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Is this a winning craps system?1 October 2016
Here is a system I am unsure of. Please let me know.
First, place table minimum bets on all numbers, say 15 is table minimum.
Next place 6 dollars each hop 7 for a total of 18 dollars. Now roll the dice.
Now put 96 dollars on come for the next roll and match the 96 with a 2 3 12 and a yo bet to obtain the 96 back in case one of those are rolled.
Add all of your numbers up 96+15+18+18+15+15= 177
Now put 177 on come. Match that with 2 3 11 and 12 prop bets to cover the 177
Continue this process and you will win every time. I play at Horseshoe and the table limit is 100,000 hop 7's pay 15-to-1
Let’s take a look at what is really happening.
This betting progression seems to have some holes in it – or at least I am unsure about some of what is to be done. When you say place a table minimum on all numbers in the first line, do you mean the “box” numbers of 4, 5, 6, 8, 9 and 10, or are you including craps (2, 3, 12) and “yo” (11) numbers also?
I don’t see them mentioned until the second and following bets. I also am assuming you are not rolling the dice as you would need a pass line bet also.
Here is my take assuming you are not rolling the dice and you are only placing standard box numbers.
If you were to just place bets on the box numbers of 4, 5, 6, 8, 9 and 10 you would have $96 placed and $18 on the 7’s hopping. I assume you work the place bets. If you hit a box number you will win from $18 to $27 depending on the number hit and you would lose the $18 on the 7’s hopping, so the worst you would do is break even.
If a 7 was rolled you would win $90 on the combination that was rolled as well as the original $6 bet on it. You would lose the $96 bet on the box numbers and the $12 bet on the other two hopping 7 bets. This results in a net loss of $12 if a 7 is rolled.
Assuming you don’t bet on the 2, 3, 11 or 12, you will lose the entire bet spread of $114 ($96 for place bets and $18 for the hopping 7’s) should one of these numbers be rolled. If you do bet on these numbers, you will not collect enough on the winning box numbers or hopping 7’s to cover the additional cost of betting them.
In short, this betting system will not work. There is no need to continue with the progressions after the first bet as you will lose money on the first bet on average.
No betting system can overcome the house edge in craps.
The casino wins when you lose by collecting all your losing bets. The casino also wins when you win by not paying true odds for the winning number. This happens with the payouts for winning box numbers and hopping the 7’s. The 6 and 8 are paid at a rate of 7-to-6. True odds are 6-to-5. The 5 and 9 are paid at a rate of 7-to-5. True odds are 3-to-2. The 4 and 10 are paid at a rate of 9-to-5. True odds are 2-to-1. The hopping 7 bets are paid at a rate of 15-to-1. True odds are 17-to-1.
Let’s face it. The casino is going to win – period. No betting system can overcome it. There is either some situation that is overlooked or you will hit the table limit with your progression. You may have good luck for hundreds and possibly thousands of rolls, but eventually you will run into the wall of the table maximum. You will lose – and you will lose big!
The only way to get an edge at craps is to learn a controlled throw. Any betting system is ultimately doomed to failure. Sorry, Rob, but that is the way it is.
May all your wins be swift and large and all your losses slow and tiny.
Jerry “Stickman” is an expert in craps, blackjack and video poker and advantage slot machine play. He is a regular contributor to top gaming magazines. He authored the video poker section of Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker! You can contact Jerry “Stickman” at firstname.lastname@example.org.
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