JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 11:22:55 AM permalink
Hello. I have been reading several thoughts on gambling and I have spent the last decade trying to come up with a decent system. This is the best I have created, but I cannot check it against a billion rolls. What little I have simulated and even used in a casino have been slightly prosperous, but I would like to know in the long run how bad it boils down to when using a large computer simulation.

Here it is free of charge.

Wait for a shooter to seven out and then make a pass line bet and a don't pass line bet of 5 dollars.

When a number is established, use the following criteria for laying odds against that number.

6 or 8 - \$102 odds
5 or 9 - \$210 odds
4 or 10 - \$300 odds."

Try to get three numbers established. If at any time one of your don't come/don't pass numbers get knocked down, take the other lay odds down and wait for the next shooter. Yes, I know you lose on the 12 every so often, but that is just part of the system.

I know this is a large odds bet, and there are only a few casinos that will allow this kind of action, but it is possible to do this "system."
(I have to find one that offers a "Hundred times odds bets.")

I am hoping for that \$20,000 which was mentioned in several "Wizard of Odds" essays. The most important thing I am seeking though is an evaluation of the system.

Another variable is to never lay odds against the 6 or 8, since these are common numbers to hit, and just use the 4, 5, 9, and/or 10. Does this change the overall evaluation much?

Any information would be appreciated.

Thanks...

Chris
ThatDonGuy
• Posts: 6543
Joined: Jun 22, 2011
September 1st, 2019 at 12:17:58 PM permalink
Quote: JORDAN1

Try to get three numbers established. If at any time one of your don't come/don't pass numbers get knocked down, take the other lay odds down and wait for the next shooter. Yes, I know you lose on the 12 every so often, but that is just part of the system.

And that's the flaw. "Every so often" is, on average, 1 in every 36 comeout rolls.

What are your target parameters - your initial bankroll, and your walkaway point, either in terms of profit or comeouts played?
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 1st, 2019 at 12:23:26 PM permalink
Quote: JORDAN1

Wait for a shooter to seven out and then make a pass line bet and a don't pass line bet of 5 dollars.

When a number is established, use the following criteria for laying odds against that number.

6 or 8 - \$102 odds
5 or 9 - \$210 odds
4 or 10 - \$300 odds."

Try to get three numbers established. If at any time one of your don't come/don't pass numbers get knocked down, take the other lay odds down and wait for the next shooter.

so, you doey/don't for each number? When one number gets 'picked off' by the shooter, you could also remove the don't come bet on that number. Is that what you really want to do?

This 'system' is in no way unique and has already been tried in, I would say, at least 1 million different configurations.
Have yet to hear of any casino losing lots at the craps table to these type of systems.

hope you had fun working on this
winsome johnny (not Win some johnny)
Lovecomps
• Posts: 427
Joined: Aug 12, 2018
September 1st, 2019 at 12:41:27 PM permalink
There are lots of permutations of this and if you should happen to run into a shooter who goes on a hot streak the whole thing falls apart.

With good money management it can work in the short term because you can enjoy yourself and maybe hit a cold period where things become profitable and you pick up a few comped drinks along the way. Speaking as a DP/DC player, just be prepared for a confused or dirty look here and there but don't let it bother you.
The best things in life are not free.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 1:05:40 PM permalink
In one out of every 36 rolls you lose just \$5, which is better to me than just laying against the number and having to pay the \$1 vig for every \$40 (4/10) \$30 (5/9) or \$24 (6/8) bet made. With heavy lay odds, the house has a low advantage, but I was wondering about what it looks like with "a billion rolls." When I set target parameters, I say if I lose three to start or three numbers overall, I stop. End of session. If I lose two and get back to even, I stop, end of session. This is me just considering a "draw occurred" and I wait later to start another session. If I get ahead about \$300 and lose one lay odds number, I stop. End of session. I either take my winnings or if I am manage to get back to even, I consider it a "draw session." I hate to win and end up in the negative at any point, so I don't let that happen. The only time that I can have a strong session is if I get ahead about a \$800 or so and then if I ever fall back to half (like \$400) I end the session. If I get ahead \$1000 then fall back to \$500 I stop, etc. The maximum hit I have found is about the \$3000 mark. If I make it there, I stop. At that point, for me it seems virtually impossible to break that number, for the many sessions that I have experimented with, that is the line that hardly ever gets crossed. Yes, most of the sessions are "draw sessions" with a few losses. The worst starting out loss is \$900. (That is laying odds against the 4 or 10 three times in a row and they are knocked down.)

Those are my parameters. My question is if there are no parameters except for stopping any wagering after one of my numbers are "knocked down", what is the house advantage (percentage wise) over a billion rolls? Is it simply somewhere in the area of .00019 as defined by the craps section in the Wizards of Odds website or is it better or worse over a billion rolls? Does "Stopping after each shooter that knocks down any one of my three odds numbers" change anything? Those are my questions.

