- Decide your unit size, starting bet, bankroll, and winning goal.
- After a loss, increase your bet by one unit.
- After a win, decrease your bet by one unit.
- Keep repeating until you have achieved your winning goal or have run out of money

I assume two exceptions:

- If the next bet would cause you to overshoot your winning goal, then drop it to whatever would result in exactly achieving it.
- If you don't have enough money to cover the next bet, drop down to betting whatever you have.

Here is a video showing an example. I'm not sure why the sound is poor, maybe because I had two fans going.

Direct: https://www.youtube.com/watch?v=jJ6NKjimHko&feature=youtu.be

My questions for now are:

1. Is my understanding of the system correct?

2. How do you pronounce D'Alembert?

3. If used in the middle of a sentence, which letters do you capitalize in D'Alembert?

Quote:Wizard

2. How do you pronounce D'Alembert?

3. If used in the middle of a sentence, which letters do you capitalize in D'Alembert?

Wikipedia describes how it's pronounced

https://en.wikipedia.org/wiki/Jean_le_Rond_d'Alembert

dæləmˈbɛər/

Whatever that means https://en.wikipedia.org/wiki/Help:IPA/English

It's small 'd' and capital A

Quote:ThatDonGuyDo you have a copy of Ethier's The Doctrine of Chances? (I think this was MustangSally's go-to book for things like this.) There is an analysis of d'Alembert on pages 289-292.

Amazon has it for $50 and up!

Quote:ThatDonGuyDo you have a copy of Ethier's The Doctrine of Chances? (I think this was MustangSally's go-to book for things like this.) There is an analysis of d'Alembert on pages 289-292.

Yes, thank you for mentioning it. I do indeed have his book. I didn't know Ethier addressed this topic.

He says stopping factors are:

1. Win on first bet.

2. Total number of wins equal total number of losses

3. Ruin

He notes if the second marker is bet, the profit will be equal to the number of wins/losses. He then goes onto giving a formula for the probability of this happening.

I have not seen another source echo this second stopping point. Usually with any system there is a specific monetary winning goal.

Wins = 1+(x-1)*x/2

Losses = x*(x+1)/2

x | Wins | Losses |
---|---|---|

1 | 1 | 0 |

2 | 2 | 3 |

3 | 4 | 6 |

4 | 7 | 10 |

5 | 11 | 15 |

6 | 16 | 21 |

7 | 22 | 28 |

8 | 29 | 36 |

9 | 37 | 45 |

10 | 46 | 55 |

11 | 56 | 66 |

12 | 67 | 78 |

13 | 79 | 91 |

14 | 92 | 105 |

15 | 106 | 120 |

16 | 121 | 136 |

17 | 137 | 153 |

18 | 154 | 171 |

19 | 172 | 190 |

20 | 191 | 210 |

21 | 211 | 231 |

22 | 232 | 253 |

23 | 254 | 276 |

24 | 277 | 300 |

25 | 301 | 325 |

26 | 326 | 351 |

27 | 352 | 378 |

28 | 379 | 406 |

29 | 407 | 435 |

30 | 436 | 465 |

Assuming the player never goes positive, the balance is a formula of the numbers of wins (W) and losses (L), as follows. Let D = L-W.

Balance = W - D*(D+1)/2

on another website a guy posted that he used Oscar's grind not because he believed it was a winning system but to milk comps out of the house

the idea was that he would last longer than flat betting

for example, if your bets range from $10 to $200 during a session will the PB really see that?

or will he rate you as a guy who sometimes bets $150 or $200

I'm not sure

I don't really know if his idea had any validity

any comments about that?

Quote:lilredroosteron another website a guy posted that he used Oscar's grind not because he believed it was a winning system but to milk comps out of the house

the idea was that he would last longer than flat betting

for example, if your bets range from $10 to $200 during a session will the PB really see that?

or will he rate you as a guy who sometimes bets $150 or $200

I don't believe either point has validity.

First, if you want your money to last as long as possible, then flat betting will achieve that. I am interested to see evidence to the contrary.

Second, the floor, in my experience, tends to estimate on the low side of your range. If your range is $10 to $200, with an average of $100, I think you would see an average bet of about $50. If your goal is comps, you should bet more on your first bet and when the floor is watching.

Quote:Wizard

First, if you want your money to last as long as possible, then flat betting will achieve that. I am interested to see evidence to the contrary.

maybe Don could do a sim - he seems to like doing them - I would do it, but it's not my thing - I would probably mess it up

Don - I hope you don't mind my asking -

compare the # of bets a guy would on average be able to make starting off betting $10 at 2 zero roulette until he taps out his whole $1,500 bankroll

compared to a guy flat betting $75

it would be fun to see the results

oscar's grind:

the very old story is that there 𝘢𝘤𝘵𝘶𝘢𝘭𝘭𝘺 𝘸𝘢𝘴 𝘢 𝘨𝘶𝘺 𝘯𝘢𝘮𝘦𝘥 𝘖𝘴𝘤𝘢𝘳

according to this link:

"The originator of the system, a weekend Craps shooter, undescribed except as Oscar, told Wilson that he had never left Las Vegas as anything but a winner.

The celebrated gambling mathematician, Dr. Allan N. Wilson, introduced the system in his immensely informative book, “The Casino Gambler’s Guide”.

