P.S. This came up in another thread. But, I never did find a formula for the D'Alembert.

wonderful. If you don't, it sucks. Just like every

other progression.

Quote:EvenBobIt's simple. If you have an edge, the d'Alem is

wonderful. If you don't, it sucks. Just like every

other progression.

The D'Alembert is the "Edge" on a 50/50, Bob. But, I was looking for a mathematical formula for calculating the EV for the D'alembert, do you have one?

PROOF?Quote:JyBrd0403The D'Alembert is the "Edge" on a 50/50, Bob.

Quote:AxelWolfPROOF?

wLwLWLWL

wwLLWWLL

wwwwLLLL

wwwwWWWW

LWLWLWLL

LLWWLLWL

LLLLWWWW

LLLL L L L L

That's the possible outcomes for 4 in a row. That's 64 trials, 32 wins/32 losses. How's the D'Alembert doing there, is it winning? How about the Marty, breaking about even? That's the "proof" that the D'Alembert beat a 50/50 game. You can do that for how ever many trials you want. D'Alembert still wins, Marty still breaks even. Or you can look at another thread where someone did a simulation for a million trials.

But, again, I'm just looking for a mathematical formula for the EV on the D'alembert. Not even sure if one exists at this point.

Your EV is simply player's edge multiplied by the total $ in action.

Spit up coke all over.Quote:RSThat's an awesome proof.

Quote:RSThat's an awesome proof.

Your EV is simply player's edge multiplied by the total $ in action.

Really, that simple, huh? Well, here's the proof for the martingale breaking even.

E[P]=∑k=1Napk+(−SN)(1−p)N=a[1−(1−p)N−(1−p)N(2N−1)]=a[1−(2(1−p))N]

Probably, pretty close to what you're talking about, but, again I was looking for a mathematical formula, not misguided sarcasm.

Quote:AxelWolfSpit up coke all over.

2+2=4

@@+ @@=4

Proof. @@@@ = 4

What's your proof? 2+2=4 and my daddy told me so, so it's true?

Martingale is break-even because it is being applied to a game with 0% house edge. It is not the system that determines the EV, it's the HE that determines the EV. Some systems will just experience more variance than others (ie: a 1-10-100-1000-10000 player will feel some variance, while the 1-1-1-1-1-1 bettor will feel very little variance).

Quote:RSThis one's got some potential. Spunky, I like that.

Martingale is break-even because it is being applied to a game with 0% house edge. It is not the system that determines the EV, it's the HE that determines the EV. Some systems will just experience more variance than others (ie: a 1-10-100-1000-10000 player will feel some variance, while the 1-1-1-1-1-1 bettor will feel very little variance).

Spunky, spoken like someone with control of the reigns, someone who doesn't know that, that's a dangerous position to be in, no? You know, like, the loftier the dream, the deeper the crater, Caesar and all :)

So, you showed the 1-10-100-1000-10000, shows a lot of variance, you also showed flat betting, kinda missed the D'alembert, though. How's that do on 0% house edge?

Sorry, so you can understand, the worse the system, the worse the EV. If I bet 1-10000-1000000-100000000-1000000000, that's worse than a marty, what if I bet a much better system, say a D'alembert, 1-2-3-4-3-4-3-4-3-2-1-2-3-2-1, How did I do there? Better system, better results, right.

But, for the 5th time, I'm just looking for a mathematical equation for the EV on the D'Alembert, that's all. If you've got one please feel free to share it, if you don't think one can ever be calculated, feel free to share that as well.

Have you found a game that has a 50/50?Quote:JyBrd0403The D'Alembert is the "Edge" on a 50/50,

does this helphttp://mathworld.wolfram.com/dAlembertsSolution.html

Quote:AxelWolfHave you found a game that has a 50/50?

Sure, a coin flip, why? You trying to make money on a coin flip? I used to know a guy... What's this got to do with a mathematical equation for the EV on A D'Alembert?

Quote:JyBrd0403Spunky, spoken like someone with control of the reigns, someone who doesn't know that, that's a dangerous position to be in, no? You know, like, the loftier the dream, the deeper the crater, Caesar and all :)

So, you showed the 1-10-100-1000-10000, shows a lot of variance, you also showed flat betting, kinda missed the D'alembert, though. How's that do on 0% house edge?

Sorry, so you can understand, the worse the system, the worse the EV. If I bet 1-10000-1000000-100000000-1000000000, that's worse than a marty, what if I bet a much better system, say a D'alembert, 1-2-3-4-3-4-3-4-3-2-1-2-3-2-1, How did I do there? Better system, better results, right.