Thanks...
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 1:16:02 PM permalink
I am only laying odds, not placing odds. I am hoping for a seven to happen all the time. If I get one number set say a 10, I lay \$300 against the number and if a seven occurs, I win \$150. If I get a 10 and a 6, and a seven occurs, I win my \$150 plus \$85. If I get three numbers set and a seven rolls, obviously I am extremely satisfied. I used to try to get them all set, but I have changed that strategy. Three don'ts numbers against one shooter is enough for me.

Thre great thing is 2s, 3s, 11s, don't help or hurt me, and I avoid the nasty seven on any roll trying to establish my don't odds lays. The 12 is the only monster, and yes, it happens once about every 36 rolls, which I lose my \$5. Every roll of the dice, a 7 can occur, which is great for any don't player.

Again, I know this is just a system like any of them, and I know they are all flawed. I was just wondering how flawed mine was? I have gotten some pretty nice results, but it takes a nasty large bankroll to do this type of gambling.
lilredrooster
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September 1st, 2019 at 1:18:37 PM permalink
Quote: JORDAN1

My question is if there are no parameters except for stopping any wagering after one of my numbers are "knocked down", what is the house advantage (percentage wise) over a billion rolls?

it's not really meaningful to consider that you lowered the house edge by making free odds bets since your expected dollar loss is exactly the same as if you hadn't made the bets at all

the calculation you are trying to make is not meaningful

what is meaningful is your expected dollar loss which can easily be calculated by multiplying how many don't pass bets you made at what dollar amount and then multiplying that number by 1.4% because that is what the house edge is on a don't pass bet

what you have done with your free odds bets is greatly increased your variance - most serious minded gamblers would agree that it is best to keep variance as low as possible

that's not to say that you won't win. you very well might win - because your free odds bets are so much greater than your don't pass bet it's almost but not quite a coin toss as to whether or not you will win or lose

however, it in no way, shape or form constitutes a winning system
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
OnceDear
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September 1st, 2019 at 1:19:17 PM permalink
Win goals mean NOTHING. NIL. NADA.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 1:20:43 PM permalink
A "hot shooter" is never going to get me because I stop as soon as he either makes the point or knocks down one of my Don't come odds numbers. I am looking for a cold table. One final deviation on this is if any one shooter makes three points in a row, I leave the table and stop the session. To me, that possibly means a hot table is coming and I don't want any part of it.

Yes, as a don't player, I keep my head down and collect my chips and move on. If everyone at a table was playing Don't then the house would be a Do player. It does not matter. I don't drink though, I need all my wits to play properly.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 1:24:43 PM permalink
The house edge on a don't pass system is 1.4 percent if no odds are taken. I am taking a large amount of odds so the overall odds are a bit different when figuring percentages.
OnceDear
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September 1st, 2019 at 1:26:13 PM permalink
Quote: JORDAN1

Again, I know this is just a system like any of them, and I know they are all flawed. I was just wondering how flawed mine was? I have gotten some pretty nice results, but it takes a nasty large bankroll to do this type of gambling.

https://wizardofvegas.com/member/oncedear/blog/5/#post1370
Risk a huge bankroll to win a little profit.
Massive probability of small success. Small probability of massive failure
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
lilredrooster
• Posts: 6894
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September 1st, 2019 at 1:27:44 PM permalink
Quote: JORDAN1

A "hot shooter" is never going to get me because I stop as soon as he either makes the point or knocks down one of my Don't come odds numbers. I am looking for a cold table.

you're assuming that a hot shooter will necessarily stay hot and that a cold table will necessarily stay cold

neither of those assumption are accurate

a hot table may stay hot or it may cool off - a cold table may stay cold or it may heat up

if you think you can't get 7 in a row of your lay bets knocked down because you are jumping around to different tables you are mistaken

if it were only that easy to win

but anyway, good luck and have fun with your system
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 2:16:48 PM permalink
I said I had parameters, and I know all parameters are jokes, but I want to know the results if I did not use them. Using my don't system, what would happen if I stayed at one table for a billion rolls? That is my question. If, "taking down all my bets if a shooter knocks one down" is too much of a parameter, then what are my results if I never, ever, in a billion rolls take any bets down. Does that change the math at all? I am still going to try establish at the most 3 don't (laying large odds) numbers. Really, I am just curious on the math side of things...
Lovecomps
• Posts: 427
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September 1st, 2019 at 2:23:14 PM permalink
"Risk a huge bankroll to win a little profit
Massive probability of small success. Small probability of massive failure."

The very definition of Martingdale.

I will confess to doing it once in a long while and have never been burned, but that's nothing to be proud of. I'm prouder of how well I've done at the slots over the years.
The best things in life are not free.
ThatDonGuy
• Posts: 6543
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September 1st, 2019 at 3:10:51 PM permalink
Quote: JORDAN1

Try to get three numbers established. If at any time one of your don't come/don't pass numbers get knocked down, take the other lay odds down and wait for the next shooter.

Am I missing something here?
How can you ever have more than one number in play at a time? Either the shooter sevens out, in which case your odds bet is paid off, or the shooter makes his point, in which case your odds bet loses.

Assuming there's only one number in play at a time, I ran 20 million sessions where you play until you either win 1000 or lose 1000; you end up with a profit 50.3% of the time, but your average loss is more than your average win to the point where your average session is a loss of about 10.

