ThatDonGuy Joined: Jun 22, 2011
• Posts: 5751
August 4th, 2014 at 5:04:36 PM permalink
Quote: JyBrd0403

Let 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: JyBrd0403

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're assuming that D'Alembert always wins. This is not the case in any remote definition of the term "always."
Quote: JyBrd0403

You 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: JyBrd0403

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.

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: JyBrd0403

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.

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: JyBrd0403

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

That's just it - you can't possibly know this for a fact.
Quote: JyBrd0403

Just 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!).
EvenBob Joined: Jul 18, 2010
• Posts: 26260
August 4th, 2014 at 5:10:46 PM permalink
A labby is a form of D'alem and works better
if you have to use a progression. It gives you
far more control.
It does not suck to be me.
JyBrd0403 Joined: Jan 25, 2010
• Posts: 548
August 4th, 2014 at 5:30:56 PM permalink
So, 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.

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 :)
thecesspit Joined: Apr 19, 2010
• Posts: 5936
August 4th, 2014 at 5:36:17 PM permalink
The return to the mean does not work like you think it does.... the split gets closer to 50/50 over time, however the actual values may still be far apart.

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.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5751
August 4th, 2014 at 6:05:39 PM permalink
Quote: JyBrd0403

So, 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: JyBrd0403

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.

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: JyBrd0403

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.

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: JyBrd0403

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.

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.
24Bingo Joined: Jul 4, 2012
• Posts: 1348
August 4th, 2014 at 8:16:58 PM permalink
Quote: JyBrd0403

So, 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: JyBrd0403

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.

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.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
HughJass Joined: Jul 20, 2014
• Posts: 40
August 4th, 2014 at 8:59:57 PM permalink
Quote: JyBrd0403

So, 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.
JyBrd0403 Joined: Jan 25, 2010
• Posts: 548
August 4th, 2014 at 9:13:16 PM permalink
Quote: HughJass

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.

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.
HughJass Joined: Jul 20, 2014
• Posts: 40
August 4th, 2014 at 9:19:26 PM permalink
It's not specifically for the D'Alembert but for all betting progressions.
24Bingo Joined: Jul 4, 2012
• Posts: 1348
August 4th, 2014 at 9:40:06 PM permalink
Quote: JyBrd0403

The 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 ak be your successive stakes.

.5*(+ak) + .5*(-ak) = .5*0 = 0 is the expected value of each trial.
0 + 0 + 0 + 0 + ... + 0 = 0.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.