# Craps Rolling Techniques

Alter then so that one die has a six on every side, and the other one has all ones and fives.

For new dice players, just rolling the dice can be a bit daunting, but once you get the hang of it, it's easy. By throwing the dice in the same manner, each time, some shooters get into a rhythm that produces monstrous rolls. The Sevens to Rolls Ratio There are 36 combinations that can be made from of a pair of dice and six ways to attain a seven.

Thanks for the kind words. No, I don't think that wishful thinking helps in the casino, all other things being equal.

The question on the dice influence is a hotly debated topic. Personally, I'm very skeptical. As I review this reply in 2013 I still have yet to see convincing evidence anybody can influence enough to have an advantage.

Dice setting and precision shooting. This IS NOT A SYSTEM, It is a skill and we can teach you. We also have a Basic Craps class. Located in Las Vegas. Classes 5 days a week! Now when you initialize your NewGames Craps object the PlayCraps method will be called as part of that initialization and the game will start. I think the other way is a little more clear and it would allow you to set Craps properties (if they ever exist) before calling your PlayCraps method, but I feel that the use of the constructor here. With Craps, the only Player who must place a bet prior to the first roll of the game is the Shooter (or person who wants to roll the dice). To become eligible to become a Shooter, place a bet on the Pass Line or Don't Pass Bar. Craps Methods of Rolling. Since craps is very dependent on dice results for winning, the most effective winning tactics has to do with dice maneuvers or throwing techniques. We have to focus a bit on craps methods of rolling the dice. Any kind of shooting style may result to a very prolonged roll round in craps.

I'm very skeptical of it. I go over some of the experiments on the topic in my craps appendix 3.

I don’t believe in it. So far I have yet to see a name I respect endorse the method, nor any evidence that it works. While I don’t entirely rule out the possibility I am extremely skeptical of it. I may live in Nevada but when it comes to things like dice setting I’m from Missouri, 'show me' it works.

With ordinary dice, the like those you get in a board game, this is true. However casino dice have inlaid spots. At the factory they drill holes for the spots then insert white colored spots into the holes, of the same density as the die itself. So the die is essentially a perfect cube. Even if they did use ordinary dice from a board game I doubt the bias would be nearly enough to overcome the house edge.

I think there is no such thing as a naturally bad shooter. With the possible exception of a few pros all dice throws can be considered truly random. There are seminars on how to overcome the house edge in craps by precession throwing but I make no claims for or against them. I have yet to see enough evidence either way.

I lost the $1800 to another gambling writer, not Stanford. I would have preferred more rolls but there was an obvious time contraint. Assuming one throw per minute it would take 34.7 days to throw the dice 50,000 times. I wasn’t the one who decided on 500 but it seemed like a reasonable compromise between a large sample size and time. You are right that 500 is too few to make a good case for or against influencing the dice, but 500 throws is better than zero.

## Craps Rolling Techniques

For large numbers of throws we can use the Gaussian Curve approximation. The expected number of sevens in 655 throws is 655 × (1/6) = 109.1667. The variance is 655 × (1/6) × (5/6) = 90.9722. The standard deviation is sqr(90.9722) = 9.5379. Your 78 sevens is 109.1667 − 78 = 31.1667 less than expectation. This is (31.1667 - 0.5)/9.5379 = 3.22 standard deviations below expectation. The probability of falling 3.22 or more standard deviations south of expectations is 0.000641, or 1 in 1,560. I got this figure in Excel, using the formula, normsdist(-3.22).

This is about controlling the dice at Craps. You previously discussed the Stanford Wong Experiment, stating, '*The terms of the bet were whether precision shooters could roll fewer than 79.5 sevens in 500 rolls of the dice. The expected number in a random game would be 83.33. The probability of rolling 79 or fewer sevens in 500 random rolls is 32.66%.... The probability of rolling 74 or fewer sevens in 500 random rolls is 14.41%.*'

The question I have about this bet is that 14.41% still isn’t 'statistically significant' [ i.e. p < 0.05 ] , which is usually taken to mean greater than two Standard Deviations from the Mean -- or a probability of less than a *combined* 5% of the event happening randomly on EITHER end of the series.

How many Sevens would have to be rolled in 500 rolls before you could say that there is a less than 2.5% chance that the outcome was entirely random (i.e. that the outcome was statistically significant) ?

Many Thanks & BTW , yours is ABSOLUTELY the BEST web site on the subject of gambling odds & probabilities that I’ve found .... keep up the good work !!!

Thank you for the kind words. You should not state the probability that the throws were non-random is p. The way it should be phrased is the probability that a random game would produce such a result is p. Nobody expected 500 rolls to prove or disprove anything. It wasn’t I who set the line at 79.5 sevens, but I doubt it was chosen to be statistically significant; but rather, I suspect the it was a point at which both parties would agree to the bet.

The 2.5% level of significance is 1.96 standard deviations from expectations. This can be found with the formula =normsinv(0.025) in Excel. The standard deviation of 500 rolls is sqr(500*(1/6)*(5/6)) = 8.333. So 1.96 standard deviations is 1.96 * 8.333 = 16.333 rolls south of expectations. The expected number of sevens in 500 throws is 500*(1/6) = 83.333. So 1.96 standard deviations south of that is 83.333 − 16.333 = 67. Checking this using the binomial distribution, the exact probability of 67 or fewer sevens is 2.627%.

