Yet Another System that Wins

I'm going to start getting children to beat these games. Here's the thing, especially playing a game like roulette, the only thing you have to beat is the vig. In roulette, it's the zero you have to beat. So, when a zero lands you have to give whatever that bet is to the House. The rest of the time you're at a 50/50 game, which we know that the D'Alembert wins against a 50/50 game. I had to teach these guys that, they swore no game would beat a 50/50 game, and I think they still do. But, it was proven to win in another post here.

So, all you have to do is beat the zero. How can you do that. First of all you have to make the progression 50/50. This got laughed at by all these so called mathematicians on this site. But, here's the thing. In a million trials for a single zero roulette game, if you played a D'Alembert the progression would be down to around 27,000 units. By, rebetting the zero, and keeping the progression 50/50 you would be + or - 300. Get that. The progression might actually be on the plus side this way. So, you pick -27,000 or + or - 300. Can you see it? Big difference, huh? You see, this way the amount of money you are now giving the Casino is guess what, a hell of a lot less.

So, now that you have the progression under control, all you have to do is make enough money to cover what you're giving the Casino. Here's the second thing, with roulette, you know exactly how much you are giving to the casino. Every time a zero comes up that's what you just gave the casino. If you're betting 17 units and a zero comes up you just gave the casino 17 units. With the D'alembert, you make 1 unit for each win, so if the progression stays under 18 units you would always profit because your giving the house less money then you're making with the D'Alembert system. So, what happens if you are betting 30 units and a zero comes up? You would be giving the house 30 units and still only making 18 units. So, what you have to do is D'alembert by 2 units at a time. this way you would give the house 30 units and the D'Alembert would profit 36 units 18*2 = 36 units. So, if you look at it you can see exactly where you have to start betting an extra unit. At 18 you have to start betting a 2 unit spread, at 36 you would spread by 3 units, at 54 by 4 units, at 72 by 5 units etc. Like this

...15-16-17-18-20-22-24-26...36-39-42-45...-54-58-62 etc.

And you just D'alembert like usual. For instance. 36-39-42-39-36-34-32-34-36-39-42-39-36-34 etc.

Simple, huh? And remember to rebet the zero, even after all these so called mathematicians tell you there's no purpose in rebetting the zero. When even a child could reason out why you have to rebet the zero.

In other words, keep working on those betting systems. Most of them work believe it or not.

-Simulated Reality Theory- If the world was a computer, you could do magic. Scientifically speaking.

*blink, blink*

I don't even know what you think you're doing.

Quote: 24Bingo

*blink, blink*

I don't even know what you think you're doing.

Well Bingo, what say you and I cash in are 401k's and hit the roulette table now that we have the holy grail.

Quote: JyBrd0403

So, all you have to do is beat the zero. How can you do that. First of all you have to make the progression 50/50. This got laughed at by all these so called mathematicians on this site. But, here's the thing. In a million trials for a single zero roulette game, if you played a D'Alembert the progression would be down to around 27,000 units. By, rebetting the zero, and keeping the progression 50/50 you would be + or - 300. Get that. The progression might actually be on the plus side this way. So, you pick -27,000 or + or - 300. Can you see it? Big difference, huh? You see, this way the amount of money you are now giving the Casino is guess what, a hell of a lot less.

You need to recheck your setup for this part. Betting the 0 will make you -27,000 per 1,000,000 spins, just like betting a 1:1 bet does. It has the same house edge as any other bet, so betting on 0 can't reduce your losses in the long run.

Quote: rdw4potus

Betting the 0 will . . . It has the same house edge as any other bet, so betting on 0 can't reduce your losses in the long run.

Ssssh!! There are certain "magic numbers" at Roulette. Zero and Seventeen are those magic numbers. Seventeen is considered the "middle" of the layout and corresponds to some theoretical "middle" of the wheel.

It is similar to "East" and "West" as cardinal directions. Lochinvar rides out of the West. Mysticism and mystery appear in the East. Nobody seems to give one whit about North or South. Its similar in Roulette wherein any "system" that involves the "magic" numbers has power to it.

