Bet Spread, for example, the Martingale can theoretically lose:

5, 10, 20, 40, 80, 160 (Win on 320)

Six times for every win, 1/6 or 16.66666667% win rate, to ultimately be successful.

Unfortunately, your system must win AS MUCH as it loses in order to succeed, which is why, I believe, you were conveniently ignoring the loss on zero.

However, not only must your system win as much, your system can also not be confronted by losing streaks (however small) that are not offset by winning streaks, or you will eventually reach a point where two losses in a row after a win would put the next bet over the table limit.

Your system has its good points, though, I'll admit. I do like how your system will capitalize on a winning streak (in terms of profit) better than will the Martingale, however that does come at increased risk of running into the table limit. I cannot reiterate enough that the goal of the Martingale is to take advantage of the spread between the minimum/maximum bets so as to ensure as many consecutive losses as possible before reaching the Table Limit.

It is for this reason, for example, that you will not see a table with a Minimum of $5 and Maximum of $500000. The percentages for this sort of thing have been fully analyzed by the casinos and the spread between Minimum/Maximum is partially based upon that.

In short, your system has a short-term rate of profit that may be higher than the Martingale's rate, but your system is also more susceptible to failure because you reach points between where there is less of a spread (in terms of doubling) between the amount bet after a win and the Table Limit.

If you would be so kind, please explain the intended purpose of ignoring the zero. I can't figure out why you would be compelled to do that.

If you can not see such an obvious thing, I doubt your ability to properly evaluate this system !!

Quote:JyBrd0403You'll have to do the math yourself to figure out why it works.

Quote:JyBrd0403I can't do the math for that.

Quote:JyBrd0403I know how it works

Case closed.

Quote:buzzpaff" If you would be so kind, please explain the intended purpose of ignoring the zero. I can't figure out why you would be compelled to do that. "

If you can not see such an obvious thing, I doubt your ability to properly evaluate this system !!

I analyzed it in the context of the Martingale, which I understand inside and out, chiefly because there isn't a whole lot to understand. I believe that the purpose of ignoring the zero is because, ignoring the zero, red and black come up evenly resulting in a 50% chance of winning.

The problem, as I have explained, is that you cannot essentially, "Ignore a loss," when playing a betting system. The reaosn why you cannot ignore a loss is because you will never bet in a fashion that makes up for the loss.

Quote:buzzpaffThe system obviously must include a reverse Martingale on 0O to counter any losses, thereby guaranteeing a winning result.

No, you eventually have to win two in a row to offset any zero. If you ignore any zero, the first win afterwards still results in an overall loss.

the casino's might take.

Quote:buzzpaff" I promise I won't let the casino's know your secret. " No worried about that as my patent will protect me from any actions

the casino's might take.

You're going to steal my game and pass it off as your own buzzpaff. :) That's mean. Why wouldn't you just kick me down some money from the winnings?

Quote:Mission146-The other reason is that I don't need your system to know a system that would work with infinite bankroll. I already know one: The Martingale. Given infinite money and infinite time, the Martingale MUST win.

The martingale breaks even with and infinite bankroll on a 50/50 game of chance. Here's the math for let's say losing 2 in a row.

Lose 1-2 = -3 You lose 2 in a row and you're at -3. On average you'll lose 2 in a row every 6 trials, so you'll have 3 wins in 6 trials, and you'll lose 2 in a row in 6 trials. You'll win 3 units ( the 3 wins) and you'll lose 3 units ( the 2 losses in a row). The system breaks even on a 50/50 game. Now, you can bet 4 units after the 2 losses in a row, but it won't help you, the game will still break even.

By the way, on a HE game the martingale loses outright.

Quote:JyBrd0403The martingale breaks even with and infinite bankroll on a 50/50 game of chance. Here's the math for let's say losing 2 in a row.

Lose 1-2 = -3 You lose 2 in a row and you're at -3. On average you'll lose 2 in a row every 6 trials, so you'll have 3 wins in 6 trials, and you'll lose 2 in a row in 6 trials. You'll win 3 units ( the 3 wins) and you'll lose 3 units ( the 2 losses in a row). The system breaks even on a 50/50 game. Now, you can bet 4 units after the 2 losses in a row, but it won't help you, the game will still break even.