*You are right that 500 is too few to make a good case for or against influencing the dice, but 500 throws is better than zero.*' Can you describe what you would require from an alleged dice influencer, in an experiment, in order for you to feel confident enough to start betting significant amounts of money on him? I ask because one billion rounds is a good benchmark for 'reliable' results in some blackjack sims. With the most efficient (i.e. requiring least amount of rolls) experimental design you can think of, how many rolls would need to be made by the shooter to be confident he is influencing the outcomes? I know the answer will depend on the skill of the shooter, but you get my drift. If you need a million rolls even under the best case scenario, it’s not going to be a worthwhile endeavor.

There is no definitive point at which confidence is earned. It is a matter of degree. First, I would ask what is being tested for, and what the shooter estimates will happen. With any test there are two possible errors. A skilled shooter might fail, because of bad luck, or a random shooter might pass because of good luck. Of the two, I would prefer to avoid a false positive. I think a reasonable test would set the probability of a false negative at about 5%, and a false positive at about 1%.

For example, suppose the claimant says he can average one total of seven every seven throws of the dice. A random shooter would throw one seven every six throws, on average. By trial and error I find that a test meeting both these criteria would be to throw the dice 3,600 times, and require 547 or fewer sevens to pass, or one seven per 6.58 rolls.

A one in seven shooter should average 514.3 sevens, with a standard deviation of 21.00. Using the Gaussian approximation, the probability of such a skilled shooter throwing 548 or more sevens (a false negative) is 5.7%. A random shooter should average 600 sevens, with a standard deviation of 22.36. The probability of a random shooter passing the test (a false positive) is 0.94%. The graph below shows the possibe results for skilled and random shooters. If the results are to the left of the green line, then I would consider the shooter to have passed the test, and I would bet on him.

The practical dilemma is if we assume two throws per minute, it would take 30 hours to conduct the test. Perhaps I could be more liberal about the significance level, to cut down the time requirement, but the results would not be as convincing. I do think the time has come for a bigger test than the 500-roll Wong experiment.

^{154}, and came up with odds of over 1.5 trillion to 1. One is about 9,000 times more likely to win the Mega Millions lottery than to pull off a 154-consecutive non-seven dice roll marathon. Given how astronomically unlikely this is, and given that people are convicted on DNA evidence that is mere billions to one against being a false match, how much would you suspect cheating, and would you offer to consult the Borgata about this? I already called them, and gave them my name, and told them to do what they want with it. I’m curious as to your thoughts.

First of all, she rolled the dice a total of 154 times, with the 154th roll being a seven out (Source: NJ.com). However, that does not mean she never rolled a seven in the first 153 rolls. She could have rolled lots of them on come out rolls. As I show in my May 3, 2003 column, the probability of making it to the 154th roll is 1 in 5.6 billion. The odds of winning Mega Millions are 1 in combin(56,5)*46 = 175,711,536. So going 154 rolls or more is about 32 times as hard. Given enough time and tables, which I think exist, something like this was bound to happen sooner or later. So, I wouldn't suspect cheating. I roughly estimate the probability that this happens any given year to be about 1%.

Also see my solution, expressed in matrices, at mathproblems.info, problem 204.

I think some of the casinos in Las Vegas are using dice that are weighted on one side. As evidence, I submit the results of 244 throws I collected at a Strip casino. What are the odds results this skewed could come from fair dice?Dice Test Data | |

Dice Total | Observations |

2 | 6 |

3 | 12 |

4 | 14 |

5 | 18 |

6 | 23 |

7 | 50 |

8 | 36 |

9 | 37 |

10 | 27 |

11 | 14 |

12 | 7 |

Total | 244 |

7.7%.

The chi-squared test is perfectly suited to this kind of question. To use the test, take (a-e)^{2}/e for each category, where a is the actual outcome, and e is the expected outcome. For example, the expected number of rolls totaling 2 in 244 throws is 244×(1/36) = 6.777778. If you don’t understand why the probability of rolling a 2 is 1/36, then please read my page on dice probability basics. For the chi-squared value for a total of 2, a=6 and e=6.777778, so (a-e)^{2}/e = (6-6.777778)^{2}/6.777778 = 0.089253802.