The fact that the wheel and the little ball don't know nuttin' 'bout East or West or Middle or Zero means nothing. Magic numbers have a power that all there own that defies logic or common-sense simply because we endow those magic numbers with that ability to defy logic and common-sense. History and physics lets us look at zillions of spins and we see some of those casinos in Monaco still in business after all those years but we still tell ourselves: Marlboro Men ride out of the West, Mystical Magic appears from the East, Seventeen and Zero are magic numbers at Roulette ... and tonight that wheel is certain to put out!!

It boggles my mind that somebody capable of writing that is at the same time capable of believing that.

(And in the spirit of the OP, I'll leave you guys to figure out what the hell the above sentence means.)

But here's the thing, you're wrong.

Quote: JyBrd0403

In other words, keep working on those betting systems. Most of them work believe it or not.
.

Most of them work. For awhile. Then they quit. For
awhile. Roulette goes in and out, like the tide. At
times it favors a system, at times it doesn't. In the end
you're losing right at the house edge, just like you're
supposed to be.

Quote: EvenBob

Most of them work. For awhile. Then they quit. For
awhile. Roulette goes in and out, like the tide. At
times it favors a system, at times it doesn't. In the end
you're losing right at the house edge, just like you're
supposed to be.

Bob, what are you doing? I thought your chump-wizing days were over?!?

Read original post...furrow brow in confusion. Reread original post. Furrow brow deeper. Wonder why I wasted 10 minutes trying to understand OP in the first place...priceless.

Quote: AcesAndEights

Bob, what are you doing? I thought your chump-wizing days were over?!?

This is just common sense. Most systems do work, when
the conditions favor it. On the roulette boards, the 'experts'
say you need a bunch of systems to switch back and forth
between. The problem is, you don't know whats happening
till its over and by then its too late.

Quote: rdw4potus

You need to recheck your setup for this part. Betting the 0 will make you -27,000 per 1,000,000 spins, just like betting a 1:1 bet does. It has the same house edge as any other bet, so betting on 0 can't reduce your losses in the long run.

You don't bet the zero, you rebet the zero when it comes up. Meaning that if you bet 10 units and a zero lands you rebet the 10 units again. That makes the progression 50/50. So, you won't ever be betting 27,000 units in 1 million spins. Now, feel free to explain how unimportant that is. Everyone here will agree with you and the mathematicians on Yahoo Answers will completely disagree with you. They're not subject to your nonsense they just give the obvious answer, it of course makes a huge difference if your playing the D'alembert to keep the progression 50/50.

edited

Quote: JyBrd0403

Everyone here will agree with you and the mathematicians on Yahoo Answers will completely disagree with you..They're not subject to your nonsense

Why aren't you and Yahoo Answers at the casino
this very minute making millions? Because it
won't work, thats why.

If this actually worked, you wouldn't be here. Of course, if this actually worked the casino also wouldn't be there...

Okay, I think I can translate it: he adheres to something called the "D'Alembert system," which is a more sensible Martingale: when you lose, increase your bet by one unit, and when you win, decrease it by one unit if it's not already at one, so that when you eventually get back to one, for every loss there will have been a win one unit greater. This of course has basically the same problem as the Martingale (lacking infinite time or money, sooner or later you'll have to leave it), but since your stake only increases polynomially, it's hidden better. For this reason, he's convinced it "beats" an even-money game.

The thing is, though, he recognizes that if you're more likely to lose than win, what's likely to happen over time is that you'll just end up betting more and more, getting less and less likely to return to your base bet. He thinks he can get around this by simply not increasing the bet when the loss is a zero, even though he recognizes that then on returning to the base bet, he may not have a unit for every win, since he figures that since you're eighteen times as likely to win as to hit a zero, he's covered as long as his progression hasn't made it to eighteen. So at eighteen, he doubles his units, triple at 36, quadruple at 54, etc. Think that's right.