By the way, on a HE game the martingale loses outright.

How would the Martingale ever break even? You profit the amount of the base bet every time the Martingale completes, (i.e. everytime you win a bet) so with infinite money, the Martingale theoretically couldn't lose because you must win a hand eventually. When you do win that hand, then you will be ahead the amount of the base unit bet plus any previous winnings.

(lose two in a row)

-1, -2 = -3

We agree so far.

Next bet

4-Win = +4 Profit = 1

Two losses in a row:

-1, -2

Next bet

4-Win = +4 Profit =1 Total Profit = 2

Where do you get breaks even?

By the way, on an HE game, the anything except for counting cards loses outright, but then you're technically not playing an HE game...at least at the time the count is in your favor.

Quote:Mission146I believe that the purpose of ignoring the zero is because, ignoring the zero, red and black come up evenly resulting in a 50% chance of winning.

By ignoring the zero you DO NOT get a 50% chance of WINNING. Ignoring the zero, makes the betting progression 50/50. If you're betting Red, and the next five rolls come out. Black-Black-Zero-Red-Red. You're D'Alembert progression would be 1-2-3-4-3. My system is 50/50. The progression would be 1-2-3-3-2-1. You give 3 units to the casino when the zero lands in this situation, but the progression stays 50/50. This is why the D'Alembert loses on an HE game.

Quote:JyBrd0403By ignoring the zero you DO NOT get a 50% chance of WINNING. Ignoring the zero, makes the betting progression 50/50. If you're betting Red, and the next five rolls come out. Black-Black-Zero-Red-Red. You're D'Alembert progression would be 1-2-3-4-3. My system is 50/50. The progression would be 1-2-3-3-2-1. You give 3 units to the casino when the zero lands in this situation, but the progression stays 50/50. This is why the D'Alembert loses on an HE game.

If there were no zero, you would be 50%/50% to win, that's my point. Your system is based off of a 50/50 chance of winning, even though such is not the case.

I understand the way your system works, and it never does anything to compensate for losses on zeroes, by design, your system fails to do this. The advantages that your system sees compared to the Martingale in being better at taking advantage of winning streaks are partially negated by the fact that you have to compensate for losses on zeroes and you are not making such compensations within the system.

I suppose not factoring the loss on zero into the Progression will help you combat the Table Limit slightly, but flat-betting would help you combat it 100% of the time.

Quote:Mission146the Martingale theoretically couldn't lose because you must win a hand eventually.

It theoretically couldn't win either, because the losses in a row break even with the total amount of wins. I'll try to explain the martingale to you. You lose 18 in a row in approx. 500,000 trials. That's $250,000 in winnings. So, you can clearly see that the martingale wins and wins and wins and eventually it will lose 18 in a row and you'll break even. It does the same thing after 6 trials. You win 3 and lose 2 in a row and you break even, it's your choice to bet 4 units or not on the next roll, but no matter what you do the martingale just breaks even. Period. The D'Alembert wins outright.

Quote:Mission146 and it never does anything to compensate for losses on zeroes

Yes,it does.

Quote:Mission146I understand the way your system works

No, you don't.

Quote:JyBrd0403It theoretically couldn't win either, because the losses in a row break even with the total amount of wins. I'll try to explain the martingale to you. You lose 18 in a row in approx. 500,000 trials. That's $250,000 in winnings. So, you can clearly see that the martingale wins and wins and wins and eventually it will lose 18 in a row and you'll break even. It does the same thing after 6 trials. You win 3 and lose 2 in a row and you break even, it's your choice to bet 4 units or not on the next roll, but no matter what you do the martingale just breaks even. Period. The D'Alembert wins outright.

What?

With an infinite bankroll and no table Maximum the Martingale would always eventually profit one unit.

Quote:Mission146What?

With an infinite bankroll and no table Maximum the Martingale would always eventually profit one unit.

Right, if you stop when you're ahead, the martingale will produce a win at some point. Problem is, once you get ahead you have to quit, and you can never, ever, play again. Otherwise, you'll just break even.

Quote:JyBrd0403Right, if you stop when you're ahead, the martingale will produce a win at some point, who knows when, but you will be ahead at some point. Problem is, once you get ahead you have to quit, and you can never, ever, play again. Otherwise, you'll just break even.

