EV for a D'Alembert
August 2nd, 2014 at 11:44:46 PM
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I've been looking for a formula to calculate the EV for the D'Alembert Progression. I haven't been able to find anything, I've checked around and all I get is that the progression makes it too difficult to calculate. I've found formulas for the Martingale, but nothing for a D'alembert. Since we already know the D'alembert wins on a 50/50 game, I was hoping to find a formula to calculate the EV for it.
P.S. This came up in another thread. But, I never did find a formula for the D'Alembert.
P.S. This came up in another thread. But, I never did find a formula for the D'Alembert.
August 3rd, 2014 at 12:33:29 AM
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It's simple. If you have an edge, the d'Alem is
wonderful. If you don't, it sucks. Just like every
other progression.
wonderful. If you don't, it sucks. Just like every
other progression.
"It's not enough to succeed, your friends must fail."
Gore Vidal
August 3rd, 2014 at 12:46:58 AM
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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?
August 3rd, 2014 at 1:43:30 AM
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PROOF?Quote: JyBrd0403The D'Alembert is the "Edge" on a 50/50, Bob.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
August 3rd, 2014 at 2:16:06 AM
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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.
August 3rd, 2014 at 3:21:40 AM
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That's an awesome proof.
Your EV is simply player's edge multiplied by the total $ in action.
Your EV is simply player's edge multiplied by the total $ in action.
August 3rd, 2014 at 3:38:56 AM
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Spit up coke all over.Quote: RSThat's an awesome proof.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
August 3rd, 2014 at 3:39:42 AM
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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.
August 3rd, 2014 at 3:58:24 AM
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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?
August 3rd, 2014 at 4:12:51 AM
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This 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).
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).
