Quote:JyBrd0403Let me try to explain, what I'm trying to show, that may help. My only point, is that the D'alembert wins

So does Martingale, if you have infinite time and money.

Quote:JyBrd0403if 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're assuming that D'Alembert always wins. This is not the case in any remote definition of the term "always."

Quote:JyBrd0403You win 1 unit per win so after a million trials, you'd win 500,000 units.

This not only assumes that 1 million trials will always have 500,000 wins and 500,000 losses, but you always had more losses than wins. Remember, in 10 trials, if you have WWWWW followed by LLLLL, you end up losing 10 units.

Quote:JyBrd0403Point 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.

And my point is, you are assuming that you will reach the point of equity, even if it takes "years." You might end up discovering that "centuries" is more like it - especially since you'll be playing a game where the probability of losing > the probability of winning.

Quote:JyBrd0403The thing with the D'Alembert is that you gain 1 unit for every win, so you're adding to your bankroll all the time.

Not with a string of losses, you don't. Remember what I said about assuming that you "will" get wins to balance them out.

Quote:JyBrd0403So, 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

That's just it - you can't possibly know this for a fact.

Quote:JyBrd0403Just keep adding to the bankroll, if you know you will eventually win and turn a profit.

Have you ever seen my "Ten Rules of Gambling"? Number 1 is, "Don't bet money you can't afford to lose under the (false) assumption that 'eventually things will even out' - in the words of Oscar Madison from The Odd Couple, 'There is no such thing as a sure thing; that's why they call it gambling.' "

As for calculating an EV, on a 50/50 game, I can't get a "formula", in part because the number of ways to have N wins and N losses such that you always have losses > wins before the (2N)th trial is (2N)P(N) / (N+1)! (where (2N)P(N) = the number of permutations of 2N items taken N at a time, which is (2N)! / N!).

if you have to use a progression. It gives you

far more control.

You're right, I didn't think about it but if you played 'til even wins and losses, the marty would work as well. It would be better to play a certain number of trials.

As, for not being able to determine whether wins and losses will eventually even out, you have to realize this is a mathematical calculation. The calculations on a 50/50 game mathematically even out. What happens when you actually play is anyone's guess, you might sit down to a 50/50 game and see that you only win 35% after a billion trials, and there's not a damn thing anyone can say or do about it. But mathematically that's impossible. Back to my point, mathematically 50/50 games have equal amounts of wins and losses, it will eventually even out, return to the mean and stuff. If you don't believe it , just figure out what the probabilities of it breaking even would be. You'll get an answer.

Anyway, thanks for nothing, Don :)

You didn't answer which EV you were looking for and under which conditions. As you don't except limits, there's no point in any calculation using infinities.

Quote:JyBrd0403So, you're saying it's not possible to calculate the EV on the D'Alembert? It doesn't surprise me that's what I keep getting is that it's too complicated to calculate. Too bad D'Alembert isn't alive, he might be the only one who could do the math for it. Anyway, that leaves the EV for the D'Alembert as an unknown.

It's almost certainly "possible," if you set a border condition (say, your initial bankroll is 1000). If you have infinite bankroll, I'm pretty sure the EV is infinite for a 50/50 game.

Quote:JyBrd0403You're right, I didn't think about it but if you played 'til even wins and losses, the marty would work as well. It would be better to play a certain number of trials.

And as I showed earlier in this thread, over a specified number of trials of a 50/50 game, the EV is zero - mainly because of the chance of having a string of losses at the end.

Quote:JyBrd0403As, for not being able to determine whether wins and losses will eventually even out, you have to realize this is a mathematical calculation. The calculations on a 50/50 game mathematically even out.

Quote:JyBrd0403

Maybe "eventually," but like I said, nobody will be around long enough to see when this "always" happens. It could take centuries.Quote:JyBrd0403Back to my point, mathematically 50/50 games have equal amounts of wins and losses, it will eventually even out, return to the mean and stuff. If you don't believe it , just figure out what the probabilities of it breaking even would be. You'll get an answer.

I already know what the probability of it breaking even "eventually" will be - pretty much 1.

Here's one for you: assume you are playing D'Alembert in roulette, and always bet on black. Also assume it's a double-zero wheel, and the numbers come up red, black, red, black, ..., red, black, and every 19th spin is green.

Your first 18 results: LWLWLWLWLWLWLWLWLWLW, so you're +9 and your bet level is 1

The next one is green, so you're +8 and the bet level is 2

The next 18 results: LWLWLWLWLWLWLWLWLW, so you're +17 and the bet level is 2 (it goes up to 3 with each L, then back down to 2 with each W)

The next one is green, so you're +15 and the bet level is 3

If your bet level at the start of a "set of 19" is N, then you will gain 9 with the next 18 spins, then lose N (and have the bet level raised to N+1) with the 19th.

After the 3rd set, you're +15 + 9 - 3 = +21 and the bet level is 4

After the 4th set, you're +21 + 9 - 4 = +26 and the bet level is 5

After the 5th set, you're +26 + 9 - 5 = +30 and the bet level is 6

After the 6th set, you're +30 + 9 - 6 = +33 and the bet level is 7

After the 7th set, you're +33 + 9 - 7 = +35 and the bet level is 8

After the 8th set, you're +35 + 9 - 8 = +36 and the bet level is 9

After the 9th set, you're +36 + 9 - 9 = +36 and the bet level is 10

After the 10th set, you're +36 + 9 - 10 = +35 (oops, it went down, didn't it?) and the bet level is 11

After the 11th set, you're +35 + 9 - 11 = +33 and the bet level is 12

After the 12th set, you're +33 + 9 - 12 = +30 and the bet level is 13

After the 13th set, you're +30 + 9 - 13 = +26 and the bet level is 14

After the 14th set, you're +26 + 9 - 14 = +21 and the bet level is 15

After the 15th set, you're +21 + 9 - 15 = +15 and the bet level is 16

After the 16th set, you're +15 + 9 - 16 = +8 and the bet level is 17

After the 17th set, you're +8 + 9 - 17 = +0 and the bet level is 18

After the 18th set, you're +0 + 9 - 18 = -9 and the bet level is 19

After the 19th set, you're -9 + 9 - 19 = -19 and the bet level is 20

The numbers only go down from there.