That's not even remotely the case with an infinite bankroll and no table Maximum. You would always add one unit to your profits for every completion of the Martingale line, and every Martinglae line would eventually complete.

Quote:Mission146That's not even remotely the case with an infinite bankroll and no table Maximum. You would always add one unit to your profits for every completion of the Martingale line, and every Martinglae line would eventually complete.

Yes, and then, you would eventually lose 18,000 in a row, and break even.

Quote:JyBrd0403Yes, and then, you would eventually lose 18,000 in a row, and break even.

You would have to stop to break even, why would you stop with infinite money? You'd just play until you finally win a hand.

Ken

Quote:mrjjjWhy do you guys argue regarding a PRETEND situation? There are max payouts at tables so.......end of story.

Ken

Very true. Unless you can find me a casino that doesn't have a table max or a person with an unlimited bankroll, then it is pointless arguing it. It is very probable that table max's were a casinos initial defense against a Martingale system, although I don't know that to be true.

I'm sure someone else has a simulator, and enough curiosity to test the system.

Quote:JyBrd0403That's alright cardshark, that's just 1 way to beat an HE game that is suppose to be impossible to do, I have 10 other ways. If you don't want to do it for free and post the results, don't bother. I do the impossible all by myself :)

I'm sure someone else has a simulator, and enough curiosity to test the system.

No worries, if you change your mind, just let me know.

Approximately :)Quote:buzzpaffI calculate your bankroll need to be $34,653.44 approximately

The following Java program uses an arbitrary precision integer class. It prints out whenever "totalWon" goes further negative than its previous most negative amount. This shows that the progression does not win in the "long run" and in fact loses arbitrarily much. The BigInteger class does not contain a "divide" method that outputs a double, so the house edge is not included in the output. It's enough to show that the player will have an increasingly negative bankroll infinitely often.

For simplicity, I assumed "red" corresponded to 1..18 and "black" to 19..36 and that the player wagered on red.

import java.math.BigInteger;

import java.util.Random;

public class roulettesim {

public static void main(String[] args) {

BigInteger wager = new BigInteger("1");

BigInteger totalWagered = new BigInteger("0");

BigInteger totalWon = new BigInteger("0");

BigInteger ZERO = new BigInteger("0");

BigInteger ONE = new BigInteger("1");

BigInteger TWO = new BigInteger("2");

BigInteger MIN = new BigInteger("0");

Random generator = new Random();

int roll;

int ctr = 0;

while(true) {

ctr++;

roll = generator.nextInt(37);

totalWagered = totalWagered.add(wager);

if (roll >= 1 && roll <= 18) {

totalWon = totalWon.add(wager);

if (wager.compareTo(ONE) == 1)

wager = wager.divide(TWO);

}

else

if (roll >= 19 && roll <= 36) {

totalWon = totalWon.subtract(wager);

wager = wager.multiply(TWO);

}

else

if (roll == 0)

totalWon = totalWon.subtract(wager);

else

System.exit(1);

if (totalWon.compareTo(MIN) == -1) {

MIN = totalWon;

System.out.println(ctr + ", " + totalWon);

}

}// while

}// main

}// class

Here is some output:

