Unbeatable Craps System
However, with your bubble craps idea, there isn't any data/information out there yet (that I know of). So perhaps you simply need the data. Go start recording throw after throw for 10k throws and make sure you're 'controlling' the throw perfectly for every single one.
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- W7, You don't say what 'Mr. Gay's 77 Craps System' is. Perhaps if you post what it is, others might reply. Have heard/read of the 'Triple 777 Craps System' by King Capper, who claims that, 'in my over 20 years of gambling experience ever seen any system that has ever beat the Zumma Craps Tester (over 300 hours of real craps decisions) as well as the 72 Hours at the craps table tester.'
- And does any of it prove beyond a shadow of a doubt that casino craps can’t be beaten? We’ll probably never have that kind of proof. But it does prove beyond a reasonable doubt that casino craps is unbeatable, and in an existential world that is enough for reasonable men. So, as for me, I’ll leave the snake oil to the.
- Winning craps players know that craps is a casino game with an inherently unbeatable house edge (unless you believe the claims of dice control advocates). They focus on having fun and enjoying their entertainment. They don't gamble with money they can't afford to lose. They don't count on craps winnings to pay their bills.
Here i am going to explain step by step a system whih i have been using for a while.
Introduction: The Martingale is the most popular betting system in the world. It is also called the “doubling-up system” and can be applied to any even moneyIs a bet with 2.00 odds. It's named like this because returns a profit equal to the stake. bet. When playing a casino game, the amount of the first bet is simply doubled after each successive loss, until one wins. After each win, the bet is reduced to its initial value and remains there until the next loss. When one wins, one wins one unit. For example, an 11-step martingale would be : 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 = 2047 units. That is a large amount to risk simply to win one unit ! Read More about Martingale System Here
If you use this strategy for a casino game such as roulette, craps or at sports betting, the danger is that eventually a long string of losses will keep increasing the amount of your bet until the table limit or the betting limit is reached.
We will use a modified martingale. We will start by betting the amount to win our target. We will be doing this until a win. If we win our first bet, we start over again. If we don’t win our first bet, we will bet the sum of the first 3 bets. So a win will erase 3 loses. Every time we lose, we keep betting for the same target. Only when we win, we will bet the sum of the first 3 bets (remaining bets). With this system, when we lose, we go one step backwards, and when we win we go 3 steps ahead.
Bankroll: I can’t tell you exactly what bankroll you need because the oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are changing everyday. But I think that a bankroll of 200 units would be reasonable.
The bet size will be calculated with the odds. There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. offer by the bookmakers.
The formula is: B = T / (IT-1)
- B = Amount to bet
- T= Target (amount you want to win)
- IT= International oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won.
Unbeatable Craps System Count
Let’s say you want to win $50 at 2.20. $50 / (2.20-1) = $50 / 1.20 = $41.66 L-L-L-W-L-L-W-WHere’s how it works: You need to bet $41.66 to win $50. $41.66 X 2.20 = $91.66 – $41.66 = $50
Let’s see what happen with a 37.5% winning rate and a target of $50.
Day 1
Write down your target of $50 you want to make. You bet $54.95 to win $50 The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 1.91 You lose. Write down the $54.95 bet next to the $50 you had written down. It should look like this. $50 + $54.95 Day 2
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 2.15 When you lose, your target still the same. So, here it’s $50 again. Your next bet is TARGET / (2.15 – 1). So it’s $50 / 1.15 = $43.48 You should bet $43.48 You lose.
