Quote:dihaigBACK FROM BILOXI. PLAYED PASS AND DON'T AT SAME TIME THEN COME AND DON'T COME UNTIL 3 POINTS WERE UP. TOOK 3 TIMES ODDS TO KEEP IN BANKROLL, AND THEN TRIED TO HIT ONE NUMBER BEFORE 7. AS SOON AS HIT ONE NUMBER TOOK DOWN ODDS BETS. VERY SLOW GRIND BUT I DID MAKE MONEY. ANYONE KNOW HOW TO FIGURE WHAT HOUSE ODDS WERE ON THAT PLAY PATTERN? THANKS

To answer your original question, here is a quote from the Wizard of Odds Craps Appendix:

Quote:Wizard of Odds Craps Appendix

Combined pass and buying odds

The player edge on the combined pass and buying odds is the average player gain divided by the average player bet. The gain on the pass line is always -7/495 and the gain on the odds is always 0. The expected bet depends on what multiple of odds you are allowed. Lets assume full double odds, or that the pass line bet is $2, the odds bet on a 4, 5, 9, and 10 is $4, and the odds on a 6 or 8 is $5.

The average gain is -2×(7/495) = -14/495.

The average bet is 2 + (3/36)×4 + (4/36)×4 + (5/36)×5 + (5/36)×5 + (4/36)×4 + (3/36)×4] =

2 + 106/36 = 178/36

The player edge is (-14/495)/(178/36) = -0.572%.

The general formula if you can take x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10 is (-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ]

Combined don't pass and laying odds

The player edge on the combined don't pass and laying odds is the average player gain divided by the average player bet. The gain on the don't pass is always -3/220 and the gain on the odds is always 0. The expected bet depends on what multiple of odds you are allowed. Lets assume double odds and a don't pass bet of $10. Then the player can lay odds of $40 for a win of $20 on the 4 and 10, $30 for a win of $20 on the 5 and 9, and $24 on the 6 and 8 for a win of $20. The average gain is -10×(3/220) = -30/220.

The average bet is 10 + 2×[(3/36)×40 + (4/36)×30 + (5/36)×24] = 30.

The player edge is (-30/220)/30 = -0.455%.

The general formula if you can buy x times odds then the house edge on the combined don't pass and laying odds is (3/220)/(1+x).

Your system is basically hedging to reduce volatility. As you have stated, it is a grind. You are trading lower risk of ruin for wins reduced by hedges. Given enough time, the inherent house advantage of every wager will assert itself (in your case, lot's of 12's on come outs, and 7's before points). Here is a response from the Wizard's Betting Systems page regarding systems similar to your's:

Quote:Wizard's Betting Systems Page

I'd like your thoughts on this craps strategy. I think it's a Patrick system for playing don't pass. Bet one unit on both pass and don't pass. Then lays odds on the don't side. You can stop here or then make a don't come bet. After the dc travels, take the odds off your don't pass bet (if you don't like to lay odds). So now you have a unit on the don't come that pretty much got there with less risk. I know you can never get the advantage over the house, but this seems like a great way to play the don't side. You eliminate the sevens on the come out roll. And only get hurt by the 12; or the 11 on your don't come bet. P.s. Your site is the greatest.

Thanks for the compliment on my site. The best thing I can say about this system is that it composed of low house edge bets. Yes, a 12 will lose the pass bet and push the don't pass on the come out roll, this is where the house edge is. By making the pass bet you are increasing the overall house edge. If you're afraid losing you shouldn't be playing at all. Never hedge your bets. So my advice is to stick to just the don't pass and laying odds. Yes, you'll lose some on the come out roll. However if you don't lose on the come out roll the don't pass bet will usually win. Jan. 2, 2002

http://wizardofodds.com/askthewizard/bettingsystems-specific.html

I can't fault your results, but hedging is generally considered a big no no around here, unless life changing amounts are on the line. All the best to you!