1, -1

2, -3

3, -7

53, -28

54, -92

58, -220

61, -348

96, -1244

442, -220663

499, -3210743

515, -10550775

516, -44105207

517, -111214071

518, -245431799

5185, -4694268046

5203, -49791424654

5221, -127100835982

5222, -676856649870

5223, -1776368277646

5224, -3975391533198

5225, -8373438044302

5226, -17169531066510

9110, -21926770405267

9113, -57111142494099

9114, -197848630849427

9115, -479323607560083

9131, -901536072626067

9143, -2590385932890003

9144, -11597585187630995

9349, -169170795587465107

9350, -745631547890888595

9855, -15802971599794892691

9856, -52696459747213995923

9859, -89589947894633099155

9860, -237163900484309512083

9861, -532311805663662337939

9862, -1122607616022367989651

11834, -9981264520366725558163

11835, -1218907084134995900264339

11836, -3636758723364254249676691

11837, -8472462001822770948501395

13202, -28368874921337306746744723

13219, -67054501149005440337342355

13225, -1304994540434385715236466579

13226, -3780874619005146265034715027

13227, -8732634776146667364631211923

13228, -18636155090429709563824205715

13229, -38443195718995793962210193299

13232, -58250236347561878360596180883

13259, -355355845776053144336385994643

13262, -672268495833110494710561795987

13268, -3207569696289569297703968206739

13269, -8278172097202486903690781028243

13270, -18419376899028322115664406671251

56624, -40821759546190969066977412375767

56625, -203081036375404332458555422663895

56626, -527599590033831059241711443240151

56627, -1176636697350684512808023484392663

56641, -1338895974179897876199601494680791

56642, -6531192832714725504730097823900887

56645, -11723489691249553133260594153120983

56646, -32492677125388863647382579470001367

56651, -42877270842458518904443572128441559

56663, -94800239427806795189748535420642519

56664, -427107238374035763415700300490728663

56665, -1091721236266493699867603830630900951

56666, -2420949232051409572771410890911245527

56667, -5079405223621241318579025011471934679

56670, -7737861215191073064386639132032623831

56671, -18371685181470400047617095614275380439

56681, -39639333114029054014078008578760893655

70385, -54020887742580753132897945067188362861

70386, -394303254663519216596272552498956574317

70387, -1074867988505396143523021767362492997229

70388, -2435997456189149997376520197089565843053

70389, -5158256391556657705083517056543711534701

70730, -8634353754033685827211085495059347644013

70731, -95746639685793932473834984997592009776749

70736, -139302782651674055797146934748858340843117

70737, -487751926378715042383642532758988989374061

70740, -836201070105756028970138130769119637905005

70741, -2229997645013919975316120522809642232028781

70742, -5017590794830247868008085306890687420276333

70752, -16167963394095559438775944443214868173266541

70764, -46831488042075166258387557068106365243989613

70765, -136034468836197658824530430158699811267911277

70766, -314440430424442643956816176339886703315754605

70767, -671252353600932614221387668702260487411441261

70768, -1384876199953912554750530653427008055602814573

70769, -2812123892659872435808816622876503191985561197

71309, -56358178925565600860139504110612352492802121325

71310, -239045883591928465635600108200147729949793689197

71311, -604421292924654195186521316379218484863776824941

71326, -1746219447089422100033150091938814593969974124141

71330, -4669222721751227936440519757371380633281839210093

71333, -7592225996413033772847889422803946672593704296045

71334, -19284239095060257118477368084534210829841164639853

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Run at your own risk.

Quote:teliotThe following Java program uses an arbitrary precision integer class. It prints out whenever "totalWon" goes further negative than its previous most negative amount. This shows that the progression does not win in the "long run" and in fact loses arbitrarily much. The BigInteger class does not contain a "divide" method that outputs a double, so the house edge is not included in the output. It's enough to show that the player will have an increasingly negative bankroll infinitely often.

Great work!

This forum is really amazing sometimes.

Thanks!

Quote:teliotThe following Java program uses an arbitrary precision integer class. It prints out whenever "totalWon" goes further negative than its previous most negative amount. This shows that the progression does not win in the "long run" and in fact loses arbitrarily much. The BigInteger class does not contain a "divide" method that outputs a double, so the house edge is not included in the output. It's enough to show that the player will have an increasingly negative bankroll infinitely often.

Here is some output:

1, -1

2, -3

3, -7

I don't understand any of that. lol I think there may be a problem in the programming, though.

I think it should be.

1, -1

2, -2

3, -4

Then you double and lose 2 on second wager so your total loss is -3.

Then you double and lose 4 on your third wager, so your total loss is -7.

I assume you won your fourth wager, then being ahead +1 after 4 rolls.

Your bankroll doesn't exceed being down 7 units until round 53, at which point you are down 28 units.

If you don't understand any of it, at least try and understand some of it.

Quote:teliotYou lose 1 unit on first wager, so your total loss is -1.

Then you double and lose 2 on second wager so your total loss is -3.

Then you double and lose 4 on your third wager, so your total loss is -7.

I assume you won your fourth wager, then being ahead +1 after 4 rolls.

Your bankroll doesn't exceed being down 7 units until round 53, at which point you are down 28 units.

If you don't understand any of it, at least try and understand some of it.

Okay, I get it now. So, did the game win outright or lose?