Of course, the same issue of you not having infinite time or money comes up, except money is now more of an issue now that your bets are increasing exponentially (albeit more slowly than in that other system). Sooner or later, you're going to go broke, and then it won't matter how much you would have had at the end of your progression, because you'll be broke. Even if you have infinite money (in which case, why are you coming up with schemes like this?), sooner or later, there will be a fire, or the dealer will have a heart attack, or a sudden Mormon revival will cause gambling to be banned while you're at the table, and at that random moment, it's quite likely that you'll be in the middle of a progression, and for that reason not unlikely you'll be well down, at least not unlikely enough for your wins to outweigh your losses reliably. And that's before you take into account the variance in your scheme to cover your losses to the zero.

Although I'm curious: say instead of the zero, before each spin, the better chose an arbitrary number on the side of the table he weren't betting, and said "if that comes up, I won't increase my bet (but I will on zero)." Would that make a difference in your world?

Quote: JyBrd0403

You don't bet the zero, you rebet the zero when it comes up. Meaning that if you bet 10 units and a zero lands you rebet the 10 units again. That makes the progression 50/50. So, you won't ever be betting 27,000 units in 1 million spins. Now, feel free to explain how unimportant that is. Everyone here will agree with you and the mathematicians on Yahoo Answers will completely disagree with you. They're not subject to your nonsense they just give the obvious answer, it of course makes a huge difference if your playing the D'alembert to keep the progression 50/50.

So if roulette is 50/50 except for the 0, then you will be even in the long run except for when the 0 hits, and then you'll lose. Then you'll play even for a while and the 0 hits again, and you lose. You can't have something be 50/50 with a condition that sometimes another factor will make you lose. The whole thing is not 50/50 then.

And what if you keep increasing your bet after each loss and then you hit the table maximum and lost that? You can't progress any higher, so you can never win that money back.

Quote: REALITY

Among the various progressive betting systems used for casino table games today, more than a few were developed by some of the greatest scientific minds of the 18th century. Of special note, French mathematicians often took the lead in analysing games of chance. They were fascinated with statistics and probability, particularly in connection with tossing coins and dice.

One of those fascinated French academics was Jean-Baptiste le Rond d’Alembert (1717~1783), whose credentials included honours in physics, philosophy and music theory as well as maths. One of his rather odd (an incorrect) theories was that the probability of a tossed coin landing “tails” would increase for every time that it came up “heads.” He referred to this as the “Law of Equilibrium.”

The mistake d’Alembert made was basing his theory solely upon observation rather than calculations. He had witnessed that whenever two events are equally likely, such as coin flips resulting in heads or tails, that they really do appear to occur in equal number over the long term. Streaks losses seem to be counterbalanced by streaks of wins.

In truth, of course, all coin flips are independent outcomes, unaffected by past results. Nevertheless, based upon his faulty logic, d’Alembert’s system of betting promoted a process of decreasing one’s bet when winning and increasing it when losing, confident that wins and losses would eventually become equal for wagers with odds of 1:1.

The d’Alembert Betting System is therefore most commonly used for “even money” bets, such as Red or Black wagers at the Roulette table, Pass or Don’t Pass wagers at the Craps table and Banker or Player wagers at the Baccarat table. With modification, the system can also be used for Blackjack as well as for sports betting where vigorish must be accounted for.

The objective of the d’Alembert Betting System is to win a single unit in profit, so the player begins by wagering one unit at Evens. Each time a bet succeeds, one unit will be subtracted from the total just wagered and the remainder will be the amount of the next bet. Whenever a bet loses, one unit is added to the total wagered for the next bet. The progression continues until the amount of the next wager becomes zero.

As an example, if £1 is the basic unit and the first bet succeeds, subtracting one unit results in a next wager of zero, so the progression ends with a profit of £1. On the other hand, if the bet loses, the wager is increased by one unit to £2. If it wins, reduce the wager by one unit back to £1. If it loses, increase it one unit to £3. Continue playing in this fashion until the required wager is zero, resulting in a single unit (£1) in profit. Then, the progression begins anew.

One aspect of the d’Alembert Betting System that sets it apart from Martingale is that it does not require risking huge amounts at unfavorable odds in an attempt to recover previously lost wagers. Also, it differs from Labouchere because strings of losses never increase the wager by more than a single unit for the next bet. The d’Alembert Betting System is thus a very slow and methodical approach to wagering, making it a less risky progression than its cousins.