On the other hand, if you play until you win or lose 5000, you end up with a profit only 48% of the time, and your average session loss is 225.

In the end, your average loss will be 5/36 x the average number of comeouts in your session - and the longer you play, the more of those there will be.
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 1st, 2019 at 3:22:22 PM permalink
Quote: JORDAN1

Try to get three numbers established.

3 numbers with Lay odds.
How exactly do you do this?
after the point is set, do you make a \$5 come and dcome bet?
If yes and the next roll is a 12
do you make another \$5 come bet to match the dcome bet trying to get to 3 don't bets with lay odds?
winsome johnny (not Win some johnny)
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 4:41:34 PM permalink
I am sorry I wasn't clear. Once a don't pass number is set, I lay the don't pass number odds, then I go through the don't come to try to get another two numbers. Once I manage to have three, I sit it out and wait until one gets knocked down or a seven rolls, which I would win all three odds bets. Of course, if the don't pass number is set and while I am trying to set the first Don't come and a seven arises, I win the don't pass. If the don't pass is set, and another number is set, and while a seven appears while I am trying to set a third, then I am happy because I have won again (one don't pass, one don't come). The SEVEN is ALWAYS my buddy which is the most common number to roll. But, at any time while I have the first don't pass or any other number set, and I get one knocked down, that is the end. I patiently wait for another roller to throw. My question is, if I sat there for a billion rolls, would I be close to breaking the 50/50 rule, solely because of the larger odds bets that I am making and I am establishing multiple don't numbers for one shooter. I know for sure, smaller lay odds, (2x/3x/5x) does not seem to work as well due to being closer to the .8 or .6 percent numbers. This works sorta on a craps table simulation that the Wizard of Odds provides. I ran \$6000 to \$20000 pretty quickly, but had a \$10000 negative after getting to the \$26000 mark. Things just settled after that. This was after about 3000 rolls or so. If there was one session where a billion were completed, where would I stand? Lets call my bank roll \$100,000 if one has to be established. Would I be broke, about even, or possibly a bit higher?

Again, here is what it looks like

First step - \$5 Pass/\$5 Don't pass
2nd step - establish a number - lay the odd amounts that I had mentioned earlier.
3rd step - go through the Don't Come by placing a \$5 chip on the Come and the Don't come.

At this point if a seven is rolls, I win the don't pass, and start over.
or
if the point come up. Stop. Wait for the next shooter.

4th step - If another number is established through the Don't come lay odds against it.
5th step - Try to set another (3rd) don't number, lay odds against it.
6th step - Wait. You have three don'ts.

At any point if a number you have established gets knocked down. Stop. Tell the dealer to take down your other lay odds. (At worst, you will only lose one don't number per shooter.) Sit on your money and wait for next shooter. He might be rolling for 30 minutes, but don't make a single bet until he is done.

Or after 1 or 2 or 3 don'ts get set and a seven happens, be happy, and rake in your chips.

Wash rinse repeat...what happens when you gamble this way for a billion rolls?

I got this idea from a woman playing craps many years ago. I modified and tweaked it a bit, but again I think it only works well with large lay odds, which most people can't do.

Thanks for any and all the information.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 1st, 2019 at 4:42:33 PM permalink
Yes, I am trying to get the other numbers through the don't come...
ChumpChange
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September 1st, 2019 at 4:54:48 PM permalink
A lot depends on your luck with the shooters. Take it day by day or week by week. Try to build up your bankroll to a few thousand units.
Tanko
• Posts: 1209
Joined: Apr 22, 2013
September 2nd, 2019 at 3:16:01 AM permalink
Quote: JORDAN1

Wait for a shooter to seven out and then make a pass line bet and a don't pass line bet of 5 dollars.

...Try to get three numbers established. If at any time one of your don't come/don't pass numbers get knocked down, take the other lay odds down and wait for the next shooter.

Your first mistake is starting with two bets that combine for a HE of 1/36= 2.77%.

Your next mistake is taking down remaining odds after a loss.

With the 4,5, and 6 working, there are twice as many losing dice combinations as winning ones.

Lose any one of those bets, and take down the odds on the remaining bets, as you do, and you are guaranteed to lose the hand.
charliepatrick
• Posts: 2997
Joined: Jun 17, 2011
September 2nd, 2019 at 3:34:56 AM permalink
Sometimes for fun I'll do three don't comes - yes it's great when the 7 comes just after you've established all the points. I can see the logic that says make and Come and Don't Come so I can get true odss on the lay bet, but it costs you \$5/36 each time. In the long run you're better just making the Don't bet. You also assumed that while making repeated DC bets a 7 is good news, as the other bets win, but an 11 is a loser, although 2 3 are winners.

Typically playing Don't can have a good result if they're aren't any hot rollers and few standoffs, howevere if you'r etaking odds then you're playing an odds-on game, winning mor eoften than not, but when you lose it's a bigger loss.

Enjoy your craps but accept it's a for fun game rather than a way to make money.
RS
• Posts: 8626
Joined: Feb 11, 2014
September 2nd, 2019 at 4:16:19 AM permalink
I don't really understand what you're asking....?