That question doesn't make sense. Sometimes you are ahead, sometimes you are behind. If you create a finite game, limited by a specific number of rolls, then it will lose at exactly the house edge. I can show you that, if you care to fix that number of rolls.Quote:JyBrd0403Okay, I get it now. So, did the game win outright or lose?

If you call it "infinite" then you can only discuss the "limsup" and "liminf". The limsup is bounded by a linear function. The liminf is bounded by an exponential function. That is, about as often as you are up 100, you will be down 2^100 (or thereabouts).

Thanks, I'll go back to college and try and learn more.Quote:JyBrd0403My guess is there's a bug in your programming.

Yes, that's why it's neither winning nor losing if you are considering an "infinite" model. It crosses the line in both directions infinitely many times. Because there is no bound on the wager size, there is no reason to expect the total win/loss to stabilize on one side or the other. Limit the wager size, or limit the number of rounds, it will settle to the H/A. Otherwise, this is the paradox the infinite model gives.Quote:JyBrd0403If it's a losing game you're losing after let's say 100,000 trials. The game is losing and will never go ahead. If it's a winning game at some point, let's say 100,000 trials, you will be winning and never go into the negative again. My guess is there's a bug in your programming.

You don't understand the results. That's not my problem.Quote:The results don't make sense.

So, you're not really making sense.

This program illustrates the oscillations, as well as the linear/exponential ideas I mentioned above.

import java.math.BigInteger;

import java.util.Random;

public class roulettesim {

public static void main(String[] args) {

BigInteger wager = new BigInteger("1");

BigInteger totalWagered = new BigInteger("0");

BigInteger totalWon = new BigInteger("0");

BigInteger ZERO = new BigInteger("0");

BigInteger ONE = new BigInteger("1");

BigInteger TWO = new BigInteger("2");

BigInteger MIN = new BigInteger("0");

BigInteger MAX = new BigInteger("0");

Random generator = new Random();

int roll;

int ctr = 0;

while(true) {

ctr++;

roll = generator.nextInt(37);

totalWagered = totalWagered.add(wager);

if (roll >= 1 && roll <= 18) {

totalWon = totalWon.add(wager);

if (wager.compareTo(ONE) == 1)

wager = wager.divide(TWO);

}

else

if (roll >= 19 && roll <= 36) {

totalWon = totalWon.subtract(wager);

wager = wager.multiply(TWO);

}

else

if (roll == 0)

totalWon = totalWon.subtract(wager);

else

System.exit(1);

if (totalWon.compareTo(MIN) == -1) {

MIN = totalWon;

System.out.println(ctr + ", " + totalWon);

}

if (totalWon.compareTo(MAX.multiply(TWO)) == 1) {

MAX = totalWon;

System.out.println(ctr + ", " + totalWon);

}

}// while

}// main

}// class

Here is a sample run:

1, -1

2, 1

5, 3

9, 7

22, 18

43, 40

67, -60

69, 132

109, 353

113, 737

133, -383

134, 1665

136, -1407

137, -5503

139, 6785

168, 13665

197, -11743

198, -44511

201, 53793

205, 119329

210, -134623

212, 258593

218, 520737

274, -533983

275, 1563169

277, 3136033

350, 6298145

577, 12775074

671, 26037922

694, 53825186

703, -42643806

704, -176861534

705, -445296990

706, -982167902

707, -1519038814

710, 360009378

716, 729108130

778, 1646612130

876, 3350793890

1003, 6788183714

1095, 13956249250

1185, 31383582370

1188, -28745959774

1189, -97465436510

1190, -234904389982

1191, -509782296926

1197, 74333255330

1208, 366391031458

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11822, -6235974108148812005000474390550913606601988

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11828, 4914398491116499565767384745773267146388220

11830, -61987837104475369858839770072171817371553028

11831, 27215143689647122707303103018421628652368636

11832, 71816634086708368990374539563718351664329468

11840, -218093053494189731849589797980710347913415940

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11850, 1922778485564750089737839156193532356660703996

11853, 4777273870976669851854411095092522629426197244

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11934, 242592420668107235033191310754614649729201353468

12143, 485436385507506455732944514884618982348575093500

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12211, -1454996055707040125058361403455060249009792863492

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