So much for your system. "Mathematically," you have 10 losing bets for every 9 winning ones. "Mathematically," you lose. Chasing losses will not change this.

Quote:JyBrd0403So, you're saying it's not possible to calculate the EV on the D'Alembert?

No.

We have calculated the EV on the D'Alembert.

On a 50/50 game, with a finite time limit or finite bankroll limit, the EV is 0.

If you had infinite time and money, there is a strategy that could give you an EV of positive infinity; however, this would also be your expected time at table.

The same would be roughly true even with a house edge, but there not only would your time at table be infinite, but there would be a non-infinitesimal chance of an eternal progression.

If you'll give me a time limit or bankroll - however high - a thousand years, a googol dollars - and the edge, I'll give you the EV. I promise you, if there's an edge, it will be negative.

Quote:JyBrd0403What happens when you actually play is anyone's guess, you might sit down to a 50/50 game and see that you only win 35% after a billion trials, and there's not a damn thing anyone can say or do about it. But mathematically that's impossible.

No. Nonononononono.

It's improbable.

It's the same difference you're missing above. A googolplex dollars is not infinite money. A billion years is not infinite time. And beating 10 sigmas is not impossible (Google's telling me about 10^(-24), but I can't find a table or calculator that goes that far).

What is impossible is getting a positive EV from a finite series of losing or even even gambles. It's not impossible to come out ahead, no matter how long you play - but it's impossible to get a positive EV.

Quote:JyBrd0403So, you're saying it's not possible to calculate the EV on the D'Alembert?

Like all betting progressions, the EV of the D'Alembert is 0 in a fair game, negative in an unfair EV- game, and positive in an unfair EV+ game. If you're up on your high school algebra, I can give you a rather lengthy mathematical proof.

Quote:HughJassLike all betting progressions, the EV of the D'Alembert is 0 in a fair game, negative in an unfair EV- game, and positive in an unfair EV+ game. If you're up on your high school algebra, I can give you a rather lengthy mathematical proof.

Yeah, maybe I didn't make myself clear. I was looking for the lengthy mathematical formula or proof. The mathematicians I talked to said it was too complicated for them, so if you've got a formula, lay it on me.

Quote:JyBrd0403The mathematicians I talked to said it was too complicated for them, so if you've got a formula, lay it on me.

For a 50/50 game?

Let a

_{k}be your successive stakes.

.5*(+a

_{k}) + .5*(-a

_{k}) = .5*0 = 0 is the expected value of each trial.

0 + 0 + 0 + 0 + ... + 0 = 0.

Quote:JyBrd0403Yeah, maybe I didn't make myself clear. I was looking for the lengthy mathematical formula or proof. The mathematicians I talked to said it was too complicated for them, so if you've got a formula, lay it on me.

Let

b_k = bet value at the kth level

p_k = probability that series terminates with a win at the kth level, having been preceded by k-1 losses in a row

n - 1 = greatest number of losses in a row that can be sustained

e = player's expectation

e = + p_1b_1 + p_2(b_2 - b_1) + p_3(b_3 - b_2 - b_1) +...

+ p_n(b_n - b_(n-1) - ... - b_1)

+ (1 - p_1 - p_2-...

- p_n)(-b_n - b_(n-1) - ... - b_1)

The terms on the first line represent products of the probability that the series will terminate with a win at each successive level times the net profit at that level. The term on the third line gives the product of the probability that the series ends in failure at the nth level times the net loss.

regroup the terms:

e = 2p_1b_1 + (2p_2 + p_1)b_2 + (2p_3 + p_2 + p_1)b_3 + ...

+ (2p_n + p_(n-1) + ... + p_2 + p_1)b_n

- (b_1 + b_2 + ... + b_(n-1) + b_n)

p_k = (1 - p)^(k-1)p, where p is the probability of a win on any individual play and 1 - p is the probability of a loss.

substituting:

e = [2p]b_1 + [2p(1 - p) + p]b_2 + [2p(1 - p)^2 + p(1 - p) + p]b_3 + ...

+ [2p(1 - p)^(n-1) + p(1 - p)^(n-2) + ... + p(1 - p)^2 + p(1 - p)^1

+ p(1 - p)^0]b_n - (b_1 + b_2 + ... b_(n-1) + b_n)

factor out p so that the kth term is rewritten as:

p[(1 - p)^(k-1) + (1 - p)^(k-1) + (1 - p)^(k-2) + ... + (1 - p)^2

+ (1 - p)^1 + (1 - p)^0]b_k

use the formula for the sum of a geometric series and rewrite the kth term:

p[(1 - p)^(k-1) + (((1 - p)^k - 1)/((1 - p) - 1))]b_k

= [(2p - 1)(1 - p)^(k-1) + 1]b_k

e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1) + 1]b_k - sum_{k = 1}^{n} b_k

e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1)b_k + sum_{k = 1}^{n} b_k - sum_{k = 1}^{n} b_k

cancel the last two summations and factor out (2p - 1):

e = (2p - 1) sum_{k = 1}^{n} (1 - p)^(k-1)b_k

Since (1 - p) is positive and b_k is positive, the summation is positive. Therefore, the sign of e depends on the sign of (2p - 1). In an even-payoff unfair game, p < .5 and (2p - 1) is negative. In a fair game, p = .5 and (2p - 1) = 0.

Quote:24BingoNo.

We have calculated the EV on the D'Alembert.

On a 50/50 game, with a finite time limit or finite bankroll limit, the EV is 0.

.

LOL. Yeah I got that part by 22 years old. What MY question is-how does D'alembert work, not on a 50/50 game but a 50.68-49.32 game-the bank bet at baccarat where over the "long term" the gambler, unlike craps, roulette, BJ, etc. CAN expect his winning bets to overwhelm his losing bets. Yes there is the little matter of the commission...