You acknowledge your system isn't a winning system (none are), but then you ask how to improve it or what's wrong with it? There's nothing wrong with it and there is no way to improve it, it does exactly what it's intended to do, AFAICT, and that is to lose. If you want to lose even more and have greater losses, then don't make the odds bet, only pass + DP at the same time. If your goal is to win money (impossible) with your system, well, that's never gonna happen.

I guess a better question is this -- what do you want to get out of your system? If you can explain that, perhaps people can help on making your system "better" at doing that, I suppose.
DeMango
• Posts: 2958
Joined: Feb 2, 2010
September 2nd, 2019 at 4:40:03 PM permalink
If you lose one half bet in 36 rolls, how is the house edge 2.77%?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
ThatDonGuy
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Joined: Jun 22, 2011
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September 2nd, 2019 at 4:48:45 PM permalink
Quote: JORDAN1

I am sorry I wasn't clear. Once a don't pass number is set, I lay the don't pass number odds, then I go through the don't come to try to get another two numbers.

This exposes more money to the house edge of 1.39% on throwing a 12 immediately after a pass & DP, or come & DC, bet.

While this doesn't make the overall house edge any higher, it does tend to burn through your money three times as fast.

On average, you will lose about 14 cents on every \$5 & \$5 pass/DP or come/DC bet, and zero on your odds bets.
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 3rd, 2019 at 6:46:20 PM permalink
Jordan1, I'm working on your computer simulation. In the meantime, ThatDontGuy is correct:

Quote: ThatDontGuy

On average, you will lose about 14 cents on every \$5 & \$5 pass/DP or come/DC bet, and zero on your odds bets.

Presidential Election tracker: https://michaelbluejay.com/election
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 4th, 2019 at 7:45:17 AM permalink
Here's a draft of the computer simulation, which runs but returns the wrong result, because of some bug(s) I haven't been able to track down.

I haven't yet coded the bit about taking Odds bets down, because I want to find and fix the existing error(s) before complicating the code further.

[Edit: I've made some progress tracking down bugs. See post(s) below about results, as soon as I have them.]
Last edited by: MichaelBluejay on Sep 5, 2019
Presidential Election tracker: https://michaelbluejay.com/election
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 4th, 2019 at 9:11:42 AM permalink
Quote: JORDAN1

My question is if there are no parameters except for stopping any wagering after one of my numbers are "knocked down", what is the house advantage (percentage wise) over a billion rolls?

The house advantage (combined edge) is the same at 1 roll or 1 billion. What changes is the 'expected loss' (# rolls * average resolved bet)

most craps players have NO clue as to what a 'house edge' is or what 'expected value' is, just their beliefs, either true, false or a combination of both.

the combined house edge is very east to calculate using the 'perfect 1980'
math will NOT be shown here
30x, 28x and 17x lay odds (4,5,6)
HE = -27/5148 = -3/5720 or about -0.052448% (-0.00052448)
'per bet resolved'
one can not just multiply that by 3 (and the avg bet) for the total ev
as 3 wagers are not always resolved at the same time.

Quote: JORDAN1

Does "Stopping after each shooter that knocks down any one of my three odds numbers" change anything? Those are my questions.

Yes, the 'expected value' because the 'handle' changes (\$\$\$ resolved wagers).
1st simulation results (only 10 million rolls - I used WinCraps)

1st simulation results (I get)
2 players at SAME TABLE with unlimited casino credit - to make any bet.

for player 'L' who does the 3 point Molly on the dont side BUT removes all the odds and refuses to wager on the same shooter once 1 bet losses. (Loser mentality)
\$ bets decided: 393,687,218
\$ ending bankroll: 746,108 (calculate the per roll loss 4 fun)
edge: -0.001895

for player 'W' who does the 3 point Molly on the dont side and cheers when he wins (Winner mentality - does not FEAR losing)
\$ bets decided: 964,729,186
\$ ending bankroll: 1,415,546
edge: -0.001467

player W lost less as a percentage but wagered more than player L.
bet more, lose more

OP wants to beat the billion rolls (computer) and show a profit
never will happen
even IF he plays that many rolls in a casino
show a profit, never will happen (His parameters just makes him FEEL better while losing. GREAT!)

enjoy
winsome johnny (not Win some johnny)
charliepatrick
• Posts: 2997
Joined: Jun 17, 2011
September 4th, 2019 at 2:44:04 PM permalink
As you say normally for each Don't bet you lose 27 / 1980. In the case given, the average total wager (including don't odds) is 26 x initial bet, hence the effective House Edge is 27 / 51480.

I ran a quick simulation of 1m shooters where the odds stayed, but any loss (except on the DC comeout roll) stopped further bets.

The first result gave Win: 153671785 Lose: 154239683 Bet: 412085504 Rolls: 6182487

The average of all 100 trials gave a House Edge of 0.048% (90% -0.07% to +0.18%); and 14 of the 100 trials gave a profit.