But, for the 5th time, I'm just looking for a mathematical equation for the EV on the D'Alembert, that's all. If you've got one please feel free to share it, if you don't think one can ever be calculated, feel free to share that as well.

Total action / rounds = average bet

(1+2+3+4+3+4+3+4+3+2+1+2+3+2+1)/15 = 2.533

EV = average bet * edge

EV = 2.5333 * 0.00

EV = 0.0

Would you like to show me a another example? As many times as I've tried, I've never been able to multiply a number by 0 and get anything other than 0.

LLLLLLLLLLLLLLLLLLLLLLLLLLLLL What now? Lest add a W then LLLLLLLLLLLLLLLLLLLLLLLLLLLL................Quote:JyBrd0403wLwLWLWL

wwLLWWLL

wwwwLLLL

wwwwWWWW

LWLWLWLL

LLWWLLWL

LLLLWWWW

LLLL L L L L

That's the possible outcomes for 4 in a row. That's 64 trials, 32 wins/32 losses. How's the D'Alembert doing there, is it winning? How about the Marty, breaking about even? That's the "proof" that the D'Alembert beat a 50/50 game. You can do that for how ever many trials you want. D'Alembert still wins, Marty still breaks even. Or you can look at another thread where someone did a simulation for a million trials.

But, again, I'm just looking for a mathematical formula for the EV on the D'alembert. Not even sure if one exists at this point.

Quote:RSTotal action / rounds = average bet

(1+2+3+4+3+4+3+4+3+2+1+2+3+2+1)/15 = 2.533

EV = average bet * edge

EV = 2.5333 * 0.00

EV = 0.0

Would you like to show me a another example? As many times as I've tried, I've never been able to multiply a number by 0 and get anything other than 0.

Kiddo, what I've been trying to tell you, and if one of these math gods attempts the EV for the D'Alembert ,will explain to you, is that average bet * edge, is complete and utter nonsense. Sorry, you've been hoodwinked, bamboozled, etc. You're missing all the complexities of mathematical probabilities, like, if I lose 1 then bet 2 and win, I just won 1 unit, as opposed to average bet of 3*0=0. I'm kinda looking for the formula that cannot be unproven within the first 10 seconds, if that helps any.

?Quote:AxelWolfLLLLLLLLLLLLLLLLLLLLLLLLLLLLL What now? Lest add a W then LLLLLLLLLLLLLLLLLLLLLLLLLLLL................

Quote:AxelWolf?

Not sure what you're going for here, but is this a fair game? or do you need to buy a revolver? Losing 25 in a row, followed by a win and then losing 25 in a row again, think you're getting cheated, if not, no worries, you'll win 25 in a row and lose 1 then win 25 in a row again in pretty short order. Making the D'alembert a winner again.

what about 30 losses in a row?Quote:JyBrd0403Not sure what you're going for here, but is this a fair game? or do you need to buy a revolver? Losing 25 in a row, followed by a win and then losing 25 in a row again, think you're getting cheated, if not, no worries, you'll win 25 in a row and lose 1 then win 25 in a row again in pretty short order. Making the D'alembert a winner again.

Quote:AxelWolfwhat about 30 losses in a row?

Fair game? or Cheated - non 50/50 game?

Blackjack basic basic strategyQuote:JyBrd0403Fair game? or Cheated - non 50/50 game?

Quote:AxelWolfBlackjack basic basic strategy, fair game

Quote:JyBrd0403Not sure what you're going for here, but is this a fair game? or do you need to buy a revolver? Losing 25 in a row, followed by a win and then losing 25 in a row again, think you're getting cheated, if not, no worries, you'll win 25 in a row and lose 1 then win 25 in a row again in pretty short order. Making the D'alembert a winner again.

Let's say we ran into 2 sessions similar to the 25 L, 1 W, 25 L....except I'll break it down to 5L 1W 5L. The second session is in reverse (5W, 1L, 5W).

Base unit: 10

5 Losses, 1 Win, 5 Losses:

-10 -11 -12 -13 -15 +16 -15 -16 -17 -19 -20

5 Wins, 1 Loss, 5 Wins:

+10 +9 +8 +7 +6 -5 +5 +4 +3 +2 +1

First: -132

Second: +50

Quote:AxelWolfBlackjack basic basic strategy

I think you would expect a 28 in a row or 29 win in a row vs. a 30 in a row loss then. Something like losing 30 in a row would occur 1 in 100,000,000 and winning 29 in a row would occur just as often 1 in 100,000,000. Something like that, right? If you can get out before those 30vs29 losses add up you might win with a d'alembert. Stop if you get ahead. 50/50 you don't have to worry about it.