Where the system fails, however, is in its basic premise of Equilibrium. An initial loss might easily be followed by series of wins and losses in equal number, never quite recovering the original wager. Indeed, the progression might never end at all.

ZCore13

"An initial loss might easily be followed by series of wins and
losses in equal number, never quite recovering the original wager."

And that is indeed what happens if you've messed
around with a d'Alem at all. You coast along
nicely and then you get behind a little and you
never get caught up again. Your bets keep getting
bigger and bigger until you run out of money. It
might take a while, but it always happens.

Quote: 24Bingo

Of course, the same issue of you not having infinite time or money comes up, except money is now more of an issue now that your bets are increasing exponentially (albeit more slowly than in that other system). Sooner or later, you're going to go broke, and then it won't matter how much you would have had at the end of your progression, because you'll be broke.

Although I'm curious: say instead of the zero, before each spin, the better chose an arbitrary number on the side of the table he weren't betting, and said "if that comes up, I won't increase my bet (but I will on zero)." Would that make a difference in your world?

I agree, bankroll is always a question. Even if you have a game that is 51% in your favor, you still need an appropriate bankroll to beat the game. You can't sit down at a $5 table with a $10 bankroll and expect to beat the game. Time is also an issue as if you play the D'alembert it can be exceedingly boring. The big disagreement is whether you will go broke or not. If you look at the D'Alembert vs. a 50/50 game, after a million trials you may be at -300 on the progression, but you still made a huge profit. In other words, the you don't have to bring the progression all the way back down to a 1 unit bet to win. You still are profitable even when you're at -300. The point being, it takes time for the progression to go to -300, around a million trials or so, during that time you're making loads of money. So, yes by increasing the bet size you are increasing the downside potential, but the profit is also increasing. Again, this is where keeping the progression under control is necessary. If you don't rebet the zero, the progression would get away from you really quickly.

As for there being a fire or the dealer having a heart attack, this could happen, but again you are in control of the game, you pick up and go to another table and start where you left off. So you don't have to quit when you're in the red, you can always just quit when you're in the black.

In my world, it makes no difference which number you decide to rebet, as long as it is a number on the side you weren't betting. It's basically 19 numbers vs. 18 numbers you can rebet any of the 19 numbers you're not betting. It's just easier for me to see that Green Zero come up.

Quote: EvenBob

"An initial loss might easily be followed by series of wins and
losses in equal number, never quite recovering the original wager."

And that is indeed what happens if you've messed
around with a d'Alem at all. You coast along
nicely and then you get behind a little and you
never get caught up again. Your bets keep getting
bigger and bigger until you run out of money. It
might take a while, but it always happens.

Not true at all on a 50/50 game. On a 50/50 game you just sit there going up and down, up and down. The only thing that causes you to get behind, and stay behind until you run out of money is the house edge (the zero).

Quote: JyBrd0403

Not true at all on a 50/50 game. On a 50/50 game you just sit there going up and down, up and down. The only thing that causes you to get behind, and stay behind until you run out of money is the house edge (the zero).

cool. And since theres a house edge on the bet on the 0, putting money on it cant help you mitigate the overall house edge on the game.

Quote: rdw4potus

cool. And since theres a house edge on the bet on the 0, putting money on it cant help you mitigate the overall house edge on the game.

Not sure what you mean. You're not placing a bet on the zero.

Quote: JyBrd0403

Not sure what you mean. You're not placing a bet on the zero.

lol. fine. using the occurrence of a 0 as a reason to manipulate your bets does nothing to mitigate the house edge on any of the bets you're placing. Without reducing the house edge, the system fails.

Quote: Zcore13

So if roulette is 50/50 except for the 0, then you will be even in the long run except for when the 0 hits,

And what if you keep increasing your bet after each loss and then you hit the table maximum and lost that? You can't progress any higher, so you can never win that money back.