Quote:HughJassLet

b_k = bet value at the kth level

p_k = probability that series terminates with a win at the kth level, having been preceded by k-1 losses in a row

n - 1 = greatest number of losses in a row that can be sustained

e = player's expectation

e = + p_1b_1 + p_2(b_2 - b_1) + p_3(b_3 - b_2 - b_1) +...

+ p_n(b_n - b_(n-1) - ... - b_1)

+ (1 - p_1 - p_2-...

- p_n)(-b_n - b_(n-1) - ... - b_1)

The terms on the first line represent products of the probability that the series will terminate with a win at each successive level times the net profit at that level. The term on the third line gives the product of the probability that the series ends in failure at the nth level times the net loss.

regroup the terms:

e = 2p_1b_1 + (2p_2 + p_1)b_2 + (2p_3 + p_2 + p_1)b_3 + ...

+ (2p_n + p_(n-1) + ... + p_2 + p_1)b_n

- (b_1 + b_2 + ... + b_(n-1) + b_n)

p_k = (1 - p)^(k-1)p, where p is the probability of a win on any individual play and 1 - p is the probability of a loss.

You see where this is wrong, right? This is for a Marty, not a D'Alembert.

Quote:JyBrd0403You see where this is wrong, right? This is for a Marty, not a D'Alembert.

It appears to apply for any method. Note that it is the expected value for playing until you either get a win or reach a loss limit (which is expressed in a number of losses, but can be converted from a dollar limit).

For Martingale, p_1 = 1, p_2 = 2, p_3 = 4, p_4 = 8, p_5 = 16, p_6 = 32, and so on.

For D'Alembert, p_1 is variable, p_2 = p_1 + 1, p_3 = p_2 + 1, p_4 = p_3 + 1, and so on.

However, it does show that EV in this case is negative (exactly how negative has to be calculated...) if the probability of winning is less than 1/2, and zero if it equals 1/2.

Quote:ThatDonGuyIt appears to apply for any method. Note that it is the expected value for playing until you either get a win or reach a loss limit (which is expressed in a number of losses, but can be converted from a dollar limit).

For Martingale, p_1 = 1, p_2 = 2, p_3 = 4, p_4 = 8, p_5 = 16, p_6 = 32, and so on.

For D'Alembert, p_1 is variable, p_2 = p_1 + 1, p_3 = p_2 + 1, p_4 = p_3 + 1, and so on.

D'Alembert doesn't start over after a win.

Quote:JyBrd0403D'Alembert doesn't start over after a win.

That's why p_1 is variable.

However, when I did some Monte Carlo analysis on D'Alembert under the stated stop condition (the first bet wins, or the number of wins = the number of losses), I did notice something:

When there is no other stop condition, EV is positive even for a 50% game - but some of the "runs" required are quite large; for example, one took three billion bets (and the bets reached almost 100,000x the original bet - have fun finding a table that allows that) before it finally evened out. (The mean number of bets needed to even out is about 200,000 - and this does include the fact that it takes only one or two bets to win 75% of the time.)

When there is a stop condition - either bankroll-based or time-based - then EV is 0.

If you remove the "total loss" condition from HughJass's calculations:

e = p

_{1}b

_{1}

+ p

_{2}(b

_{2}- b

_{1})

+ p

_{3}(b

_{3}- b

_{2}- b

_{1})

+ ...

+ p

_{n}(b

_{n}- b

_{n-1}-...- b

_{1})

e = b

_{1}* (p

_{1}- p

_{2}- p

_{3}-...- p

_{n})

+ b

_{2}* (p

_{2}- p

_{3}- p

_{4}-...- p

_{n})

+ b

_{3}* (p

_{3}- p

_{4}- p

_{5}-...- p

_{n})

+ ...

+ b

_{n-1}* (p

_{n-1}- p

_{n})

+ b

_{n}* p

_{n}

e = b

_{1}* (p - p (1-p) - p (1-p)

^{2}-...- p (1-p)

^{n-1})

+ b

_{2}* (p (1-p) - p (1-p)

^{2}-...- p (1-p)

^{n-1})

+ b

_{3}* (p (1-p)

^{2}-...- p (1-p)

^{n-1})

+ ...

+ b

_{n-1}* (p (1-p)

^{n-2}- p (1-p)

^{n-1})

+ b

_{n}* p (1-p)

^{n-1}

e = b

_{1}* p * (2 - 1 -(1-p) - (1-p)

^{2}-...- (1-p)

^{n-1})

+ b

_{2}* p * (2(1-p) - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-1})

+ b

_{3}* p * (2(1-p)

^{2}- (1-p)

^{2}-...- p (1-p)

^{n-1})

+ ...

+ b

_{n-1}* (2 (1-p)

^{n-2}- (1-p)

^{n-2}- (1-p)

^{n-1})

+ b

_{n}* p * (2 (1-p)

^{n-1}- (1-p)

^{n-1})

e = b

_{1}* p * (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-1})

+ b

_{2}* p (1-p) * (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-2})

+ b

_{3}* p (1-p)

^{2}* (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-3})

+ ...

+ b

_{n-1}* p (1-p)

^{n-2}* (2 - 1 - (1-p))

+ b

_{n}* p (1-p)

^{n-1}* (2 - 1)

e = b

_{1}* p * (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-1})

+ b

_{2}* p (1-p) * (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-2})

+ b

_{3}* p (1-p)

^{2}* (2 - 1 - (1-p) - (1-p)

^{2}-...- (1-p)

^{n-3})

+ ...

+ b

_{n-1}* p (1-p)

^{n-2}* (2 - 1 - (1-p))

+ b

_{n}* p (1-p)

^{n-1}* (2 - 1)

e = b

_{1}* p * (2 - [(1-p)

^{n}- 1] / [(1-p) - 1])

+ b

_{2}* p (1-p) * (2 - [(1-p)

^{n-1}- 1] / [(1-p) - 1])

+ b

_{3}* p (1-p)

^{2}* (2 - [(1-p)

^{n-2}- 1] / [(1-p) - 1])

+ ...