Wilson reported that Julian Braun, the foremost computer analyst of gambling probabilities, had found that a player who used the system on even-money Craps wagers with a betting unit of $1 would risk reaching a $500 house limit no more often than once in 4,250 sessions."

very, very interesting................to me anyway

https://www.casinoencyclopedia.com/oscars_grind_for_bj_etc/

before anybody jumps to play this system -

they player will still lose more than he won if he gets into what is generally referred to as 𝘵𝘩𝘦 𝘭𝘰𝘯𝘨 𝘳𝘶𝘯

Quote:lilredroostermaybe Don could do a sim - he seems to like doing them - I would do it, but it's not my thing - I would probably mess it up

Don - I hope you don't mind my asking -

compare the # of bets a guy would on average be able to make starting off betting $10 at 2 zero roulette until he taps out his whole $1,500 bankroll

compared to a guy flat betting $75

it would be fun to see the results

Double Zero Roulette:

Flat Bet: 380 bets

D'Alembert: 290 bets

Single Zero Roulette:

Flat Bet: 740 bets

D'Alembert: 477 bets

The flat bet values are calculated from the Gambler's Ruin formula: mean number of bets = initial bankroll / (1 - 2p), where p is the probability of winning a bet.

For double zero roulette with a bet of 75 and a bankroll of 1500 (i.e. 20 bets), the number = 20 / (1 - 2 x 9/19)) = 20 / (1/19) = 380.

Note that flat betting 1 will last longer than a D'Alembert starting with 1, for a very simple reason; you will expose more money to any house edge with D'Alembert than you would with a flat bet equal to the initial D'Alembert bet.

Quote:ThatDonGuy

Note that flat betting 1 will last longer than a D'Alembert starting with 1, for a very simple reason; you will expose more money to any house edge with D'Alembert than you would with a flat bet equal to the initial D'Alembert bet.

yes, but I was referring to Oscar's grind - not the D'Alembert

also, I was postulating that with Oscar you would start betting with $10 and tap out when you have lost your $1500

but flat betting you would average things out and bet $75 every time - not starting at the lowest bet when playing Oscar which is $10

or maybe it's more realistic for the flat bet to be $50 - not sure

I was wondering who would last longer if they both had roughly 𝘵𝘩𝘦 𝘴𝘢𝘮𝘦 𝘢𝘮𝘰𝘶𝘯𝘵 𝘰𝘧 𝘮𝘰𝘯𝘦𝘺 exposed to the house edge.

Quote:ThatDonGuyCorrect me if I'm wrong, but, assuming the first bet is a loss and the number of losses exceeds the number of wins until the last bet of the run, if the player has a run of N losses and N wins in some order, won't the profit be N?

That's correct. Do you feel it contradicts something I said?

Quote:WizardThat's correct. Do you feel it contradicts something I said?

No - I think I misread what you were describing in your table.

Everyone also agrees that to win one must bet more at times. When to do it, is what not all of us agree on.

Quote:ThatDonGuyI 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

Another way to denigrate D'Alembert is to start at one unit. I suggest 5 units to negate the effects of early success.

Hi TDGQuote:ThatDonGuyI 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 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.

Quote:OnceDearHi 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.

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.Quote:ThatDonGuyQuote:OnceDearHi 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.

https://wizardofvegas.com/member/oncedear/blog/5/#post1370

Quote:OnceDearAre 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.

Quote:ThatDonGuyYou 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.

Quote:OnceDearFor 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.

Actually, for flat betting, there is a formula for the probability of reaching the target before losing the entire initial bankroll:

(1 - ((1-p)/p)^B) / (1 - ((1-p)/p)^(B+T))

where p is the probability of winning the initial bet (assuming p < 1/2), B is the initial bankroll (in terms of the number of bets), and T the target profit (again, in terms of the number of bets).

For p = 1/2, the probability = B / (B + T), as you state in your blog post.

Quote:USpapergamesCan someone help me? I just want to post my own form questions but I don't know how.

1. Click on "Forums" in the menu bar at the top of each page.

2. Scroll down until you see the list of topics - for example, this thread is in "Betting Systems" under "Gambling".

3. Click on the topic name to bring up that topic's page.

4. Click on the "New Thread" button at the top of the topic's page.

Direct: youtu.be/ucvz-pDgv50

I'm not saying it's a long run winner

I mean for playing for fun

as far as I know it doesn't have a name

I think it's somewhat similar to what John Patrick (no I don't recommend him) calls 𝙪𝙥 𝙖𝙣𝙙 𝙥𝙪𝙡𝙡 but it is not the same

the progression is 4 bets and if you lose any one or win all you return to the beginning

assuming you win your first 3 bets the progression is:

𝐛𝐞𝐭 𝟏............𝐛𝐞𝐭 𝟑............. 𝐛𝐞𝐭 𝟐...................𝐛𝐞𝐭 𝟒

the worse thing (other than long losing runs) is if you lose your 2nd bet - 3 - which means in the first 2 bets you will lose 2 instead of breaking even if you flat bet

if you win the 2nd bet and lose the 3rd you will win 2 - whereas if you flat bet you would win only 1

if you win the 3rd bet but lose the 4th you will win 2 - which is the same as if you flat bet

if you win all 4 bets you will win 𝟏𝟎 𝐮𝐧𝐢𝐭𝐬...........................𝘵𝘩𝘢𝘯𝘬 𝘺𝘰𝘶 𝘑𝘦𝘴𝘶𝘴...................which means you've won 6 more than if you flat bet