1) Place an even amount on pass and don't pass lines. And place $1 bets on C&E, and any 7. This is a total of $13 on a $5 min table. The idea is to make it pass the come out roll and get a point established.

2) After established point, place a don't come bet (still playing with the amount to minimize loss, but I think it should be enough to cover your pass line bet and the inside numbers of the iron cross and the money lost for established point on come out), so we'll say $25. That's $5 on pass, $3 on C&E, and any 7, $17 on inside numbers. Now place a bet on Yo-leven to cover don't come if eleven rolls out. So in this case $2 eleven. Now place odds on don't pass line that will cover your don't come bet in the event they 7-out. Here we'll place $50 odds on don't pass line. How many times has someone rolled seven right after point established? If it happens now, we'll lose $25 don't come, $2 eleven, $5 pass = $32 loss, but we gain $30 in don't pass line with odds for a total loss of $2.

3) After don't come point is established, remove odds on don't pass line, set inside numbers $17 across 5,6,8 and $5 on come and field. The idea is to hit on any roll going forward. Like anything else, it works really well when you have a shooter that can keep hitting numbers without 7-outting. The set up is costly if the 7-out afte the bets. Pass and don't pass are a wash. Come and field are a wash, $17 from place bets, $2 eleven, $3 C&E for a total of $22 if they 7-out on third roll. If the hand continues, I like fill the numbers with the come bets. If 5,6,8 hits, take the odds on the come bet and wait until it hits enough to cover your original bets ($47) Ofcourse, this changes if you decide to place odds on pass or don't pass lines. Replenish the field and come for every roll. The only flaw is if you lose your don't come bet, then you to decide if your going to place that bet again or just ride that session out.

So the question is,1) if the average number of rolls is 8.something, what are my best options for making a sound decision for which bets to make? The nice thing about this "system" is you have a few options to place odds on on pass if you think they'll make the point or don't pass if you think statistically a 7-out is about to come to try and minimize your loss until you find that "hot" streak. 2) what is the minimum number of rolls needed to see any kind of profit?

Thanks for your comments in advance.

It's not necessary to pick it apart... Yet you can still use mathematics:

1) DC7 is a system that has been proven to have a House Edge in favor of the house.

2) Iron Cross is a system that has been proven to have a House Edge in favor of the house.

3) Field/Come bets have also been mathematically shown to have a house edge (as one would expect).

Any combination of 3 systems that all contain house edges is impossible to come up with a new system that does not contain a house edge.

If you want a deeper mathematical analysis, I actually did completely tear apart the DC7 system to find the flaw and ultimately the house edge in my craps thread here.

So the only "math" piece you should be missing is why the Iron Cross is of negative expectation to the player, which given the DC7 breakdown, shouldn't be too hard to do, if you really want to do it =). That's why I did the DC7 system. I knew it was flawed but wanted to know why, so I did the math for fun. If that's what you want to do, by all means, but if you want to just come with a mathematically sound conclusion... All 3 are house systems, thus you can't combine them in any way/shape/or form to make anything other than another house system.

Quote:CuriousOne2) what is the minimum number of rolls needed to see any kind of profit?

Technically, the answer is 1 - if you roll, say, 11, your Pass and DP wash, and you gain 14 on the 11 while losing 2 on the Craps and 7 bets.

As for #1, your system is too confusing - for example, in step 2, you say to bet $25 on DC, but then you list three different bets (including Pass - and pardon me for asking, but how do you bet Pass when there's a point in play?) for the 25.

Example: $5 pass and DP, $3 on C&E, 7s. Roll is 5, the point is 5. Place $25 don't come, $50 odds on DP, $2 eleven. 2nd roll is a 9. Take down odds on DP, place $6 on 6 and 8 each. $5 come and $5 field. 3rd roll is 6. Place $5 odds on 6 with winnings and take down place bet on 6. So I have $5 come bet on 6 with $5 odds on that bet. Place $5 on come and field. 4th roll is 10. Come bet goes to 10, winnings from the field goes to next come bet and take same bet on the field. 5th roll is 6. Lost the $5 field bet, replenish field with winnings, pocket the few dollars left over. 6th roll is 11. Win on come and field. Pocket $10. 7th roll is a 7. Pass and DP wash, come and field wash, lost all bets, but won $25 don't come bet. So let's add this example up... $3 after 5th roll + $10 after 6th roll + $25 when 7-out equals $38 - $12 original place bets - $10 original come/field bet - $10 come/field when the 6 hit on 3rd roll equals $38 - $32 for $6 profit during that hand.

Math is involved in the thought process. The best bets are the pass, DP, come, and don't come DC. The way I see it, you're hoping to hit a few numbers and then decide if you want to place odds on the pass or DP. The best scenario is when the point is 4 or 10, then I would keep the odds on DP. The worst case is when you lose your DC and the odds on DP. Is this too risky?? Or is there a way figure how many rolls before deciding on which odds to take, Pass or DP?

If a 5 comes up, you gain 7 on the 5, but lose 17 on the 6, 8, and Field; loss of 10 on the roll.

If a 6 comes up, you gain 7 on the 6, but lose 16 on the 5, 8, and Field; loss of 9 on the roll.

If an 8 comes up, you gain 7 on the 8, but lose 16 on the 5, 6, and Field; loss of 9 on the roll.

If a 4 or 10 comes up, you gain 5 on the Field, but lose 17 on the 5, 6, and 8; loss of 12 on the roll.