+ b

_{n-1}* p (1-p)

^{n-2}* (2[(1-p)

^{2}- 1] / [(1-p) - 1])

+ b

_{n}* p (1-p)

^{n-1}* (2 - [(1-p)^1 - 1] / [(1-p) - 1])

e = b

_{1}* p * (2 + [(1-p)

^{n}- 1] / p)

+ b

_{2}* p (1-p) * (2 + [(1-p)

^{n-1}- 1] / p)

+ b

_{3}* p (1-p)

^{2}* (2 + [(1-p)

^{n-2}- 1] / p)

+ ...

+ b

_{n-1}* p (1-p)

^{n-2}* (2 + [(1-p)

^{2}- 1] / p)

+ b

_{n}* p (1-p)

^{n-1}* (2 + [(1-p)^1 - 1] / p])

e = b

_{1}* (2p + [(1-p)

^{n}- 1])

+ b

_{2}* (2p (1-p) + (1-p) [(1-p)

^{n-1}- 1])

+ b

_{3}* (2p (1-p)

^{2}+ (1-p)

^{2}[(1-p)

^{n-2}- 1])

+ ...

+ b

_{n-1}* (2p (1-p)

^{n-2}+ (1-p)

^{n-2}[(1-p)

^{2}- 1])

+ b

_{n}* (2p (1-p)

^{n-1}+ (1-p)

^{n-1}[(1-p) - 1])

e = b

_{1}* (2p + (1-p)

^{n}- 1)

+ b

_{2}* (2p (1-p) + (1-p)

^{n}- (1-p))

+ b

_{3}* (2p (1-p)

^{2}+ (1-p)

^{n}- (1-p)

^{2})

+ ...

+ b

_{n-1}* (2p (1-p)

^{n-2}+ (1-p)

^{n}- (1-p)

^{n-2})

+ b

_{n}* (2p (1-p)

^{n-1}+ (1-p)

^{n}- (1-p)

^{n-1})

e = b

_{1}* ((2p-1) + (1-p)

^{n})

+ b

_{2}* ((2p-1) (1-p) + (1-p)

^{n})

+ b

_{3}* ((2p-1) (1-p)

^{2}+ (1-p)

^{n})

+ ...

+ b

_{n-1}* ((2p-1)(1-p)

^{n-2}+ (1-p)

^{n})

+ b

_{n}* ((2p-1) (1-p)

^{n-1}+ (1-p)

^{n})

If p = 0.5, then all of the 2p-1 terms = 0, and e = 0.5^n * (b

_{1}+ b

_{2}+ ... + b

_{n})

When you start, b

_{k}= (k-1) + b

_{1}, which is not necessarily 1 (e.g. if your first five bets are LLLLW, the first bet of the next "sequence" is 4, not 1)

e = 0.5

^{n}* (b

_{1}n + (n-1) n / 2) = (b

_{1}n + (n-1) n / 2) / 2

^{n}

Then again, as n approaches positive infinity, this approaches (b

_{1}+ n) / (ln 2 * 2

^{n}) by L'Hopital's Rule; in turn, this approaches 1 / (ln 4 * 2

^{n}), which approaches zero...assuming, of course, that b

_{1}is finite. That's what happens when you deal with "infinity."

Let's take a 1:1 game where p is your chance of winning.

Your credit card company is having a special where they'll lend out all the money you need, no fees, no questions asked... but only for one [T]. After that, you've got nothing but a bill.

That is to say, infinite bankroll, but you only have T hands.

So let's go back to the reasoning from before. First the trivial case: W and you're out, 1 unit up. So we add p to the expected value. (We won't be considering pairs of hands this time, so there's no point singling out LW.)

Now, again we'll get recursive, but unlike before, we're taking it one hand at a time, and again unlike before, we have a counter; when it reaches 0, game over.

So we've got the function f(d, t), and it runs as follows:

if (t == 0 || d == 1) return 0;

else return (p*(f(d-1,t-1) + d) + (1-p)*(f(d+1,t-1) - d));

And our ultimate answer will be p + (1-p)*f(2,T-1).

Now this... technically... gives you a way to calculate the expected value, but since it'll take you a billion iterations for a time limit of thirty hands, I wouldn't recommend it.

I'll work on a closed-form representation tomorrow night.

EDIT: Frankly... screw it.

Quote:ThatDonGuyWhen there is no other stop condition, EV is positive even for a 50% game - but some of the "runs" required are quite large; for example, one took three billion bets (and the bets reached almost 100,000x the original bet - have fun finding a table that allows that) before it finally evened out. (The mean number of bets needed to even out is about 200,000 - and this does include the fact that it takes only one or two bets to win 75% of the time.)

Ahh, thank you ThatDonGuy, you just showed that the D'Alembert wins on a 50/50 game. Now feel free to take that back and state with 1000% certainty that there is no Betting System that Works on an HE or 50/50 game. The crap you have to go through to get to the obvious, huh.

See, either the D'alembert wins on 50/50 or flat betting 50/50 is a 1000% LOSING PROPOSITION, because you'll be down -5000 units on a 50/50 game. So, if you can say flat betting will break you even, then you can say the D'alembert will win. How you work that out with all the BS math, and I do mean the math is wrong, is up to you, but it's obvious the D'Alembert wins.

Quote:JyBrd0403Ahh, thank you ThatDonGuy, you just showed that the D'Alembert wins on a 50/50 game. Now feel free to take that back and state with 1000% certainty that there is no Betting System that Works on an HE or 50/50 game. The crap you have to go through to get to the obvious, huh.

Why would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.

I think I have been able to calculate the EV for an infinite D'Alembert under the given stop conditions.

It appears that there are (2n)! / (n! (n+1)!) ways to have n wins and n losses such that the first result is a loss and losses > wins until the (2n)th bet.

