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.