When I changed to odds to always win \$100 (i.e. 200/150/12) there were similar ideas a House Edge of 0.093% (90% within -0.04% to +0.22%); and 17 made a profit.

Has anyone else done simulations where, over such a long time, the player can come out ahead? My personal guess is that the bets on 4 and 10 are larger if you lose, so the results might be more affected by how these perform.

Note: For number of rolls, where all the DC bets had lost the shooter immediately stopped rolling and the next shooter started their comeout roll.

Note: Subsequently running a simple the same with no Odds gives a House Edge between 1.273% and 1.433% (av 1.331%). So I guess large Odds make the variance so big that sometime people can come out ahead over long runs.
Last edited by: charliepatrick on Sep 4, 2019
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 4th, 2019 at 5:57:34 PM permalink
Quote: MichaelBluejay

The problems:

(1) Line Bets Placed (including Don't Come) should approximate Line Bets Won, but it's a fair bit off.

(1) Odds Bets Placed should be close to Odds Bets Won, but it's even further off.

I suggest changing the bankroll to something way lower just to test.

I changed the bankroll to \$10 and when a 5 rolled it gave me \$10 lay odds when I had \$0 bankroll.
You must have something set twice for that to happen.

I also have seen stuff like this
Results
Pass/Don't bets: 110 • Won = 130 • (house take = -18.18%)

The Pass/Don't (Doey/Dont) can't show a profit, never
so your code must have that in more than one place or
your video poker style of accounting is not right for the 12 roll

I only looked at it quickly, but have no more time this week to look

too much Football coming this weekend. Gots to get ready...
winsome johnny (not Win some johnny)
Lovecomps
• Posts: 427
Joined: Aug 12, 2018
September 4th, 2019 at 5:58:51 PM permalink
If there was a true winning system then I would have figured it out by now. That's the one truth that I know :-)

My own personal "systems" are

A) Just play the DP/DC straight up and lay max odds

B) Go on the DP on the come out roll and then lay odds while placing an equal sum of the total bet on the 6 and 8.

C) Staying off the line completely and placing the 6/8.

Once in awhile I'll make a field bet or a hardway bet just for kicks and giggles.

To reiterate, if the perfect system is out there I'd have found it by now.
The best things in life are not free.
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 5th, 2019 at 4:49:17 PM permalink
Okay, I got the Pass/Don't Pass to return the proper ~-1.4%. But now I'm trying to run a single Don't Come per round (no Pass, no Don't Pass, no Odds), and I'm getting ~-0.7% over a million spins. This should be dead simple but I can't find the logic error.

Jordan1, since you were the one who requested a test of your system, and since I'm trying to work on it for you, I'd like to ask you to go through the play-by-play on my [url redactedl] page to try to find the error. Update: I'm writing a sim from scratch that does Don't Come only to keep it simple and make sure I've coded it correctly. Will post results.
Last edited by: MichaelBluejay on Sep 6, 2019
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ThatDonGuy
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September 5th, 2019 at 5:25:37 PM permalink
I notice one error: when your first roll on a DC bet is a 7, it should be a loss (and you are counting 11s as a loss, and 2s and 3s as a win, correctly), but you are counting them as wins.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 5th, 2019 at 7:59:14 PM permalink
Thank you for trying to help. It seems you are rewarding 2's and 3's and punishing 7's and 11's, which is normal during a Don't session. I am playing only to have the 12 hurt me on any don't pass or don't come roll. Even if it comes, I am out only \$5. That should only happen once in 36 rolls (average). With large odds bets, and the ability to take them down when I lose one is what my "system" has going for it.

I will try to clarify how I like to play.

COME OUT ROLL (1)
Place \$5 on the pass line and \$5 on don't pass line.
Say a 10 rolls...

Lay huge odds behind/against the 10 (\$300)

NEXT ROLL (2)
Place a \$5 chip in the come and a \$5 chip on the don't come.
Say an 8 rolls...

Lay huge odds behind/against the 8 (\$102)

NEXT ROLL (3)

Place a \$5 chip in the come and a \$5 chip in the don't come.
Say a 4 rolls...

Lay huge odds behind/against the 4 (\$300)

Now wait...
If a 7 rolls you win your laying odds bets. (Congratulations)

If at any time a 4 or 10 or 8 shows. Take the other two bets down (the laying odds) not the \$5 don't that was used to establish the don't number, and wait for another shooter to throw dice. (Hard to do if you the only one at the table.)

Now, If you have only established one don't number (the first 10 for example) and a 7 rolls you win!

Or,
If the 10 is rolled (you lose. boo), at that point you just wait until that shooter Sevens out before you make any bets again.

Same goes for getting two number set. If one of the two is knocked down take the other odds down and wait for the next shooter.

