MDawg
MDawg
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July 29th, 2020 at 7:48:04 PM permalink
Everyone with any sense agrees that any mechanical progression or system of betting based on the system itself fails. Martingale, D'Alembert, Oscar's Grind, whatever - all fail.

Everyone also agrees that to win one must bet more at times. When to do it, is what not all of us agree on.
I tell you itís wonderful to be here, man. I donít give a damn who wins or loses. Itís just wonderful to be here with you people.
ThatDonGuy
ThatDonGuy
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July 30th, 2020 at 8:14:26 AM permalink
I am trying to find some general solution for the probability of reaching a target amount before losing an entire bankroll based on the target and initial values, but I'm not entirely sure one exists, especially if an additional condition that the bet is capped by both the remaining bankroll and the amount needed to reach the target. It's more of a Markov Chain problem, which can be solved with simultaneous equations; the problem is, if the initial bankroll is 200 and the target is 300 (i.e. making a profit of 100 before losing the entire initial 200), there are around 6500 separate equations. Even with initial 20 and target 30, there are 163 equations.
Wizard
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Wizard
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July 30th, 2020 at 8:42:19 AM permalink
Quote: ThatDonGuy

I am trying to find some general solution for the probability of reaching a target amount before losing an entire bankroll based on the target and initial values, but I'm not entirely sure one exists, especially if an additional condition that the bet is capped by both the remaining bankroll and the amount needed to reach the target. It's more of a Markov Chain problem, which can be solved with simultaneous equations; the problem is, if the initial bankroll is 200 and the target is 300 (i.e. making a profit of 100 before losing the entire initial 200), there are around 6500 separate equations. Even with initial 20 and target 30, there are 163 equations.



This is a telescoping sums problem, assuming flat betting. Let's say the question is...

A player is playing double-zero roulette, making even money bets, flat betting one unit at a time. He brings 200 units to the table. His goal is to win 100 units (for a total of 300). He plays until he achieves that goal or goes bust trying. What is his probability of success?


((10/9)^220 - 1) / ((10/9)^300 - 1) = 0.0002184745
It's not whether you win or lose; it's whether or not you had a good bet.
DeMango
DeMango
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July 30th, 2020 at 9:03:06 AM permalink
Wish we could get away from double zero Roulette. As far as I know the D'Alembert was designed for even money bets with single zero+en prison rule. EV roughly equal to a line bet position at a craps table and player bet at Baccarrat. Anybody serious about D'Alembert is not going to play bets with over 5% -ev.

Another way to denigrate D'Alembert is to start at one unit. I suggest 5 units to negate the effects of early success.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
OnceDear
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OnceDear
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July 30th, 2020 at 10:00:47 AM permalink
Quote: ThatDonGuy

I am trying to find some general solution for the probability of reaching a target amount before losing an entire bankroll based on the target and initial values, but I'm not entirely sure one exists, especially if an additional condition that the bet is capped by both the remaining bankroll and the amount needed to reach the target. It's more of a Markov Chain problem, which can be solved with simultaneous equations; the problem is, if the initial bankroll is 200 and the target is 300 (i.e. making a profit of 100 before losing the entire initial 200), there are around 6500 separate equations. Even with initial 20 and target 30, there are 163 equations.

Hi TDG
This is almost exactly described in my blog post.
I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
ThatDonGuy
ThatDonGuy
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July 30th, 2020 at 10:25:27 AM permalink
Quote: OnceDear

Hi TDG
This is almost exactly described in my blog post.
I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.


I don't see any formula, or even a way that the probability is calculated - just a single run.

I also noticed this:
"I assigned a starting Bankroll of $1,000,000 ( You can vary that in the model. )
I decided an initial wager of $10 ( You can vary that in the model. )
I decided a progression increment of $1. ( You guessed it )
Whenever a bet won, the next bet was $1 less than the previous bet, or $1, whichever was the greater.
Whenever a bet lost, the next bet was $1 more than the previous bet, or total remaining bankroll, whichever was the smaller.
The model simulated potentially 50,000 spins."
You then noted that you never doubled the bankroll. This is not surprising.

A run of N wins and N losses that starts with a win does not change the bankroll unless you happen to reach the target within the run
A run of N wins and N losses that starts with a loss adds N to the bankroll.
However, you need to add 1,000,000 to the bankroll in 50,000 spins.

10 + 11 + 12 + ... + (10 + (N-1)) = 1,000,000
10 N + (1 + 2 + ... + (N-1)) = 1,000,000
10 N + (N-1) N / 2 = 1,000,000
(20 + (N-1)) N = 2,000,000
N^2 - N - 1,999,980 = 0
N = about 1414

This means the wins need to exceed the losses by 1414 at some point in 50,000 spins.
OnceDear
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OnceDear
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July 30th, 2020 at 10:30:34 AM permalink
Quote: ThatDonGuy

Quote: OnceDear

Hi TDG
This is almost exactly described in my blog post.
I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.


I don't see any formula, or even a way that the probability is calculated - just a single run.

Are you seeing the same post as me? Oncedear's rule of thumb. For some reason, I get scrolled forward to the appropriate post. So scroll down till you see it, while I wander off to amend the link.
https://wizardofvegas.com/member/oncedear/blog/5/#post1370
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
ThatDonGuy
ThatDonGuy
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July 30th, 2020 at 10:41:13 AM permalink
Quote: OnceDear

Are you seeing the same post as me? Oncedear's rule of thumb. For some reason, I get scrolled forward to the appropriate post. So scroll down till you see it, while I wander off to amend the link.
https://wizardofvegas.com/member/oncedear/blog/5/#post1370


You mean, "Oncedear's Rule - What's the Chances?"
All that says is, the probability of success can never exceed the starting bankroll divided by (the starting bankroll + the target profit).
I was referring to calculating a specific probability of success.
Wizard
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Wizard
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July 30th, 2020 at 11:24:27 AM permalink
My page on the d'Alembert is ready to preview. I welcome all comments.
It's not whether you win or lose; it's whether or not you had a good bet.
OnceDear
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OnceDear
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July 30th, 2020 at 11:50:17 AM permalink
Quote: ThatDonGuy

You mean, "Oncedear's Rule - What's the Chances?"
All that says is, the probability of success can never exceed the starting bankroll divided by (the starting bankroll + the target profit).
I was referring to calculating a specific probability of success.


For a zero edge game...
Probability of Success = (Starting Bankroll)/(Starting Bankroll + Target Profit)

For a low house edge game, it gives a decent probability estimation if you wager a modest proportion of bankroll on each wager, especially so with a progressive such as Marty or d'Alembert

I acknowledge that it's far more complex, and beyond me to derive an equation where there is a house edge.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

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