ZCore13

Not true if roulette is 50/50 except for the 0, and you play the D'Alembert system you will be Winning in the long run except for when the 0 hits. You just have to win more then you give the house when a 0 hits, but you're not breaking even on the 50/50, you're winning on the 50/50.

Again the table maximum doesn't matter. You can always go to another table and pick up where you left off. It's a non factor.

Quote: JyBrd0403

Not true at all on a 50/50 game. On a 50/50 game you just sit there going up and down, up and down. The only thing that causes you to get behind, and stay behind until you run out of money is the house edge (the zero).

Obviously you've never played in real time. You do NOT
keep going up and down, up and down, that would be
the ideal scenario. What happens is, you hit a bad sequence,
and you never recover from it. You get back to being almost
even, only to fall even farther behind. Soon the bets are
so large you run out of money. Don't take my word for,
do it yourself. I messed with this progression for years, the
only way to get ahead is to have a good bet selection. And
then you wouldn't need it.

Quote: EvenBob

Obviously you've never played in real time. You do NOT
keep going up and down, up and down, that would be
the ideal scenario. What happens is, you hit a bad sequence,
and you never recover from it. You get back to being almost
even, only to fall even farther behind. Soon the bets are
so large you run out of money. Don't take my word for,
do it yourself. I messed with this progression for years, the
only way to get ahead is to have a good bet selection. And
then you wouldn't need it.

But, Bob, are we talking a straight 50/50 game or a HE game. The D'alembert on a 50/50 game of chance wins. That's on another post here. I can look it up for you, but the D'alembert on a 50/50 game wins. You can't change that. So, I'm not sure what your point is. If you have a bad sequence let's say that takes you to betting 50 units, you will either start winning or just sort of sit there at around 50 units. Either way your going make a profit. Now, this is where my lack of advance mathematics fails me, but there is a reason the 50/50 game doesn't just run off to down 27,000 units. I don't know what the advance mathematical terms are for that, but I do know the 50/50 game just sits there between + and - 300 or so in a million trials. The progression just sits there and you keep making you're 1 unit profits for a million trials.

Quote: JyBrd0403

But, Bob, are we talking a straight 50/50 game or a HE game. The D'alembert on a 50/50 game of chance wins. That's on another post here.

First, no, it doesn't win. It breaks even. Given unlimited time and unlimited money, with 0% HE all systems break even. No betting system can alter the house edge, so that's the only possible outcome.

Second, this isn't a 50/50 game. Every bet you're making is exposed to the house edge, and your system can't possibly change that fact. So, over time, you will lose money.

Quote: JyBrd0403

The progression just sits there and you keep making you're 1 unit profits for a million trials.

This isn't true. The longer you play, the further out, overall, your progression will have been. This is roughly proportional to the square root, so since the amount you're stuck for at any given time is quadratic in it (since you've bet 1 + 2 + 3 + 4...), they cancel each other out.

EDIT: Overall, not on average, sorry.

Quote: rdw4potus

First, no, it doesn't win. It breaks even. Given unlimited time and unlimited money, with 0% HE all systems break even. No betting system can alter the house edge, so that's the only possible outcome.

Second, this isn't a 50/50 game. Every bet you're making is exposed to the house edge, and your system can't possibly change that fact. So, over time, you will lose money.

Hey, a trick question. Given unlimited time you would lose an infinite amount in a row at some point during that infinite timeline, therefore draining any infinite bankroll. I think that's called singularity. But in a million or a billion trials, the game wins. Popcan did a simulation on a billion coin flips showing the D'Alembert winning approx 500 million which is what I would expect it to win. So, yes indeed it does win, in a specified finite period of time. Let's say a 100 years or so.

No, rebetting the zero does not cause zero's to stop coming up 1/37 of the time. But, it does keep the progression at 50/50 and under my control.

Quote: JyBrd0403

So, yes indeed it does win, in a specified finite period of time. Let's say a 100 years or so.

No, rebetting the zero does not cause zero's to stop coming up 1/37 of the time. But, it does keep the progression at 50/50 and under my control.

No, and no.