If p is the probability of winning a particular bet, then the stop conditions are:

(a) the first bet is a win (probability p, result +1)

(b) the first two results are LW (probability (1-p) p, result +2)

(c) the first two results are LL, then n Ls and Ws that do not result in the number of Ws = the number of Ls (including the first 2) at any point, then WW (probability (2n)! / (n! (n+1)!) * p

^{n+2}* (1-p)

^{n+2}, result (n+2))

For a 50/50 game, this appears to diverge

For double-zero roulette, it appears to converge at around 5.210526

For a 25% game, it appears to converge at 0.625

Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.

Quote:ThatDonGuyWhy would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.

For a 50/50 game, this appears to diverge

For double-zero roulette, it appears to converge at around 5.210526

For a 25% game, it appears to converge at 0.625

Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.

You're getting close to understanding it Don. If we play until it "diverges" that would be the finite. You just can't determine when that would be, stinking probabilities :)

Now the funny thing is Flat betting would break even when it "diverges" and the Marty would, and maybe this is what is giving the problems, the Marty would also just break even when it "diverges". So you've got a finite timeline, and a winning system right there in front of you.

So, like I said, if you can say a 50/50 flat bettor will break even, then you can say a D'Alembert player will win. Simple as that. Otherwise, you have to say that a Flat bettor would lose and never break even, and that's just silly.

Quote:JyBrd0403You're getting close to understanding it Don. If we play until it "diverges" that would be the finite.

If the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.

Quote:JyBrd0403So, like I said, if you can say a 50/50 flat bettor will break even, then you can say a D'Alembert player will win. Simple as that.

Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.

Quote:ThatDonGuyIf the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.

Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.

Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.

Also, you're now going back on the math you did. The game doesn't end on a losing streak, so you can't end on a losing streak. The second point is that the D'Alembert doesn't have to go all the way back to even to win. You can be betting 100 units and still be positive on your bankroll. So, maybe, it would be easier to understand if you stop when you have won money, as opposed to breaking even on wins and losses. So, if you're point was flat betting you go up +100 units and then - 100 units that you're even on wins/losses, and the D'alembert would be betting 100 units even though the wins/losses are even. I still don't think that would matter, but you can just play until the bankroll is positive that way you don't have to break even, or go back to +100 to stop the progression. I don't know why that would matter since the longer you play, according to you guys, the higher you go up. Eventually, you'll be able to go right past +100. But, if that's giving you trouble, this stopping on a losing streak, you can just stop when you're bankroll is positive.

But, the whole point, which you seemed to agree with is that the D'Alembert eventually wins, flat betting and marty break even.

Quote:JyBrd0403Quote:ThatDonGuyIf the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.

Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.

Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.

Just as EVENTUALLY, six monkeys in front of six typewriters can write the complete works of Shakespeare.

Quote:JyBrd0403Also, you're now going back on the math you did. The game doesn't end on a losing streak, so you can't end on a losing streak.

It can if you drop dead during one, or the casino shuts down during one.

Quote:JyBrd0403But, the whole point, which you seemed to agree with is that the D'Alembert eventually wins, flat betting and marty break even.

Wrong. Marty "eventually wins" just as much as D'Alembert.

And the phrase "D'Alembert eventually wins" is meaningless when you realize that "eventually" includes "millions of years from now."

Quote:JyBrd0403Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.

Where do you get that marty players break even? Eventually, with infinite time and bankroll, they win, too. One unit per win, with no regard to losses, just like D'Alembert players. In fact, that's what they're ahead by after every single win, unlike D'Alembert players, who only get there by resetting their progression. In fact, also unlike D'Alembert players, the expected time until their progression ends is finite. In their case, it's only the expected nadir of their progressions that's infinite.

Quote:ThatDonGuyEventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.

Don't forget that the game has to have no edge, or the probability of an eternal progression converges to a nonzero value.

Quote:24BingoWhere do you get that marty players break even? Eventually, with infinite time and bankroll, they win, too. One unit per win, with no regard to losses, just like D'Alembert players. In fact, that's what they're ahead by after every single win, unlike D'Alembert players, who only get there by resetting their progression. In fact, also unlike D'Alembert players, the expected time until their progression ends is finite. In their case, it's only the expected nadir of their progressions that's infinite.

Right, I'm trying to explain that the D'Alembert will produce a profitable game. By this, I mean that eventually, you will be playing with the houses money. All the losses you sustain will be covered by the money you won, and you will just keep adding to your profit. How is this possible? Because the profits will eventually outpace the loss limits. You may be betting 5000 units, but still playing on the houses money after 100 million trials. My point is that not only will you be winning at that point, but you have the benefit of being assured that you will come back to a 1 unit bet eventually. This is what is apparently being disputed, and maybe I didn't explain it well enough, but the point is that the D'Alembert profits will outpace the loss limits.

One of my ways of showing this is with the Marty. The marty will always break even. The wins will equal the losses. So, while you win 100 million dollars, a losing streak hits and you lose 100 million dollars. Then you have to pony up a 200 million dollars from you're own money and make a 200 million dollar bet. You never are in the Black you always have to pony up your own money. I can show you quickly that the wins and losses are equal for the marty. You lose 4 in a row once every 30 trials, that's 1-2-4-8 = 15 unit loss. In 30 trials you get 15 wins. 15 losses = 15 wins. This holds true for all cases. 10 in a row 18 in a row 30 in a row etc. You break even, and you are never in the black, you always have to pony up you're own money to cover. So your bankroll goes up +100 million then down -100 million.

The D'Alembert Bankroll should be in the Black at some point, going up let's say +100 million down let's say to +85 million, staying in the black. The reason this is better than the Marty is obvious. Losing 18 in a row in a million trials doesn't hurt the D'Alembert at all! The losses in a row don't push the game down like it does for a marty. If you aren't in the Black after a million trials, you know what the big worry is, you would now lose 19 in a row. The D'Alembert should easily outpace the loss limits on a 50/50 game.