Quote:RSQuote:JyBrd0403Not sure what you're going for here, but is this a fair game? or do you need to buy a revolver? Losing 25 in a row, followed by a win and then losing 25 in a row again, think you're getting cheated, if not, no worries, you'll win 25 in a row and lose 1 then win 25 in a row again in pretty short order. Making the D'alembert a winner again.

Let's say we ran into 2 sessions similar to the 25 L, 1 W, 25 L....except I'll break it down to 5L 1W 5L. The second session is in reverse (5W, 1L, 5W).

Base unit: 10

5 Losses, 1 Win, 5 Losses:

-10 -11 -12 -13 -15 +16 -15 -16 -17 -19 -20

5 Wins, 1 Loss, 5 Wins:

+10 +9 +8 +7 +6 -5 +5 +4 +3 +2 +1

First: -132

Second: +50

That's not a D'Alembert, that's whatever you're doing. D'alembert would win 21-20-19-18-17 etc. not 10-9-8-7-6 etc.

Would never try the D'Alembert with black jack, sure path to ruin because of the 47% win rate.

What would be interesting, if someone could program this, playing Banker in Bac, and somehow adding in the 5% commission.

According to ACE he lost 30 in a row. Assuming he didn't make a mistake, you can see where one can lose his ass on any system. There is nothing to say losing 40 in a row cant happen. Improbable? Yes I'm sure someone can calculate what the maximum amount of losses in a row has occurred.Quote:JyBrd0403I think you would expect a 28 in a row or 29 win in a row vs. a 30 in a row loss then. Something like losing 30 in a row would occur 1 in 100,000,000 and winning 29 in a row would occur just as often 1 in 100,000,000. Something like that, right? If you can get out before those 30vs29 losses add up you might win with a d'alembert. Stop if you get ahead. 50/50 you don't have to worry about it.

Quote:JyBrd0403Kiddo, what I've been trying to tell you, and if one of these math gods attempts the EV for the D'Alembert ,will explain to you, is that average bet * edge, is complete and utter nonsense. Sorry, you've been hoodwinked, bamboozled, etc. You're missing all the complexities of mathematical probabilities, like, if I lose 1 then bet 2 and win, I just won 1 unit, as opposed to average bet of 3*0=0. I'm kinda looking for the formula that cannot be unproven within the first 10 seconds, if that helps any.

You are getting expected value and actual value confused here. The expected vale over two bets for three units is 0 in this case. The actual value, as say, for the results you got is 1.

Are you looking for the EV of the complete D'alembert run, rather than the individual bets? You will need a loss limit, or you'll have problems with infinity or limits.

(The EV of a complete Marty run is calcuable. I am not sure the D'alembert is)

Quote:JyBrd0403The D'Alembert is the "Edge" on a 50/50, Bob.

Ya, it's a miracle worker if you can play

50/50 in the short term. But nobody

does, they get 8 out of 10 wrong at

times and there goes the progression.

Quote:thecesspit(The EV of a complete Marty run is calcuable. I am not sure the D'alembert is)

Thanks, Cess. That's what I've been running into, people saying the EV for the D'Alembert would be too complicated to do. I was hoping someone just had a formula available, like for the marty, but, noone seems to be able to calculate the EV for the D'Alembert. Kinda odd really, the D'Alembert has been around forever, and it was developed by a mathematician, you'd think someone would have made a formula for it, but maybe it's just not calculable.

Quote:JyBrd0403noone seems to be able to calculate the EV for the D'Alembert..

The EV is negative, just like with all progressions.

Unless you have the edge, then it will be positive.

All a progression does is forestall the inevitable,

that you will lose. If you have the edge, it enhances

your positive EV.

Just don't bet the 0-00-1-2-3 combination (7.89% edge).

Quote:JyBrd0403I've been looking for a formula to calculate the EV for the D'Alembert Progression.

For a 50/50, the EV is zero.

Proof by induction:

Let N be the number of games played.

If N = 1, EV = 1/2 x (+1) + 1/2 x (-1) = 0.

Assume the EV = 0 for N = some number X.

The EV for X+1 = the EV for X + the EV for the next game, which is always 0 regardless of the bet size, so the EV for X+1 = 0 + 0 = 0.