It might win on some billion-coin trials, but it will lose on some as well. Overall, you should expect your dollars won on a 50/50 game to be $0.

How does rebetting on a 0 keep the progression at 50/50? You're still more likely to lose than win on every bet you place, because you haven't done anything to tackle the house edge on each spin.

Quote: rdw4potus

No, and no.

It might win on some billion-coin trials, but it will lose on some as well. Overall, you should expect your dollars won on a 50/50 game to be $0.

How does rebetting on a 0 keep the progression at 50/50? You're still more likely to lose than win on every bet you place, because you haven't done anything to tackle the house edge on each spin.

Absolutely not. You seem to think that it's possible for you to be -27,000 on a 50/50 game in a million trials. On a 50/50 there's a range of possilbe outcomes. It's basically + or - 300 on average. It's called the Standard Deviation. IF you get a fair 50/50 game the D'Alembert will ALWAYS win there's nothing to cause the betting to outpace the profits.

The trick question (unlimited time) negates this.

Rebetting the zero keeps the progression itself at 50/50. In other words the betting is only subject to the + or - 300 that you get with a 50/50 game, not the -27,000 you would get with a 48.5% chance of winning. You can control the progression, not the house edge.

Quote: JyBrd0403

Absolutely not. You seem to think that it's possible for you to be -27,000 on a 50/50 game in a million trials. On a 50/50 there's a range of possilbe outcomes. It's basically + or - 300 on average. It's called the Standard Deviation. IF you get a fair 50/50 game the D'Alembert will ALWAYS win there's nothing to cause the betting to outpace the profits.

You really need to look up standard deviation. Like, alot. Let's say +-300 is the first standard deviation from the mean ($0) on a 50/50 game. I think it's much wider than that over 1,000,000 runs of 1,000,000 trials, but let's say 300 is right. The first standard deviation accounts for about 67% of all possible outcomes. It does not guarantee that -27,000 cannot happen. In fact, -27,000 is a possible, though highly unlikely, outcome.

The catch is that at this point the amount you'll be betting will be equal to the number of losses minus the number of wins that were not sustained when you were already at your base bet. Obviously, the average difference between the total losses and the total wins is zero and, not by coincidence, the expected number of times you'll have been at your base bet and won is proportional to the square root of number trials. Since you will have bet every number from one to this number, the amount you'll have won on average will be equal to the total number of wins (obviously, roughly half of all trials) minus (n^2+n)/2, where n is your bet... oh dear. Yep, you can see where this is going.

Incidentally, the standard deviation - just of trials - over a million trials of a coin flip is 1000 up or down, not 300. (Variance is mean of square 1 minus square of mean 0 = 1, multiply by a million and take the square root to get 1000.)

EDIT: Right, I was thinking of it as a Bernoulli trial (1/0, mean 0.5, var 0.25), taking into account that it's +1/-1 this is a habit (idiot) math corrected.

Quote: JyBrd0403

But, Bob, are we talking a straight 50/50 game or a HE game. The D'alembert on a 50/50 game of chance wins. That's on another post here. I can look it up for you, but the D'alembert on a 50/50 game wins.
.

Yes, thats right. IF and thats IF you have an unlimited
amount of money. So it wins eventually, so what. You
have to play in the real world with a real BR and real
casino limits. You'll have a losing streak where you
get far behind with huge bets, and if you play long
enough it will right itself. On paper, in theory. Not in a
real casino with real money. And thats with no zeros
involved.

Quote: 24Bingo

Since you will have bet every number from one to this number, the amount you'll have won on average will be equal to the total number of wins (obviously, roughly half of all trials) minus (n^2+n)/2, where n is your bet... oh dear. Yep, you can see where this is going.

Incidentally, the standard deviation - just of trials - over a million trials of a coin flip is 1000 up or down, not 300. (Variance is mean of square 1 minus square of mean 0 = 1, multiply by a million and take the square root to get 1000.)

EDIT: Right, I was thinking of it as a Bernoulli trial (1/0, mean 0.5, var 0.25), taking into account that it's +1/-1 this is a habit (idiot) math corrected.