Here's the problem, the math these guys are giving are saying that the losses somehow magically keep up with the wins the D'Alembert produces. I haven't a clue how they think that's possible. The only thing I can see is they are basing the math on streaks ( losses in a row). The losses in a row, can grow the more trials you have, but there's no reason why the total losses would grow any higher than the losses in a row. I mean if the 50/50 game is on the positive side of wins and losses you can play a million trials and only go down -18 units. Again, my only point is that the D'Alembert wins should easily outpace the losses eventually, because the losses in a row are nothing to the D'Alembert. And, that means that the player is winning, or always in the BLACK, always playing on the houses money eventually. I can see no other possibility. This what will eventually happen with the D'Alembert, the Marty will always break even, and the Player will always be playing with his own money.

To try to clarify, the total losses should reach a limit over the losses in a row. Let's say you're at -5000 at some point. That -5000 is going to be the limit for Billions of trials. It was the worst run possible for the 50/50 game, and nothing will push that further down, because the losses in a row are only at 40 in billions of trials. So, there should be a definitive loss limit, that the D'alembert should eventually, easily, overcome.

That's how the casino's can make money, Buzz. There's a LIMIT on how much a player can win on a 50/50 or 49% or 47.5% game. Also, a LIMIT on how much the Casino can lose on a 51%, 52.5% game. It's not Infinity.

Easy Babs, finger off that trigger. Lots of stores on that show !

Quote:JyBrd0403Here's the problem, the math these guys are giving are saying that the losses somehow magically keep up with the wins the D'Alembert produces. I haven't a clue how they think that's possible. The only thing I can see is they are basing the math on streaks ( losses in a row). The losses in a row, can grow the more trials you have, but there's no reason why the total losses would grow any higher than the losses in a row. I mean if the 50/50 game is on the positive side of wins and losses you can play a million trials and only go down -18 units. Again, my only point is that the D'Alembert wins should easily outpace the losses eventually, because the losses in a row are nothing to the D'Alembert.

Nope. Nothing to do with streaks. Remember a few simple facts (all these on a 50/50 game):

1. You will not necessarily be up, except at the end of each progression.

2. Although the chance of an individual progression lasting forever diverges to zero, the length has a nonzero chance of being arbitrarily high. A million, a billion, A(g

_{64},g

_{64}), doesn't matter, it has a finite chance of happening.

3. In fact, the chances of large progressions decrease slowly enough that the expected length diverges - it's infinite.

4. Because it's infinite, a stop condition of "I'll play until my progression ends!" plugged into a calculation gets you a dope slap from Athena.

5. If instead you're stopped after an arbitrary number of hands t, the expected value of the bet you were about to make will be 1/2 + sqrt(t + 1/4).

So let's say, for the sake of argument, you can only play 1,332 hands. You happen to be having perfectly average luck as concerns both the game itself and the formula in 5, so you've won 666 hands and lost 666 hands, but of those 666 hands you won, 36 were on the first bet of their respective progressions, so not having counted them in your D'Alembert, you're about to bet 37 units. Tell me... when you get the tap on the shoulder, how much have you won?

Quote:24BingoNope. Nothing to do with streaks. Remember a few simple facts (all these on a 50/50 game):

1. You will not necessarily be up, except at the end of each progression.

2. Although the chance of an individual progression lasting forever diverges to zero, the length has a nonzero chance of being arbitrarily high. A million, a billion, A(g_{64},g_{64}), doesn't matter, it has a finite chance of happening.

3. In fact, the chances of large progressions decrease slowly enough that the expected length diverges - it's infinite.

4. Because it's infinite, a stop condition of "I'll play until my progression ends!" plugged into a calculation gets you a dope slap from Athena.

5. If instead you're stopped after an arbitrary number of hands t, the expected value of the bet you were about to make will be 1/2 + sqrt(t + 1/4).

So let's say, for the sake of argument, you can only play 1,332 hands. You happen to be having perfectly average luck as concerns both the game itself and the formula in 5, so you've won 666 hands and lost 666 hands, but of those 666 hands you won, 36 were on the first bet of their respective progressions, so not having counted them in your D'Alembert, you're about to bet 37 units. Tell me... when you get the tap on the shoulder, how much have you won?

Were you able to understand why the Marty is not considered a winning game? Was I able to explain that so you understand?

One question for you, if streaks don't matter, what exactly pushes the losses lower? If you're only losing 18 in a row in a million trials, how far do you think you can actually go down in a million trials? In other words, what's the loss limit for a million trials? If you're suppose to only lose 18 in a row in a million trials, how much do you think the total losses are suppose to be? What's mathematically pushing the game down? For losses in a row it's simply double the trials add 1 loss in a row. How do you think the total losses are mathematically determined?

Quote:JyBrd0403Were you able to understand why the Marty is not considered a winning game? Was I able to explain that so you understand?

I understand that you understand how quickly Martingale progressions increase, and correctly recognize that this means you could be playing quite a long time, and still be wiped out very quickly by a streak of losses. But a single win will still put you back on your merry way, rather than the slog (of mean length ∞) you'll need to get back to the point where you're genuinely ahead if you do fall behind on the D'Alembert. In fact, unlike the D'Alembert, "I'll play until I win" is a strategy with a finite expected time at table (but an infinite expected nadir to your progression).

Quote:JyBrd0403One question for you, if streaks don't matter, what exactly pushes the losses lower?

As I said, wins on the first bet of your progression.

Quote:JyBird0403If you're only losing 18 in a row in a million trials, how far do you think you can actually go down in a million trials?

If you're stopped after a million trials, sight unseen, you will on average be stuck in your progression for 500,000 units, which is how much you'd be stuck for after about 1,000 losses in excess of wins not sustained on the first bet of a progression. As for how much you "can" go down - well, ignoring wins on the first bet of your progression, you'd have a standard deviation of 1,000 out from the origin, so about a 1 in 6 chance of having taken more than 1,000 more losses than wins - about a 1 in 40 chance of having sustained 2,000. You might notice those numbers are a bit larger than 18. Unlike the Martingale, it doesn't matter if your losses are lined up in a neat little row, only that they're not offset by enough wins.

