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

Quote:JORDAN1Try 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?

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?Quote:JORDAN1Wait 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.

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

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.

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...

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.

Quote:JORDAN1My 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

Win goals mean NOTHING. NIL. NADA.

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.

Quote:JORDAN1Again, 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

Quote:JORDAN1A "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

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.

Quote:JORDAN1Try 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.

3 numbers with Lay odds.Quote:JORDAN1Try to get three numbers established.

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?

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.

Quote:JORDAN1Wait 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.

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.

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.

Quote:JORDAN1I 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.

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

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.]

The house advantage (combined edge) is the same at 1 roll or 1 billion. What changes is the 'expected loss' (# rolls * average resolved bet)Quote:JORDAN1My 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?

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.

Yes, the 'expected value' because the 'handle' changes ($$$ resolved wagers).Quote:JORDAN1Does "Stopping after each shooter that knocks down any one of my three odds numbers" change anything? Those are my questions.

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

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.

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

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.

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.

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.

so, it appears you do not fear the come out roll 12.Quote:JORDAN1I 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).

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)

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

Old Thread Discussion

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.

Quote:MichaelBluejayOkay, 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.

Quote:mipletIt 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.

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.

very common coding mistake I make and have seen others make too.Quote:MichaelBluejayThanks, miplet and ThatDontGuy, you guys nailed it. Once I fixed that, the program returns the proper -1.4% over 5 million come-out rolls.

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.

now that sounds coolQuote:MichaelBluejayMy 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.

Thanks!

(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.

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.

very well wordedQuote:MichaelBluejayOkay, 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.

"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)

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

You need a large enough sample to ensure the result you get isn't just luck.Quote:7crapswhy is a larger simulation more important than a shorter one?...

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.

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.

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.

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:

I call heads.

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.