Also, in all of those cases, your $5 Come bet is now more likely to lose than win - in the case of 4 or 10, you lose 2/3 of the time.

4, 5, 6, 8, or 10 will come up 11/18, or slightly more than 60%, of the time.

Quote:wilbsmittMy brain hurts reading this example, so I'm not sure if you're accounting for the $3 you lost on C/E, any 7 on the come out roll of 5.

+1

If your goal is to minimize the house edge, bet pass (or dp), take max odds, then bet come (or dc) as many times as you like, and take max odds every time.

As other posters have said above, any other bet has a higher house edge, and taking any combination of bets with higher house edges just gives the house a higher edge.

Not sure if you'll appreciate the analogy, but a simple one that I think might resonate is betting 1-35 in roulette, leaving 36, 0, and 00 open. You have SO MANY WAYS TO WIN (indeed, 35/38 times, or 92.1% of the time, you'll be a winner), but you'll only win $1 each time. 3/38 times you'll be losing $35 (when the 36, 0, or 00 hits).

Your craps system effectively does the same thing - you give yourself a ton of ways to win a very small amount of money, but sometimes you'll lose quite a bit. It's easier to see how flawed the system is in roulette because the odds for every bet are effectively the same, but the basic reasoning above holds: take any combination of bets, each having a house edge, and the combination must necessarily have a higher house edge than the single bet you make with the lowest house edge. That's not an artfully-worded phrase, and I'm a bit too tired to make it make more sense, so I'll just give you an example:

System "Winning" combines three bets - A, B, and C. Let bet A have a house edge of 5%, B a house edge of 7%, and C a house edge of 10%. It doesn't matter how much money you put on each bet...your system will have a combined house edge of more than 5%.

If you like to have a lot of bets out there (which it seems like you do), go for the pass/come OR dp/dc, all with max odds, and bob's your uncle.

Quote:CuriousOneI think this has the best opportunity to minimize the losses.

I'm not trying to sound sarcastic, but the best way to minimize losses is not to play. I think most would say they are playing what they realize is a negative expectation game while attempting to maximize their chance of winning. That doesn't require tying yourself in knots with betting systems.

Quote:CuriousOneIs this too risky??

That one we all have to answer for ourselves.

That said, craps is about fun. Give it a try and let us know how it comes out.

Yes, in the meantime you're enjoying yourself and conversing with fellow players. I'm not a math genius and I seen someone said that the example made their head hurt, but I'd like to figure this out mathematically. It may be a waste of time...

there is no discussion of weighted probabilities. It's designed to confuse the brain. Even people who would be able to dissect it just get a headache looking at that stuff.

To the degree that the hedging reduces risk, the risk hedging always reduces is the risk that the House might have a losing session. In other words, you are killing your variance, and with negative expectation it was your only hope.

Did someone give you this system, or did you come up with it by yourself?

Quote:CuriousOneI'll leave this "system" on the simulator for now....

good

and don't hedge!

Quote:CuriousOneYes, in the meantime you're enjoying yourself and conversing with fellow players. I'm not a math genius and I seen someone said that the example made their head hurt, but I'd like to figure this out mathematically. It may be a waste of time...

If you want it figured out mathematically, you need to be very specific as to how each bet is made. Things like "it's up to you to decide" or even "maybe" make calculations impossible. Even "wait until it hits enough to cover your original bets ($47)" is not specific enough.

Here's what I am reading:

Come-out: 5 on Pass, 5 on DP, 1 each on Craps, 11, and 7

Second roll: 25 on DC, 2 on 11, 50 odds on your DP (this assumes you can get 10x odds on DP)

Third roll: remove DP odds; bet 5 each on Come, 5, and Field, and 6 each on 6 and 8

Fourth and subsequent rolls: 5 each on Come and Field; if 5, 6, 8 was rolled on the previous roll, take down the place on that number and bet the winnings as odds on the previous roll's Come bet

Am I missing anything?

Quote:ThatDonGuyI think your main problem is the $17 bets on 5, 6, 8 plus the $5 on Field.

If a 5 comes up, you gain 7 on the 5, but lose 17 on the 6, 8, and Field; loss of 10 on the roll.

If a 6 comes up, you gain 7 on the 6, but lose 16 on the 5, 8, and Field; loss of 9 on the roll.

If an 8 comes up, you gain 7 on the 8, but lose 16 on the 5, 6, and Field; loss of 9 on the roll.

If a 4 or 10 comes up, you gain 5 on the Field, but lose 17 on the 5, 6, and 8; loss of 12 on the roll.

Also, in all of those cases, your $5 Come bet is now more likely to lose than win - in the case of 4 or 10, you lose 2/3 of the time.

4, 5, 6, 8, or 10 will come up 11/18, or slightly more than 60%, of the time.

OOPS - of course you don't lose the 5, 6, or 8 unless you roll a 7.

What it should be is more like:

1/36 - roll 2: +10

1/36 - roll 12: +15 (assume 12 pays triple on the field)

14/36 - 3, 4, 9, 10, 11: +5

14/36 - 5, 6, 8: +7 on the placed number, -5 on the field = +2

6/36 - the dreaded 7: all four bets lose = -22

Total = 1/36 x (10 + 15 + 14 x 5 + 14 x 2 - 6 x 22) = -1/4.

This doesn't take your Come bets into account, all of which lose on that 7.