After all said and done you only lose \$5 every time a 12 is rolled, but you are not getting hammered by a potential 7,7,11,7,11 or something similar on don't pass and don't come bets. Yes, a large bankroll is needed. That is another problem for most. This just does not work that well when you only lay double or triple odds. You . play like this on any of the sit-down computer gambling games in a casino either, it has to be done at a table. And, worse yet, most casinos (in Vegas) don't let you lay over 5 times odds. I think I see why due to no house advantage on an odds bet.
For me, its fun, and sometimes rewarding. I have a lot of "Draw" sessions with this due to my not wanting to have to come back from a massive failure. I don't chart tables, I don't watch and analyze any shooter, I don't hedge against any numbers, and I stay away from any proposition bets.
Thanks for any input.
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 5th, 2019 at 8:20:50 PM permalink
Quote: JORDAN1

I am playing only to have the 12 hurt me on any don't pass or don't come roll. Even if it comes, I am out only \$5. That should only happen once in 36 rolls (average).

so, it appears you do not fear the come out roll 12.
over 1 billion rolls, just for the dpass, about 296,296,297 come out rolls.
1 in 36 on average you lose the \$5
296,296,297/36=8,230,453 times \$5 = \$41,152,265 LOSS (it could be more or less)

now we won't even count the dcome bets that lose.

How do U plan on winning that much over what you would win to show a profit?
you have a special lucky charm you have not mentioned yet?

I doubt you know
as you see that \$5 loss as insignificant. (in-sign-if-I-cant)
winsome johnny (not Win some johnny)
Tanko
• Posts: 1209
Joined: Apr 22, 2013
September 6th, 2019 at 1:55:22 AM permalink
Quote: JORDAN1

...Even if it comes, I am out only \$5. That should only happen once in 36 rolls (average).

COME OUT ROLL (1)
Place \$5 on the pass line and \$5 on don't pass line.
Say a 10 rolls...

Lay huge odds behind/against the 10 (\$300)

The DD is a bet the house can only win, and never lose. Which is why they allow it. You can’t win or lose the DP, so all the HE is on the PL. 1/36=2.77%

Sim it with odds on the DP and see what it will cost you, compared to betting the DP alone.

WinCraps sim 35,000 Zumma rolls:

Dooey-Don’t \$10 DP/ \$10 PL 5X odds. 10K Bankroll

High: 10,812
End: 8,855
Low: 4,740

\$10 Dont Pass Only 5X odds 10K Bankroll

High: 11,075
End: 10,685
Low: 5,170

MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 6th, 2019 at 2:41:55 PM permalink
Okay, I'm back to soliciting help (especially from Jordan1) in reviewing the play-by-play to try to find my error(s). I rewrote the whole thing from scratch, having it make only a Don't Come bet to make sure I got the Don't Come working correctly, but it's showing a player edge of ~6% instead of a house edge of ~1.4%. My eyes are glazing over. I'm not sure I've ever had this much trouble with a programming problem, this stuff is usually pretty easy.

Here's the page with the test. Thanks very much to all who looked at the old version, and to any willing to have a gander at the new one.
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miplet
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September 6th, 2019 at 6:42:09 PM permalink
Quote: MichaelBluejay

Okay, I'm back to soliciting help (especially from Jordan1) in reviewing the play-by-play to try to find my error(s). I rewrote the whole thing from scratch, having it make only a Don't Come bet to make sure I got the Don't Come working correctly, but it's showing a player edge of ~6% instead of a house edge of ~1.4%. My eyes are glazing over. I'm not sure I've ever had this much trouble with a programming problem, this stuff is usually pretty easy.

Here's the page with the test. Thanks very much to all who looked at the old version, and to any willing to have a gander at the new one.

It looks like if you have an active DC on the come out, you are paying them on a 7, but not taking them down if the come out roll is one of the the active DC.
“Man Babes” #AxelFabulous
ThatDonGuy
• Posts: 6543
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September 6th, 2019 at 7:14:35 PM permalink
Quote: miplet

It looks like if you have an active DC on the come out, you are paying them on a 7, but not taking them down if the come out roll is one of the the active DC.

Miplet is right - search from the top for "active come points = 10"; the third match (Bankroll \$4975) shows an active come point of 10 that does not lose when your come-out is also a 10, and in fact the DC on 10 wins when a 7 is then rolled.
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 6th, 2019 at 7:35:13 PM permalink
Thanks, miplet and ThatDontGuy, you guys nailed it. Once I fixed that, the program returns the proper -1.4% over 5 million come-out rolls.

Next step is to code in Jordan1's system, but I have some other projects that are pressing. I'll get to it as soon as I can and post here about it.

In the meantime, Jordan1, the house edge doesn't change just because you hedge your bets or place them at different times. Over the long run, the loss on your system will be:

Total Pass / Don't / Come / Don't Wagers X 1.4%

My simulator will show that's actually true for *all* betting systems (at least those based on 1.4%-edge bets), once I get the systems plugged in.
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7craps
• Posts: 1977
Joined: Jan 23, 2010
September 7th, 2019 at 7:46:49 AM permalink
Quote: MichaelBluejay

Thanks, miplet and ThatDontGuy, you guys nailed it. Once I fixed that, the program returns the proper -1.4% over 5 million come-out rolls.

very common coding mistake I make and have seen others make too.

I was looking at your 2nd version (the 3rd one-quickly) and I think that works well, with the corrections.

