Craps - Betting Systems
Have you ever heard of the craps pro? I can honestly say that I have been playing this craps method for a year now and over the long run I have been winning 60% of my bets with the casino paying me odds on the place bets. I'm one of the crazies out here that believe that this works. I have tried many systems and spent thousands of dollars believing there must be a way to beat the casino over the long haul. Nothing worked until I purchased this method. I'm a true believer and my wallet shows it, or maybe hundreds of hours at the table is not enough time to prove it to be true I'll let you know how I do over the next year. If you haven't heard of this method than research it and prove me wrong.
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.
Let me begin by saying that of all the gambling related sites, the Wizard of Odds is by far the best. My question has to do with a betting strategy for craps. My question has to do with what some people have termed variance. As you state in your Ten commandments, the house has an edge in the long term, but there are short term fluctuations.
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.
As a moderate craps player who is of course interested in receiving every "comp" possible, could you give advice on the best betting (craps) strategy. Tried to find it in your GREAT web site.
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.
I like your site very much. It is very informative. Thanks for putting out your thoughts. I noticed a betting strategy for craps suggested at Crappers Delight called "classic regression". In it he suggests, placing a 6 and 8, after a point is established. Then taking it down after one of them is hit. He said there are 10 combined ways to make the 6 and 8, but only 6 combined ways to make the 7. It sounds logical, but I've seen where you are able to show, that what appears logical on the surface is not so bright once it is analyzed. What are your thoughts on this strategy and what would the true odds be, if you did take the bets down after one hit?
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.
What is the better system, or which gives me the better chance to win on craps? On the come out roll, I bet $10 on the don’t and $10 on the do, and then when a point comes out I lay full odds against the number. Or is it better to just play the don’t pass, and then lay the odds. I think getting passed the come out roll will increase my chances of winning.
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.
I played craps for the first time the other night and went from $70 to $700 with small bets on the pass odds and field bets. I then lost it all down to $6 because my bets were too large (by the dealers suggestion), and gained it back to $1000 after slowing down. For this being the first time it seems like a very easy game to win if you have patience, was it beginners luck?
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.
In craps, could one gain an advantage over the house by making both a Pass and Don't Pass bet (one unit each) and then playing the Don't Pass odds? Although the occasional 12 would steal one unit here and there, it seems that the seven would have an advantage over the point. At triple odds one could take 3x on the 4&10, 2x on the 5&9 and 1x on the 6&8.
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.
I love to play craps and would like your opinion on a conventional method of play. Pass line and two come bets with full double odds or with one come bet? Does having three different bets working superior to two?
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.
In craps, does the house edge change if you make a don't pass bet then remove it if the point is 6 or 8? What if you remove it if the point is 6,8,5,or 9?
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%.
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.
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.
The Kelly strategy for betting requires a positive edge to be effective. I play craps and I give the house less than a 1% edge. Once a week I get comps of $62. I gamble only 1 1/2 hours and my total betting doesn’t reach $3000. Theoretically I earn approximately $30 per session. Would the Kelly strategy be helpful to me?
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
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).
What is the best way to make money at craps consistently?
Own the casino dealing the game.
Is the "parity hedge" myth in craps true?
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.
I have a question about a series of bets in craps. The strategy is called the "Iron Cross." It involves a bet on the 5, 6, 8, and the field. I read up on this, and found that this particular bet will pay on every roll that is not a 7. I was told that this gives you the lowest house edge. What are all the various odds and what-nots to go along with it?
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%.
What is your opinion of the 5-Count strategy in craps?
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.
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.
Imagine a craps player who takes maximum odds, say 10x, on his pass line and come bets reducing the house edge to 0.18%. He avoids other bets that give the house a bigger edge. He is an "astute" right bettor in every way except this: he is determined to lose. Through bankroll management, and a determined effort to only leave the table a "loser," he hopes he can look back on his years of craps playing and say, "I wagered $1 million at the craps table over the years and gave back $50,000 to the house; because of my ’skill,’ I left 5% at the table." Is he deluding himself? Is he doomed, in spite of his efforts to leave the table a loser every time, to only give the house roughly 0.18% or $1800?
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.