You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

Quote:JyBrd0403Let me ask you this. If the downside is Infinite, wouldn't the Upside be Infinite as well? We used to always say for +/- 300 or +/- 500 for 50% games. Meaning you might be up +300 or -300. All, you stated above is that you could win for Infinity or lose for Infinity +/- Infinity. You seem to only look at the loss side and ignore that you will also be up on a 50/50 game. IF you would take into account the upside you would see that a 50/50 game would have a range of how high or low it will go.

You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

You're deluding yourself.

Quote:JyBrd0403Let me ask you this. If the downside is Infinite, wouldn't the Upside be Infinite as well? We used to always say for +/- 300 or +/- 500 for 50% games. Meaning you might be up +300 or -300. All, you stated above is that you could win for Infinity or lose for Infinity +/- Infinity. You seem to only look at the loss side and ignore that you will also be up on a 50/50 game. IF you would take into account the upside you would see that a 50/50 game would have a range of how high or low it will go.

I literally don't know what you're trying to say here. All I'm saying is that the expected length of each progression is infinite. I don't know what you mean by "win for Infinity or lose for Infinity +/- Infinity." I don't know what the "upside" is, unless you mean that that means that the expected return at the end of that progression is also infinite. That's true, but that's only a further illustration of how impossible your strategy is.

Quote:JyBrd0403You stated that you could be down -1000 after a million trials. That's fine.

No, I said you were reasonably likely to have 1000 or more losses than wins, or 1000 more wins than losses. I said you were likely to be stuck for an expected value of 500,000 units, i.e., as though you'd just lost a 999-unit bet and were about to make a 1,000-unit one (or won a 1,000-unit bet and about to make a 999-unit one).

Quote:JyBrd0403But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001.

That's how it works when you're talking about wins vs. losses, but I wasn't talking about wins vs. losses, I was talking about the amount you were likely to be stuck by if you played exactly that number of hands, then got the tap on the shoulder.

The bet you were about to make after two million, that's 1400.

The bet you were about to make after four million, that's about 2000.

Of course, they could be much higher or much lower, but that's the average.

It keeps going up because every time you get back to the beginning of your progression, you have a 50/50 chance of winning, and those wins don't count. Over time, they add up.

Quote:JyBrd0403In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

What the hell is a "loss limit," anyway?

Quote:JyBrd0403You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

I just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).

At one point, I got a run that consisted of the following before it eventually turned a profit:

621,937,312 bets

Max bet was 70,120x the initial bet

Lowest loss point was 2,147,438,594 initial bets

So if the minimum bet is $1, you had to have had a bankroll of at least $2.147 billion - otherwise you had to stop with an overall loss at some point.

Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet $5; maximum bet $350,000). I haven't seen many of those around.

Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Quote:ThatDonGuyI just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).

At one point, I got a run that consisted of the following before it eventually turned a profit:

621,937,312 bets

Max bet was 70,120x the initial bet

Lowest loss point was 2,147,438,594 initial bets

So if the minimum bet is $1, you had to have had a bankroll of at least $2.147 billion - otherwise you had to stop with an overall loss at some point.

Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet $5; maximum bet $350,000). I haven't seen many of those around.

Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Legitimate claim that the D'Alembert always wins. Remember you said that.

The D'Alembert doesn't have to be used for gambling alone. If I run simulations and it always wins, then the D'Alembert always eventually wins.

Congratulations, Don. You just proved every mathematician on the planet WRONG. How does it feel?

Also, I finally broke down and got one of these simulator things that I hate for myself. Already got the D'Alembert programmed in. I'll run some simulations tonight or tomorrow.

Quote:JyBrd0403Quote:ThatDonGuyI just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).

At one point, I got a run that consisted of the following before it eventually turned a profit:

621,937,312 bets

Max bet was 70,120x the initial bet

Lowest loss point was 2,147,438,594 initial bets

So if the minimum bet is $1, you had to have had a bankroll of at least $2.147 billion - otherwise you had to stop with an overall loss at some point.

Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet $5; maximum bet $350,000). I haven't seen many of those around.

Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Legitimate claim that the D'Alembert always wins. Remember you said that.

I don't remember ever saying that it didn't.

Quote:JyBrd0403Congratulations, Don. You just proved every mathematician on the planet WRONG. How does it feel?

Name one mathematician, much less "all of them," that ever said that D'Alembert doesn't always "eventually" win given infinite time and bankroll.

What I want to know is, how is the claim "D'Alembert always wins eventually" useful in any way, shape, or form in real life? Certainly not in gambling.

Quote:ThatDonGuyQuote:JyBrd0403Quote:ThatDonGuyI just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).

At one point, I got a run that consisted of the following before it eventually turned a profit:

621,937,312 bets

Max bet was 70,120x the initial bet

Lowest loss point was 2,147,438,594 initial bets

So if the minimum bet is $1, you had to have had a bankroll of at least $2.147 billion - otherwise you had to stop with an overall loss at some point.

Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet $5; maximum bet $350,000). I haven't seen many of those around.

Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Legitimate claim that the D'Alembert always wins. Remember you said that.

I don't remember ever saying that it didn't.

Name one mathematician, much less "all of them," that ever said that D'Alembert doesn't always "eventually" win given infinite time and bankroll.

What I want to know is, how is the claim "D'Alembert always wins eventually" useful in any way, shape, or form in real life? Certainly not in gambling.

Go, through some of my old threads, if you want to see the mathematicians say the D'Alembert Breaks even. You yourself, just gave 3 conditions that had to be met in order to say the D'Alembert wins. But, to make the point clear, when I say it eventually wins, it means it goes into the Black and NEVER goes into the red at some point. Not at all like the Marty that goes up and down.

It's useful because it's the only system that I know of, that actually wins. It's useful because it does what every mathematician says is impossible, the betting system changes the EV, if not the HE. So, that's every mathematician on the planet you just proved WRONG. Congrats, again.