Write down the $43.48 bet next to the previous two bets. Now it looks like this. $50 + $54.95 + $43.48 147
Day 3
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 1.83 You lose. Write down the $60.24 bet next to the previous three bets. Now it looks like this. $50 + $54.95 + $43.48 + $60.24 Our target is the same until a win. Your next bet is TARGET / (1.83 – 1). So it’s $50.00 / 0.83 = $60.24 You should bet $60.24
Day 4
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 2.10 Your next bet is TARGET / (2.10 – 1). So it’s $50.00 / 1.10 = $45.45 You should bet $45.45 You win. Cross out the first target you have written down. You do this (cross out the target) only after a win. That leaves you with the $54.95 bet, the $43.48 bet and the $60.24 bet. Now it looks like this. $50 + $54.95 + $43.48 + $60.24
Day 5
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 1.99 After the first win, the new target is the sum of the first three numbers we have written down. In our case we have $54.95, $43.48 and $60.24. So, our next target is $158.67. The next bet is TARGET / (1.99 – 1). So in our case it’s $158.67 / 0.99 = $160.27 Your next bet should be $160.27 You lose. Write down this bet next to the previous three bets. It looks like this. $54.95 + $43.48 + $60.24 + $160.27 148
Day 6
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 1.75 Your next bet is the TARGET / (1.75 – 1). So it’s $158.67 / 0.75 = $211.56 Your bet should be $211.56 You lose. Write down this bet next to the previous four bets. $54.95 + $43.48 + $60.24 + $160.27 + $211.56
Day 7
The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 2.45 Your next bet is the TARGET / (2.45 – 1). So it’s $158.67 / 1.45 = $109.43. Your bet should be $109.43. You win. Cross out the first three bets (OUR TARGET). That leaves us with… $54.95 + $43.48 + $60.24 + $160.27 + $211.56 Day 8 The oddsThe odds for an event.There are three types of odds:American, fractional and Decimal. They determine the profit that can occur if the bet won. are 1.90 You win. The session is over. You have recovered from all your loses and made a $50 profit. Your next target the sum of the two remaining bets / (1.90 – 1). Since we don’t have 3 numbers written down, we bet on only the sum of the two remaining bets. So it’s $371.83 / 0.90 = $413.14 You should bet $413.14
You won 3 bets and lost 5 bets. Winning rate: 37.5%
Do you like this. You are welcomed to try. If you have any question please comment and reply.
You can find more systems here.
Although I have tested a lot of systems, I don't need to test all of them to know they are all worthless. No system can ever pass the test of time. It is not unusual to win for a while with a system, but if you keep playing the odds will eventually catch up to you and you will fall behind.
For more information about the futility of betting systems, please see The Truth about Betting Systems.
A casino I played at had the 3,4,5 odds system where you were allowed 3x on the 4 and 10, 4x on the 5 and 9 and 5 x on the 6 and 8. I feel that with this 'system' of placing odds, you reduce the fluctuations (with respect to standard 5x odds on all numbers) in your bankroll, and change the distribution of net gain/loss per session, i.e. you would produce a sharper peak located slightly more to the loss side than with 5x odds. Is this so, and could you put some numbers to it?
That is known as 3-4-5X odds, and is now pretty common. The following table shows all the possible outcomes, for the pass and odds combined, with full odds.
Return Table with 3-4-5X Odds
| Event | Pays | Probability | Return |
|---|---|---|---|
| Pass line win | 1 | 0.222222 | 0.222222 |
| Pass line loss | -1 | 0.111111 | -0.111111 |
| Point of 4 or 10 & win | 7 | 0.055556 | 0.388889 |
| Point of 4 or 10 & lose | -4 | 0.111111 | -0.444444 |
| Point of 5 or 9 & win | 7 | 0.088889 | 0.622222 |
| Point of 5 or 9 & lose | -5 | 0.133333 | -0.666667 |
| Point of 6 or 8 & win | 7 | 0.126263 | 0.883838 |
| Point of 6 or 8 & lose | -6 | 0.151515 | -0.909091 |
| Total | 1.000000 | -0.014141 |
The standard deviation per pass line bet is 4.915632.
Unlike most gambling writers, I don't put much emphasis on betting strategies. Assuming the same game and bet, there is no one right or wrong strategy. They all behave differently in the short run, but in the long run you will give the house the same percentage of total money bet.
