vox
Joined: Jun 24, 2013
• Posts: 11
April 14th, 2018 at 10:37:38 PM permalink
I'm curious to know if anyone has tried using the D'A'embert system on the free odds bet only. (100x odds needed.)

For example, place \$10 on pass and don't pass. When a point is established, take free single odds. If you lose, on the next point, take double odds, then triple if another loss, etc, just a traditional D'Alembert. Stop whenever a profit is shown for the series, not when the system returns to single odds, so in a series with the points 6-8-10, and the results being L-L-W, you would stop after the third bet and start a new one on the next roll.
helpmespock
Joined: Mar 6, 2010
• Posts: 386
April 15th, 2018 at 7:47:04 AM permalink
Sorry I'm too lazy to run all the numbers and there are people much better at it than me here, but I believe the answer lies along these lines.

On a doey-don't you lose \$10 every time a 12 is rolled on the come-out.

There is no house edge on the free odds bet once the point is established so ultimately neither side is winning or losing money after a large number of trials.

You're more likely to lose than win once the point is established on the pass line. What happens if you've lost 10 times in a row and now you've hit max odds?

According to the Wizard's Craps Appendix 1 the probability of winning once the point is established is 9648/35640 so the probability of losing is 25992/35640 so the probability of losing 10 times in a row (25992/35640)^10 is 0.04256 or just under 1 in 24 times.

--helpmespock
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4463
April 15th, 2018 at 7:48:43 AM permalink
Without detailed analysis, I can see two problems with this.
First, 12s on the comeout.
Second, the fact that DP odds are less than even money.

Question: do you take odds on both sides (i.e. if it is double odds, 20 on the pass bet and another 20 on the don't)?
If that's the case, then assume the point is 4, and you have 20 with 40 odds on pass and 20 with 40 odds on don't.
Roll a seven; the pass bet loses \$60, but the don't gains only 40 (20, plus 20 odds at 1-2).

I assume in your system, you apply D'Alembert by increasing after a point is missed and decreasing after a point is made?
helpmespock
Joined: Mar 6, 2010
• Posts: 386
April 15th, 2018 at 8:04:07 AM permalink
I assumed this was all being done on the pass line.

As ThatDonGuy points out you could also try this on the don't pass, but as he points out you're betting more to win less. Again lay odds on don't pass have no house edge either so over a large number of trials neither side is winning money. The problem of hitting max odds after too many losses in a row also applies to the don'ts.

--helpmespock
Ace2
Joined: Oct 2, 2017
• Posts: 773
April 15th, 2018 at 8:45:50 AM permalink
I rarely bet the don't but I think that with a \$10 line bet and a point of 6 the minimum single odds bet is \$12.

So you would push every time except for 12 on the come out. You’d never make a profit just very low variance.
It’s all about making that GTA
troopscott
Joined: Apr 3, 2017
• Posts: 394
April 15th, 2018 at 8:57:46 AM permalink
A \$5 table would be less outlay
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4463
April 15th, 2018 at 9:04:02 AM permalink
I am trying to simulate this, but invariably, I get to a point where you get a run of 1 million or more comeouts and are still well behind, with your odds bets maxed out. Is there a bankroll and/or time limit of any sort?
odiousgambit
Joined: Nov 9, 2009
• Posts: 8407
April 15th, 2018 at 9:05:34 AM permalink
If you've ever bet big, 10x or more, with free odds you will soon come to the realization what happens on the pass line is near immaterial to your results over any period of time you can relate to. So don't waste your time with doey don't.

a progressive system using free odds up to 100x raises the variance to the moon.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4463
April 15th, 2018 at 9:35:56 AM permalink
I ran a simulation of 100 million sessions under the following conditions:
The pass and DP bets are 30 each - this is the only way to get true odds on both sides on all of the numbers, since the 6 and 8 pay 6-5 odds on the pass and 5-6 on the don't.
I put a bankroll cap of 600,000 (which would be 10,000 at a \$5 table, but \$5 odds on the don't with a point of 6 only pays 4-5).
Also, when the odds bets exceeded 100x the pass bet, I kept the session going with the maximum odds bet.

Of the 100 million sessions, 10,000 of them busted the bankroll. Yes, that's only one out of every 10,000 sessions, but most sessions will make far, far less money.

36,000 sessions (about 1 out of every 2800) required 10,000 or more comeouts to either become profitable or bust. Are you willing to grind it out that long?

EDIT: After 200 million sessions, changing the cap from being behind 600,000 to having the session last 1 million comeouts, 1 out of every 137 sessions reached the 1 million comeouts point, and the largest bankroll deficit was 177,540 times the pass bet value (so even on a \$5 table, that's \$887,700).
Last edited by: ThatDonGuy on Apr 15, 2018
vox
Joined: Jun 24, 2013