Come bets and DCome bets are 'down' after they win.
basic craps rules that many do not know about because they know the 'place bet rules' (a win and they are still up to win again' and just think Come/DCome bets are the same, having never made the bets themselves.

Quote: MichaelBluejay

My simulator will show that's actually true for *all* betting systems (at least those based on 1.4%-edge bets), once I get the systems plugged in.

now that sounds cool
Thanks!
winsome johnny (not Win some johnny)
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 7th, 2019 at 12:09:18 PM permalink
Jordan1, here's my understanding of your system. Is this right?

(1) Bet \$5 each on Pass / Don't Pass.

(2) Lay Odds on the Pass point (\$310 on 4/10, \$210 on 5/9, \$102 on 6/8).

(3) If no point has hit yet, then bet \$5 each on Come/Don't Come until 3 #'s established (1 Pass point + 2 Come points).

(4) Lay Odds on each new Come point set, using the same schedule as for the Pass point.

(5) If a # hits, take down the remaining Odds bets. Don't make any new Come/Don't Come bets.

(6) Repeat the whole sequence on each new Come-out roll.

I think that's what you described, but even if so, what happens when Come Point is hit on a Come-Out Roll? Do we take odds down on the other Come Point if there is one? Do we lay odds on the new Pass Point that has just been set? Do we try to establish 3 points during this Come-Out round, or wait for the next come-out round?

BTW, here's how you could figure the average loss for a Pass or Don't Pass on billion come-out rolls:

1B x \$5 wager x 1.41%

It's that simple. For betting Pass / Don't Pass together, double the loss, since you're doubling your bet. The Odds bets have no house edge, so we ignore them in the calculations. Yes, the Odds bets dilute the house edge %, but they don't dilute the overall loss in dollars.
Last edited by: MichaelBluejay on Sep 8, 2019
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MichaelBluejay
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September 8th, 2019 at 11:53:14 AM permalink
Okay, I took my best guess at the system rules (see questions noted above), and finished the simulation. Here it is.

Jordan1, you asked for a billion rolls. That many would choke some web browsers, so I did 2 million come-outs, which is nearly 7 million rolls total. That returns a player loss, and trust me, it won't get better by increasing from 7 million rolls to a billion.
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7craps
• Posts: 1977
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September 8th, 2019 at 7:59:25 PM permalink
Quote: MichaelBluejay

Okay, I took my best guess at the system rules (see questions noted above), and finished the simulation. Here it is.

Jordan1, you asked for a billion rolls. That many would choke some web browsers, so I did 2 million come-outs, which is nearly 7 million rolls total.

very well worded
"the system hasn't even put a dent in the house edge: It's still the expected ~-1.39% for the regular bets, and ~0% on the Odds bets."
A++

I just looked and extended that sim to 100 million rounds (about 337 million rolls)
and got about the same results

photo below (I changed a few things in the code on my machine only)

winsome johnny (not Win some johnny)
Boz
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September 8th, 2019 at 8:23:39 PM permalink
Solved @https://easy.vegas/
7craps
• Posts: 1977
Joined: Jan 23, 2010
September 9th, 2019 at 7:37:53 AM permalink
why is a larger simulation more important than a shorter one?
some would think one way and others another
to some, a simulation of any size is proof their system works

the latest sim is for only 1000 rounds (a large sim to many)

and some sessions easily do show a profit and that easily leads to one believing that their system is a winner to whomever plays it correctly. (a false belief)

photo
winsome johnny (not Win some johnny)
charliepatrick
• Posts: 2997
Joined: Jun 17, 2011
September 9th, 2019 at 7:52:17 AM permalink
Quote: 7craps

why is a larger simulation more important than a shorter one?...

You need a large enough sample to ensure the result you get isn't just luck.

For instance I could try suggesting that betting Red always wins at roulette. If I only ran a simulation of 1000 spins (for simplicity the single zero causes a loss) the average #reds is 486.5 (np) but the sd is 15.8 (SQRT(npq)) - there's a reasonable chance of being ahead by betting red. However if you ran simulations of 1m then the average is 486k but the sd is 499.8, so there's not a ghostly chance of any simulation coming out ahead.

Very similar logic applies for Craps due to it's either win 1 or lose 1 property.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 10th, 2019 at 7:00:17 PM permalink
Hello. I first want to thank you for all of your mathematical evaluations for my "system", but there is one thing that it seems you might have figured incorrectly, and it might be my fault for not being clear. I am "laying odds" not "placing odds". Laying odds is betting a 7 will show before the pass or come number that has been established is made. "Placing odds is betting a number will be hit before the dreaded 7 shows. If I was placing odds, and established three numbers, the hateful 7 knocks down all my bets, which is pretty bad. My thinking is that I always have the 7 working for me, which is the most common number to be rolled on a pair of dice. So when I get one two or three "don't"s set (with the laying of odds) a seven will be always be my hero! If at any time one of my numbers is rolled, by any particular shooter, I take all the "laying odds bets" down and wait for another shooter (or if I am the only one at the table, when I seven out) and start the process over.