This is similar to a question I got last week. Yes, it is true that there are ten ways to roll a 6 or 8, and six ways to roll a 7. However, one must not look at the probabilities alone, but weight them against the payoffs. The place bet on the 6 and 8 pays 7 to 6 odds when fair odds would pay 6 to 5. By making six unit place bets on the 6 and 8, and taking the other down if one wins, the probability of winning 7 units is 62.5% and the probability of losing 12 units is 37.5%. If the player must cover both the 6 and 8, then the place bet is the way to go. This rate of return isn't bad but could be better. For the player who puts a priority on minimizing the overall house edge, the best strategy is to make combinations of pass, don't pass, come, and don't come bets, and always take the maximum allowable odds.
The better system is to bet on the don't pass only and take full odds. Yes, betting on both does increase you chances of winning on any one bet. However you are suffering a higher combined house edge by betting on both the pass and don't pass and it will cost you in the long run.
Yes, it was luck. It helped that you stuck to the low house edge bets. However, next time, make the line bets with odds only, and don't bet the field, especially if it pays 2 to 1 only on both the 2 and 12.
No combination of bets can give the player an advantage. In your example you would lose one unit for every 12 on the come out roll. You don't make up for it laying the odds. While you usually win laying the odds, you have to risk more. In the end, laying the odds has zero house edge.
As long as you are backing up your pass and come bets with full odds, it doesn't make any difference how many come bets you make. However, it does reduce the overall house edge to keep the odds on your come bets working on the come out roll.
You should never remove a don't pass bet after a point is made! Once a point is made of 6 or 8 the don't pass has equity of 9.09% of the bet amount, which you would be throwing away by taking the bet down. The equity of a don't pass bet on a point of 5 or 9 is 20%, and on a 4 or 10 is 33.33%.
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.
I am a novice, just starting to play. My question concerns the 'Five Count Doey/Don’t' System. The way I understand the system:- Wait until the shooter establishes a point.
- Play both come/don’t come (same amount). Until you have a maximum of four numbers
- After the shooter has rolled five times without rolling a 7, take odds on all your numbers on the front side.
The rationale: Limit your exposure until you find a 'qualified' (five rolls without a 7) shooter. Only betting the odds so there is no 'house edge'! Can you compare this system with just playing pass/come and taking the odds?
As I stated in the other craps strategy question you are only mixing another house edge bet into the game by betting on both the pass and don’t pass, or come and don’t come. It is also not going to help to wait until a shooter hits five points. The probability of making a point is the same for me and you as it is for somebody who just threw 100 points in a row. In other words, the past does not matter. As I stated to the person who asked the other question (whom I think may also be you) don’t make opposite bets, just stick to either the do or don’t side and always back up your bets with the odds.
Unless bankroll preservation is very important to you then Kelly betting won’t help. I would just flat bet. Nice strategy to milk the comp system.
The American Mensa Guide to Casino Gambling has the following 'anything but seven' combination of craps bets that shows a net win on any number except 7. Here's how much MENSA advises to bet in the 'Anything but 7' system:- 5- place $5
- 6- place $6
- 8- place $6
- field- $5
- total= $22
Unbeatable Craps System
They claim the house edge is 1.136%. How is that possible if every individual bet made has a higher house edge?Good question. To confirm their math I made the following table, based on a field bet paying 3 to 1 on a 12. The lower right cell does shows an expected loss of 25 cents over $22 bet. So the house edge is indeed .25/22 = 1.136%.
Mensa Anything but Seven Combo
| Number | Probability | Field | Place 5 | Place 6 | Place 8 | Win | Return |
|---|---|---|---|---|---|---|---|
| 2 | 0.027778 | 10 | 0.000000 | 0.000000 | 0.000000 | 10 | 0.277778 |
| 3 | 0.055556 | 5 | 0.000000 | 0.000000 | 0.000000 | 5 | 0.277778 |
| 4 | 0.083333 | 5 | 0.000000 | 0.000000 | 0.000000 | 5 | 0.416667 |
| 5 | 0.111111 | -5 | 7 | 0.000000 | 0.000000 | 2 | 0.222222 |
| 6 | 0.138889 | -5 | 0.000000 | 7 | 0.000000 | 2 | 0.277778 |
| 7 | 0.166667 | -5 | -5 | -6 | -6 | -22 | -3.666667 |
| 8 | 0.138889 | -5 | 0.000000 | 0.000000 | 7 | 2 | 0.277778 |
| 9 | 0.111111 | 5 | 0 | 0.000000 | 0 | 5 | 0.555556 |
| 10 | 0.083333 | 5 | 0.000000 | 0.000000 | 0.000000 | 5 | 0.416667 |
| 11 | 0.055556 | 5 | 0 | 0.000000 | 0.000000 | 5 | 0.277778 |
| 12 | 0.027778 | 15 | 0.000000 | 0.000000 | 0.000000 | 15 | 0.416667 |
| Total | 1 | -0.25 |
The reason the overall house edge appears to be less than the house edge of each individual bet is because the house edge on place bets is generally measured as expected player loss per bet resolved.