Thus, since it is true for N = 1 and (if it is true for N = X, then it is true for N = X + 1), it is true for all positive integers N.

Quote:WizardThis has hopefully been said already but the house edge on the D-Alembert, and every other betting system, is exactly the same as the house edge of the game being played. For example, every possible method of betting on double-zero roulette has a house edge of 5.26%. It is mathematically impossible to dent that, in either direction.

Just don't bet the 0-00-1-2-3 combination (7.89% edge).

Yes, but do you know of a formula that calculates the EV on the D'Alembert? I think it's possible to have a negative or zero HE, and a positive EV, no? I find formulas for the Marty, but nothing for the D'Alembert. Do you know if it is even possible to calculate?

Here's the formula I found for the Marty. They seem to have it readily available. Nothing for the D'Alembert, though.

E[P]=∑k=1Napk+(−SN)(1−p)N=a[1−(1−p)N−(1−p)N(2N−1)]=a[1−(2(1−p))N]

(And also, what do those parameters represent, exactly...?)

But that's sort of a waste of time because I can tell you right now that on the 50/50 progression you insist you have beat, the EV is exactly 0, since the EV of every bet you will make is 0. Further I can tell you that on roulette, with one or more zeroes, the EV will be negative, since every bet you make will have a negative EV, and EV always adds. Yes, even if the events aren't independent.

Quote:JyBrd0403Here's the formula I found for the Marty. They seem to have it readily available. Nothing for the D'Alembert, though.

E[P]=∑k=1Napk+(−SN)(1−p)N=a[1−(1−p)N−(1−p)N(2N−1)]=a[1−(2(1−p))N]

Could you define what the terms (k, N, a p, S) mean? Also, I think that some of those terms (especially a lot of the times N appears) should be exponents.

For a Martingale where you stop after a win or after N consecutive losses, if the probability of a win is p, the EV is 1 - (2(1-p))

^{N}.

D'Alembert EV can probably be calculated, if the conditions of stopping were known - for example, "Stop if the first game is a win, or when the number of wins = the number of losses, or when the amount behind is 100 or more." What conditions do you want applied?

Quote:ThatDonGuyD'Alembert EV can probably be calculated, if the conditions of stopping were known - for example, "Stop if the first game is a win, or when the number of wins = the number of losses, or when the amount behind is 100 or more." What conditions do you want applied?

Yeah, that would be fine, stop if the first game is a win, also, stop when wins = losses. If it can be calculated that would be great.

I'm not sure what that Marty formula is, I was looking around for a D'Alembert EV formula, and all I could find was the marty formula LOL. Couldn't tell you if it's accurate or not.

Quote:24BingoBy that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

Quote:JyBrd0403Yes, but do you know of a formula that calculates the EV on the D'Alembert?

Yes: house edge = 2/38.

Quote:WizardYes: house edge = 2/38.

I'll take that as a no. I don't get it, I've asked around and can't find any formula for it, weird.

Quote:JyBrd0403Quote:24BingoBy that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

Two problems with that statement.

First, "well before infinity" is pretty much meaningless, given that infinity is very, very, very, veryveryvery, very, did I mention very, very large. "You'd reach equal number of wins and losses" may take years.

Second, you would also reach a loss limit well before infinity.

For limited plays, I can show that EV is zero. For example, if you stop after three plays, there are 8 possible results:

WWW: +3

WWL: +1

WLW: +2 (+1, -1, +2)

WLL: -2 (+1, -1, -2)

LWW: +2 (-1,+2, +1)

LWL: 0 (-1, +2, -1)

LLW: 0 (-1, -2, +3)

LLL: -6 (-1, -2, -3)

In a 50/50 game, each is equally likely, so the EV = 1/8 x (the sum of the 8 values) = 0.

I am trying to see if I can work out an EV given the conditions (win the first game or eventually reach wins = losses) over certain loss limits (as well as one with infinite bankroll and time), but there are quite a few states to consider. For example, if you lose your first 2 bets, you are at 2 more losses than wins and bankroll -3, but a win and loss (in either order) results in still having 2 more losses than wins but bankroll -2.

Quote:ThatDonGuyQuote:JyBrd0403Quote:24BingoBy that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

Two problems with that statement.

First, "well before infinity" is pretty much meaningless, given that infinity is very, very, very, veryveryvery, very, did I mention very, very large. "You'd reach equal number of wins and losses" may take years.

Second, you would also reach a loss limit well before infinity.