Quote:JyBrd0403Quote:ThatDonGuyName one mathematician, much less "all of them," that ever said that D'Alembert doesn't always "eventually" win given infinite time and bankroll.

Go, through some of my old threads, if you want to see the mathematicians say the D'Alembert Breaks even.

The only ones I saw that claim that it breaks even assume there's a real world stop condition. When there is a stop condition - either time or budget - then the EV is zero for a 50/50 game. However, your continual use of the word "eventually" means that these don't apply.

Quote:JyBrd0403But, to make the point clear, when I say it eventually wins, it means it goes into the Black and NEVER goes into the red at some point. Not at all like the Marty that goes up and down.

What does "goes into the black and never goes into the red at some point" mean? Martingale played under the same "play until you eventually reach the stated goal" condition as D'Alembert always ends with a return of +1, so when does it ever "go into the red"?

Quote:JyBrd0403Quote:ThatDonGuyWhat I want to know is, how is the claim "D'Alembert always wins eventually" useful in any way, shape, or form in real life? Certainly not in gambling.

It's useful because it's the only system that I know of, that actually wins.

Not under real life conditions, it doesn't - at least not every time. Prove otherwise. Remember, "real life conditions," including limits on betting amounts and how much money you have. Claims that I already proved this are false.

Quote:DeMangoCan we run a sim and get over this thread?

You don't think facts will change his opinion, do you ?

So, after 30 million trials, I can only ask myself, WTF have you guys been talking about?????

Remember, in the real world, some arbitrary point will stop you. So... 28 million trials? Where were you after EXACTLY 1 million... 2 million... 3... 4... 5... 6... 7...?

Quote:24BingoWell? Let's see it. The full sequence.

Remember, in the real world, some arbitrary point will stop you. So... 28 million trials? Where were you after EXACTLY 1 million... 2 million... 3... 4... 5... 6... 7...?

This real world stuff is something you guys started. I simply said the D'alembert wins. Wins, meaning it goes into the Black and doesn't come back, just like a 51% game would do.

Actually, with these RNG's, I think the D'Alembert needs 10 to 20 million to get into the clear. I'm still running the simulation, but I can see that it's just going to go UP DOWN UP DOWN, as expected.

The bets will go up and go down I mean, 1000+ units back down to 150 units. Up Down Up Down.

Quote:JyBrd0403So, I sit down and start fiddling with this simulator. 28 million trials so far, profit of 14 million dollars so far. Some notes. The game has been in the Black for the last 20 million trials or so. At the +10 million mark, the bet was 1,668 units (the starting bet is 1 unit). What did surprise me a bit was the loss limit appears to be around 2 million dollars, of course, we're up 15 million now so that doesn't really matter. Most units bet so far was 3862. A quick observance and you can see that what happens is that the betting goes up to 1000+ units, and surprise surprise surprise it comes back down to 150 or so units, then it goes back up to 1000+ units and back down to 150 or so. UP DOWN, UP DOWN, UP DOWN, kinda what you'd expect from a 50/50 game.

So, after 30 million trials, I can only ask myself, WTF have you guys been talking about?????

Keep playing.

I ran some trials as well.

After 247 million trials, I was up 113 million.

After 299 million trials, I was down 29 million.

Of course, this works both ways:

After 345 million trials, I was down 135 million.

After 366 million trials, I was up 23 million.

Over 1 billion trials, the largest loss point was 256 million (between 844 million and 845 million - and by 860 million, the losses were wiped out and I was back on the plus side) and the largest bet was 36,847.

You will keep going back and forth. Your periods between losses will almost certainly get longer and longer, but "eventually" your total profit goes back down below zero - and "eventually" it goes back up above any point you were at in the past. However, you will never reach a point where you will always be ahead.

It's just that "eventually" is meaningless if you aren't immortal and you can't cover every possible overall loss level. You have to stop sometime. No bank is going to lend you billions of dollars if you tell them, "Don't worry; D'Alembert guarantees that you will get it back, with interest!"

Why don't you compare D'Alembert against, say, a modified Martingale, where your initial bet is 1, then, after each win, it resets to twice the previous starting point bet minus 1? For example, your initial bet is 1; after your first win, reset to 3 (then bet 6, 12, 24, etc. after losses until you win again); after your second win, reset to 5; after your third win, reset to 7, and so on. After your Nth win, your profit is N

^{2}.

Quote:JyBrd0403This real world stuff is something you guys started.

Damn that reality!

Quote:JyBrd0403I simply said the D'alembert wins. Wins, meaning it goes into the Black and doesn't come back, just like a 51% game would do.

You'd have to define that, but if we're going with the definition that there comes a point where your chance of never again falling behind converges to a non-zero value, nope. TheDonKing explains better than I could.

But also you asked for the EV. That has a specific definition, and under that definition, the expected values of multiple trials always add, so if the expected value of each trial is zero, it will be zero. When you introduce infinite parameters into the mix, things get a bit fuzzy, for the same reason limits that approach 0*∞ will sometimes converge to nonzero values, but you have to understand that the parameters do truly have to be infinite, not "very large."

Quote:JyBrd0403I'm still running the simulation, but I can see that it's just going to go UP DOWN UP DOWN, as expected.

The bets will go up and go down I mean, 1000+ units back down to 150 units. Up Down Up Down.

...what exactly do you think we were expecting? To this point, you'd been saying "up up up up."

If the probability of winning an even-money bet is p and we let q = 1 - P, then, if the system ends only after a loss of an N-unit bet, then the EV for the entire run is:

(p / (q-p))

^{2}* (1 - (p/q)

^{N}) - N p / (q-p) + N (N+1) / 2

For example, with the pass line on craps, if the stop condition is 20, then the EV is -37.78765, and the average length is 349.365 bets.

Yes, I do realize that the OP had a different stop condition - "keep betting until the bet drops back down to 1 (i.e. you win a bet of 2)".