However, in this case the player is only keeping the place bets up for one roll. This significantly reduces the house edge on the place bets from 4.00% to 1.11% on the 5 and 9, and from 1.52% to 0.46% on the 6 and 8.
For you purists who think I am inconsistent in measuring the house edge on place bets as per bet resolved (or ignoring ties) then I invite you to visit my craps appendix 2 where all craps bets are measured per roll (including ties).
Craig from Los Angeles
No. I had to Google this to find out what this is. This appears to me to be an amusing urban legend about some young scientists who developed a winning craps system. The story is told at Quatloos. I would file this under other fictional stories that have become mistaken for fact, like Joshua’s missing day. As I have said hundreds of times, not only can betting systems not beat games like craps, they can’t even dent the house edge.
If the player bets $5 on the field and 5, and $6 on the 6 and 8, then he will have a net win of $2 on the 5, 6, and 8, $10 on the 2, $15 on the 12, and $5 on the other field numbers, assuming that the 12 pays 3 to 1 on the field. The player will lose $22 on a 7. On a per roll basis, the player can expect to lose 25 cents compared to $22 in bets, for a house edge of 1.136%.
This begs the question, why is this lower than the individual house edge of each bet made? It’s not. The reason it seems that way is the result of comparing apples to oranges. The house edge of place bets is usually expressed as the expected loss per bet resolved. Looking at the individual bets on a per-roll basis, the house edge on the 5 is 1.11%, and on the 6 and 8 is 0.46%, according to my craps appendix 2. Comparing apples to apples, the house edge is a weighted average of the house edge on the field, 5, 6, and 8, on a per-roll basis, or (5/22)×2.778% + (5/22)×1.111% + (6/22)×0.463% + (6/22)×0.463% = 1.136%.
For the benefit of other readers, the 5-Count is a method of slow-playing craps, as discussed in ’Golden Touch Dice Control Revolution’ by Frank Scoblete and Dominator. As the book states, it is a way of betting nothing on some rolls, reducing your expected loss on random shooters, while still getting the full comp value of table time.
The way the 5-Count works is you start counting rolls as soon as a new shooter throws any point number. When you get to five rolls after you start counting, the shooter is deemed worthy, and you start betting. However, you if the 5th roll is not a point number, it doesn’t count.
The book says you will only be betting 43% of the time, which I agree with. It is common for craps players to not bet, bet small, or bet the don’t pass on new shooters, as a way to qualify him. Once a shooter has made a point, or thrown lots of point numbers, the other players will gain confidence in him, and start betting with him. So, this kind of strategy seems natural. When casinos rate your average bet, they don’t lower the average for betting nothing some of the time. However, sometimes they will dock your time, especially if you are betting big.
Unbeatable Craps System
An alternative strategy is to wait until the shooter makes a point. Under this strategy you will only be betting 40.6% of the time, less than the 43.5% with the 5-Count.
Unbeatable Craps System Reviews
Yes! I’ve said many times that betting systems not only can’t beat a house edge game, they can’t even dent it. That includes denting it in the house’s favor. In other words, even if he tried to lose, he still only gives up 0.18% over the long-run, under your assumptions. Over a shorter time, he probably could do this, but not over 'years.' Some might argue that to deliberately lose, the player should do an anti-Martingale, where the player kept pressing his bets until he lost. However, a problem there is that a winning player will eventually reach the table maximum, which is rather low in craps. It just goes to show how futile betting systems are.