So, -27,000 on a 50/50 game is ridiculous. You're buddy just said that that's a possibility and that's 1 reason the D'Alembert won't win. Did you hear that? Mind telling that guy that's nonsense. Guys this one has been done D'Alembert produces a massive win on a billion trials. I'll give you the .000000000000000001% chance that you could actually produce a loss on a million trials. But, that means 99.9999999999999999% of the time it wins. I just like to say the D'alembert wins on a 50/50 game of chance. Hey, you lose a million in a row and you'd lose too.

This is getting as silly as it always gets around here. Now, if I ask a real mathematician, what do you think they'll say about this nonsense.

Quote: EvenBob

So it wins eventually, so what. You
have to play in the real world with a real BR and real
casino limits. You'll have a losing streak where you
get far behind with huge bets, and if you play long
enough it will right itself. On paper, in theory. Not in a
real casino with real money. And thats with no zeros
involved.

Thank you, Bob. We now agree in principle. Have you notice we are now at odds with these guys who say that it doesn't win. WTF is that all about.

Quote: JyBrd0403

Thank you, Bob. We now agree in principle. Have you notice we are now at odds with these guys who say that it doesn't win. WTF is that all about.

They only care about it winning in real play. That
it wins on paper is useless. A Martingale wins on
paper just fine.

Quote: JyBrd0403

So, -27,000 on a 50/50 game is ridiculous. You're buddy just said that that's a possibility and that's 1 reason the D'Alembert won't win. Did you hear that? Mind telling that guy that's nonsense. Guys this one has been done D'Alembert produces a massive win on a billion trials. I'll give you the .000000000000000001% chance that you could actually produce a loss on a million trials. But, that means 99.9999999999999999% of the time it wins. I just like to say the D'alembert wins on a 50/50 game of chance. Hey, you lose a million in a row and you'd lose too.

This is getting as silly as it always gets around here. Now, if I ask a real mathematician, what do you think they'll say about this nonsense.

It is both ridiculous and possible. They aren't mutually exclusive. It's possible for me to win the Mega Millions tomorrow and the Powerball on Wednesday. The odds are so long that it's a ridiculous idea, but that doesn't mean it can't happen.

Now, one idea that is both ridiculous and impossible is creating a betting system that overcomes the house edge in roulette over the long run.

Quote: rdw4potus

Now, one idea that is both ridiculous and impossible is creating a betting system that overcomes the house edge in roulette over the long run.

Note that I never said a d'Alem beats the HE. Just that
it beats 50/50 in the long run on paper. Just as the
Marty does.

I gotta ask…

JyBrd, your entire existence here (almost completely) has been arguing about 50/50 chances, the D’Alembert, and whether .999…= 1.

Do you just enjoy challenging math-based beliefs?

Quote: EvenBob

Note that I never said a d'Alem beats the HE. Just that
it beats 50/50 in the long run on paper. Just as the
Marty does.

OK. I will let this go and ridicule you for other reasons. Like...hmmm...your state of residency? :-P

Quote: Face

I gotta ask…

JyBrd, your entire existence here (almost completely) has been arguing about 50/50 chances, the D’Alembert, and whether .999…= 1.

Do you just enjoy challenging math-based beliefs?

Let me put it this way. The thing I love about numbers is that they are 100%, you can't argue with it. 2 is 2, 3 is 3 etc. What I like about art is that it's all subjective. Put the number .999... on a canvas and hang it in a museum, and you can argue that it's actually a choo choo train, and nobody can say anything about it. As a matter of fact, if you smoke the right amount, you can start seeing a choo choo train. That's Art. When you start arguing that a game that wins 500 million units after a billion trials is a losing game, that's harder for me to swallow. Just trying to establish that point, so we can move on to beating the HE game. As you can see, if they give me that point, they already know the HE game is beatable. So, they refuse to give me the point. Or, they give it to you and then take it back. You can't do that with math. That's art.

