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JyBrd0403
JyBrd0403
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August 3rd, 2014 at 3:34:23 PM permalink
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.
EvenBob
EvenBob
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August 3rd, 2014 at 3:51:31 PM permalink
Quote: JyBrd0403

noone 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.
"It's not enough to succeed, your friends must fail." Gore Vidal
Wizard
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Wizard
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August 3rd, 2014 at 4:09:58 PM permalink
This 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).
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy 
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August 3rd, 2014 at 4:46:31 PM permalink
Quote: JyBrd0403

I'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.
JyBrd0403
JyBrd0403
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August 3rd, 2014 at 4:52:21 PM permalink
Quote: Wizard

This 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]
24Bingo
24Bingo
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August 3rd, 2014 at 5:23:16 PM permalink
I would need a rigorous algorithm for your D'Alembert. What happens when you get to 1 unit and win? How many units do you start with? Give me that, and I'll give you a formula, although you won't like the results.

(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.
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.
ThatDonGuy
ThatDonGuy 
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August 3rd, 2014 at 5:39:21 PM permalink
Quote: JyBrd0403

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]


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?
JyBrd0403
JyBrd0403
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August 3rd, 2014 at 6:22:34 PM permalink
Quote: ThatDonGuy

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?



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.
24Bingo
24Bingo
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August 3rd, 2014 at 6:30:22 PM permalink
By that strategy, technically, the EV will be positive, but your expected time at the table will be infinite. In other words, the strategy is impossible to follow in all circumstances. Put a finite time limit or bankroll limit on it, no matter how high, and it will turn negative.
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.
JyBrd0403
JyBrd0403
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August 3rd, 2014 at 7:45:12 PM permalink
Quote: 24Bingo

By 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.

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