According to the Wizards of Odds website a "single" laying or placing of odds bets the formula is (3/220)/(1+X). X is the amount of multiples the lay bet is compared to the original line bet. For example of 100 times odds. \$5 don't pass, \$500 odds = (3/220)/(1+100) this comes to a very small house advantage of .000135. My bets are not at 100 times odds, but let's say I have the funds to make such bets. I was just wondering if anything improves by setting multiple don'ts instead of just the one don't and adjusting to any loss by taking down odds bets that I have not lost. Probably not. Let's say that I try to establish even more lay numbers than 3. What changes on getting possibly 4, 5 or an amazing 6 numbers, from any particular shooter (or whatever one would call the roller's time throwing dice on a come out roll) and lay 100 times odds against all of them. Does that hurt or help my case or does it all just stay at the .000135? Or would I be doing a bunch more harm due to having 26 ways to lose and 6 ways to win if I did manage to establish all six numbers, even though if a seven shows I win 6 bets at once?

Dunno. Up to three works for me. I know gambling is a fool's gambit. I have been working on ways to win for years, and it is fun to think about, but yes I understand the house always has the edge.

Thanks again for everyone's input.
MichaelBluejay
• Posts: 1636
Joined: Sep 17, 2010
September 10th, 2019 at 8:41:11 PM permalink
Hi Jordan. First, please know that I do understand craps. My How to Play Craps article has been on my site for nearly 20 years, and in it I explain the difference between taking odds after Pass/Come bets vs. laying them after Don't Pass/Don't Come bets. My knowledge is also how I knew what the exact results of the simulation would be even before I programmed and ran it.

So, rest assured that my simulation does indeed lay odds, not take them. You can see that in the play-by-play. (Not that it matters. The house take % is the same whether laying or taking.)

As you saw from the Wizard's site, making big odds bets dilutes the house edge, but doesn't erase it. Think of it this way: The average of zero plus any set of negative numbers is always negative. For example, the average of 0, -5, -3, -8, -9, -7, -2 is negative. The odds bet has no house edge (0), and every other bet in craps does have a house edge (negative), so whatever combination of bets you make, however and whenever you make them and how much you bet, your expectation will always be negative. Whatever modifications you think you might make to your system, it's always gonna be the average of zero plus negative numbers, which is why it can't ever win.

Rather than looking at the combined house edge, split the bets up. Your expectation on Pass / Don't Pass / Don't Come together is -1.39%, and your expectation on the Odds bets is 0%. And that's exactly what the simulation shows, as expected.
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ChumpChange
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Joined: Jun 15, 2018
September 10th, 2019 at 9:31:31 PM permalink
Every Passline point number loses on average. Odds bets can win or lose, and maybe make up for the losing average of Passline point numbers. Probably need triple odds minimum.
JORDAN1
• Posts: 13
Joined: Sep 1, 2019
September 11th, 2019 at 5:38:00 PM permalink
I saw your computer simulation, and it is close to what I do, but still there is one variant that, to me seems to be part of the program, which probably doesn't matter. If a shooter makes any of my don't numbers, I stop betting against that shooter, take all of the odds on all the numbers I have established don't bets with, and wait until another shooter turn arrives. The shooter that knocked down one of my don't numbers might roll for 30 minutes or more, but I don't dare make another bet. Your simulation seems to think that I going to bet on the same shooter if he makes a point, and try to start betting don't numbers again once he made it. Again it probably doesn't matter though. I guess the \$5 I lose every 36 rolls or so is what cost me so much in the long run. If odds bets have no house advantage, then in a million rolls the eventual result would be a tie, but \$5 times 1,000,000/36 is where the problem lies with my system.

Another question would be just a single coin toss bet. No house advantage, just a 50/50 bet. Is there a way to come out ahead on such a game? I am not playing for a million flips in a row, but something to the effect of sessions. Everyone hates sessions, but don't they deserve some merit, especially in a game of coin toss? What would happen if you played a coin toss game with four simple rules.

1 -Whenever you lose three coin tosses in a row, stop a session. (Could be a hundred flips occur but you only stop when you lose three in a row)

2 - If you lose three in a row starting out a session, stop the session.

3- Lose a total of three tosses (like this -1, 0, -1, -2, -1, -2 ,-3) stop the session.

4 - If you get down to -2 and get the losses back to 0 (even) stop the session.

Is there any way that one could turn this into a winning game?

Example:

First session:
Lose three in a row (-3) (Stop)
Second session:
Win 20 out of 30 (+10) (Stop because I lost three in a row)
Third session
Win 6 out of 10 (+4) (Stop because I lost three in a row)
Forth session:
Lose three in a row to start (-3)
Fifth Session:
Lose 3 out of 6 (-3)
Sixth Session
Lose the first two flips (-2) but manage to get back to even (0). Stop the session.

These sessions would add up to 14+/-9 for a net of +5.

The reason behind my question if the house odds can get down to .0004, isn't the game as close to a coin toss as possible? So if it is, then couldn't sessions be used to possibly obtain a possible (ever so slightly) advantage? That simulator you could be used to figure this (with max lay odds on don'ts). Probably wishful thinking, but it is thinking....

Thanks again for all of your help. Sorry to be so persistent.