For limited plays, I can show that EV is zero. For example, if you stop after three plays, there are 8 possible results:

WWW: +3

WWL: +1

WLW: +2 (+1, -1, +2)

WLL: -2 (+1, -1, -2)

LWW: +2 (-1,+2, +1)

LWL: 0 (-1, +2, -1)

LLW: 0 (-1, -2, +3)

LLL: -6 (-1, -2, -3)

In a 50/50 game, each is equally likely, so the EV = 1/8 x (the sum of the 8 values) = 0.

I am trying to see if I can work out an EV given the conditions (win the first game or eventually reach wins = losses) over certain loss limits (as well as one with infinite bankroll and time), but there are quite a few states to consider. For example, if you lose your first 2 bets, you are at 2 more losses than wins and bankroll -3, but a win and loss (in either order) results in still having 2 more losses than wins but bankroll -2.

Let me try to explain, what I'm trying to show, that may help. My only point, is that the D'alembert wins, if it takes years for it to win it take years, but I hope you also realize that after years of waiting, the payoff will be HUGE. You win 1 unit per win so after a million trials, you'd win 500,000 units. Point being, you'd end up making the same amount of money, whether it takes years to complete 1 sequence or you complete 100 sequences daily.

As for bankroll, that's always a question, even on a +HE game you have to worry about the bankroll. The thing with the D'Alembert is that you gain 1 unit for every win, so you're adding to your bankroll all the time. So, with the D'alembert the only thing that really hurts is sharp downturns, you need a bunch of trials or a sharp upturn to make up for the losses. In reality, though, if you know you will eventually win, you can always go back and put more money together if you lose your bankroll while playing. Just keep adding to the bankroll, if you know you will eventually win and turn a profit.

As you can probably tell, I try to do most of this in my head, so I just try to see the big picture (like you'll have equal wins/losses before reaching Infinity). Doing all the technical stuff is just not as easy for me.

Quote:JyBrd0403Quote:24Bingo

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

In one sense, that's true. In a 50/50 game (although NOT one with an edge), you have probability 1 of ultimately coming to that point. And, of course, hand #googolplex is just as long "before infinity" as the fourth.

However, just like the return of the St. Petersburg game, that doesn't mean the expected number of hands you'll need is finite.

First, the easy part: half of all progressions will be over in one hand, and another quarter will be over in two. If we were to keep a running total, it would be at 1 hand, but that doesn't really matter for reasons you'll soon see.

It makes sense to look at the remaining quarter of progressions as pairs of trials, with the first pair fixed at LL. So starting at -1, with each step bringing us up 1 (.25), down 1 (.25) or standing still, what's the expected number of steps it'll take us to get to 0? Let's call it x (so EV = (2x+2)/4 + 1).

To "answer" the question, we'll do a bit of recursion. If you drop down, the expected number of steps it'll take you to get back to where you were for the first time will be the same as the expected number of steps it'll take you to then get back up to the top. Likewise, if you stay where you are, the expected number of steps will increase by 1. So let x be the expected number of steps.

x = .25*1 + .5*(x+1) + .25*(x+x)

x = x + .75

In short, there can be no finite x.

I'm sure I could come up with a less hackish demonstration that x blows up, but it's 1 AM and I've been drinking.

In fact, it occurs to me that your expected winnings, if you could follow the strategy you describe, would be not only positive, but infinite. However, you can't follow such a strategy, because no matter how trusting your credit card company is, sooner or later you'll keel over, and whether it takes a hundred years, or a googol, or even 10^10^10^10^10^1.1 years, that's still finite, and any finite limit of time or funds is -EV.

Quote:RSI don't think you understand how math works, OP.

What are you questioning?

Your sanity.Quote:JyBrd0403What are you questioning?

I guess the real question is what game are you trying to beat and how? Perhaps someone will then show you how or why its not possible. Not that it would help you anyways but do you ave a game that's a true 50 /50?

I just don't know why people believe they are smarter then EVERYONE. Years and years of very intelligent people have been analyzing table games and systems.

Nothing works unless you have something more then just some betting system. IE card counting

Quote:JyBrd0403What are you questioning?

In a perfect world, the D'alem works perfectly. In a

real casino the pit smiles while they take your money.

Maybe they'll comp you a pack of smokes for

your trouble.

In a perfect world no one would be looking for easy money betting schemes.Quote:EvenBobIn a perfect world, the D'alem works perfectly. In a

real casino the pit smiles while they take your money.

Maybe they'll comp you a pack of smokes for

your trouble.

BOB did you see the new Episode of Heel on wheels yet? I started it but feel asleep .