Quote: JyBrd0403

Let me put it this way. The thing I love about numbers is that they are 100%, you can't argue with it. 2 is 2, 3 is 3 etc. What I like about art is that it's all subjective. Put the number .999... on a canvas and hang it in a museum, and you can argue that it's actually a choo choo train, and nobody can say anything about it. As a matter of fact, if you smoke the right amount, you can start seeing a choo choo train. That's Art. When you start arguing that a game that wins 500 million units after a billion trials is a losing game, that's harder for me to swallow. Just trying to establish that point, so we can move on to beating the HE game. As you can see, if they give me that point, they already know the HE game is beatable. So, they refuse to give me the point. Or, they give it to you and then take it back. You can't do that with math. That's art.

Yep, math is absolute. And 48.65/51.35 is not equal to 50/50. In roulette every bet you make has a 48.65% chance of winning. Your system attempts to overcome that by making additional bets with different dollar values. But they all also have a 48.65% chance of success. It's not hard to see how and why this will fail.

Quote: rdw4potus

Yep, math is absolute. And 48.65/51.35 is not equal to 50/50. In roulette every bet you make has a 48.65% chance of winning. Your system attempts to overcome that by making additional bets with different dollar values. But they all also have a 48.65% chance of success. It's not hard to see how and why this will fail.

Now we're back to an HE game. I thought you just said it fails because the D'alembert doesn't even beat a 50/50 game?

Quote: JyBrd0403

Now we're back to an HE game. I thought you just said it fails because the D'alembert doesn't even beat a 50/50 game?

I think it breaks even in a 50/50 game. Every bet has even odds of winning and losing in a 50/50 game. .5*infinity+(-.5*infinity) is 0.

But you've come here to show a system that you claim can beat roulette. Roulette is not a 50/50 game, and it's silly to pretend that it is.

Quote: rdw4potus

.5*infinity+(-.5*infinity) is 0.

Ah, that was my original point. The wins and losses will even out at some point in a 50/50 game. Then we had to argue standard deviation, and how you could lose a billion in a row etc. When the whole point is the wins and losses will even out at some point. That makes the D'alembert a huge winner on a 50/50 game. You're making 1 unit every win lose 18 win 19 win 1 unit. Only way the game loses is if there's a HE pushing the betting to INfinity.

Quote: JyBrd0403

... When the whole point is the wins and losses will even out at some point. That makes the D'alembert a huge winner on a 50/50 game. ...

Yes, I think this summarizes JyBrd's position: the wins and losses even out, so you are a huge winner. (Is there an appropriate smiley to post here?)

I suppose the strategy has to be to bet bigger when the next spin is going to be a winner and bet smaller when the next spin is going to lose.


I suspect that there is no one who can actually believe the arguments being presented in the OP of this thread and that once again we have someone who just likes to make outrageous statements to see how much he/she can irritate/stir up the rational-thinking set of folks here, knowing that they can't stand to ignore irrational gibberish. It's been tried a few times before, even by this same poster, I think.

Quote: Doc

Yes, I think this summarizes JyBrd's position: the wins and losses even out, so you are a huge winner. (Is there an appropriate smiley to post here?).

I'm sorry, doc. I forgot you can't take a pen and pencil and go 1-2-3-4-5-4-3-2-1 and figure out how much you won there. As a matter of fact I know you can't. That's why we're sitting here.

Quote: JyBrd0403

I'm sorry, doc. I forgot you can't take a pen and pencil and go 1-2-3-4-5-4-3-2-1 and figure out how much you won there. As a matter of fact I know you can't. That's why we're sitting here.

LOL. that's an odd number of events. How is that proof that things even out? We can all cherry pick results, and we can all make half-assed slanted arguments. The difference here is that you're the only person in this conversation who is actively trying to do those things.

Quote: JyBrd0403

You're making 1 unit every win lose 18 win 19 win 1 unit.

I agree. But the expectation is to win 18 and lose 18, win 0 units.

Quote: JyBrd0403

Only way the game loses is if there's a HE pushing the betting to INfinity.

Doesn't the game lose if you lose 19 and win 18? You're -1...

The house edge just makes it worse. Much, much worse. speaking of...let's get back to your original point. How does this system that exposes you to a house edge on every bet help you overcome the house edge on a series of bets?