### Chi-Squared Results

Dice Total | Observations | Expected | Chi-Squared |

2 | 6 | 6.777778 | 0.089253 |

3 | 12 | 13.555556 | 0.178506 |

4 | 14 | 20.333333 | 1.972678 |

5 | 18 | 27.111111 | 3.061931 |

6 | 23 | 33.888889 | 3.498725 |

7 | 50 | 40.666667 | 2.142077 |

8 | 36 | 33.888889 | 0.131512 |

9 | 37 | 27.111111 | 3.607013 |

10 | 27 | 20.333333 | 2.185792 |

11 | 14 | 13.555556 | 0.014572 |

12 | 7 | 6.777778 | 0.007286 |

Total | 244 | 244 | 16.889344 |

Then take the sum of the chi-squared column. In this example, the sum is 16.889344. That is called the chi-squared statistic. The number of 'degrees of freedom' is one less than the number of categories in the data, in this case 11-1=10. Finally, either look up a chi-squared statistic of 10.52 and 10 degrees of freedom in a statistics table, or use the formula =chidist(16.889344,10) in Excel. Either will give you a result of 7.7%. That means that the probability fair dice would produce results this skewed or more is 7.7%. The bottom line is while these results are more skewed than would be expected, they are not skewed enough to raise any eyebrows. If you continue this test, I would suggest collecting the individual outcome of each die, rather than the sum. It should also be noted that the chi-squared test is not appropriate if the expected number of outcomes of a category is low. A minimum expectation of 5 is a figure commonly bandied about.

Whether or not it is called a valid roll depends on where you are. New Jersey gaming regulation 19:47-1.9(a) states:

A roll of the dice shall be invalid whenever either or both of the dice go off the table or whenever one die comes to rest on top of the other. -- NJ 19:47-1.9(a)

Pennsylvania has the exact same regulation, Section 537.9(a):

A roll of the dice shall be invalid whenever either or both of the dice go off the table or whenever one die comes to rest on top of the other. -- PA 537.9(a)

I asked a Las Vegas dice dealer who said that here it would be called a valid roll, if it was otherwise a proper throw. Although he has never seen it happen, he said if it did the dealers would simply move the top die to see what number the lower die landed on. However, one can determine the outcome of the lower die without touching, or looking through, the top die. Here is how to do it. First, by looking at the four sides you can narrow down the possibilities on top to two. Here is how to tell according to the three possibilities.

- 1 or 6: Look for the 3. If the high dot is bordering the 5, the 1 is on top. Otherwise, if it is bordering the the 2, the 6 is on top.
- 2 or 5: Look for the 3. If the high dot is bordering the 6, the 2 is on top. Otherwise, if it is bordering the the 1, the 5 is on top.
- 3 or 4: Look for the 2. If the high dot is bordering the 6, the 3 is on top. Otherwise, if it is bordering the the 1, the 4 is on top.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.

First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.

Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.

The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.

So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.

Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.

What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:

pr(A or B) = pr(A) + pr(B) - pr(A and B)

You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,

pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3) - pr(4 before 2 and 3) = (3/4)+(3/5)-(3/6) = 0.85.

The probability of not getting the four along the way to the two and three is 1.0 - 0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15*12 = 43.8.

Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.

What is the probability of getting the five before achieving the two, three, or four? The general rule is:

pr (A or B or C) = pr(A) + pr(B) + pr(C) - pr(A and B) - pr(A and C) - pr(B and C) + pr(A and B and C)

So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)-pr(5 before 2 and 3)-pr(5 before 2 and 4)-pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)-(4/7)-(4/8)-(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1 - 83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)*9 = 44.5.

Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.

Here is the general rule for pr(A or B or C or ... or Z)

pr(A or B or C or ... or Z) =

pr(A) + pr(B) + ... + pr(Z)

- pr (A and B) - pr(A and C) - ... - pr(Y and Z)*Subtract the probability of every combination of two events*

+ pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z)*Add the probability of every combination of three events*

- pr (A and B and C and D) - pr(A and B and C and E) - ... - pr(W and X and Y and Z)*Subtract the probability of every combination of four events*Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.

The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.

### Expected Number of Rolls Problem

Highest Number Needed | Probability | Expected Rolls if Needed | Probability not Needed | Probability Needed | Expected Total Rolls |
---|---|---|---|---|---|

2 | 0.027778 | 36.0 | 0.000000 | 1.000000 | 36.000000 |

3 | 0.055556 | 18.0 | 0.666667 | 0.333333 | 42.000000 |

4 | 0.083333 | 12.0 | 0.850000 | 0.150000 | 43.800000 |

5 | 0.111111 | 9.0 | 0.922222 | 0.077778 | 44.500000 |

6 | 0.138889 | 7.2 | 0.956044 | 0.043956 | 44.816484 |

7 | 0.166667 | 6.0 | 0.973646 | 0.026354 | 44.974607 |

8 | 0.138889 | 7.2 | 0.962994 | 0.037006 | 45.241049 |

9 | 0.111111 | 9.0 | 0.944827 | 0.055173 | 45.737607 |

10 | 0.083333 | 12.0 | 0.911570 | 0.088430 | 46.798765 |

11 | 0.055556 | 18.0 | 0.843824 | 0.156176 | 49.609939 |

12 | 0.027778 | 36.0 | 0.677571 | 0.322429 | 61.217385 |

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

The Wizard says that website sounds like a lot of ranting and raving with no credible evidence whatsoever to justify the accusation. I'd be happy to expose any casino for using biased dice, if I had any evidence of it.

If anybody has legitimate evidence of biased dice, I'd be happy to examine it and publish my conclusions. Evidence I would like to see are either log files of rolls or, better yet, some actual alleged biased dice.

Furthermore, if the casinos really were using dice that produced more than the expected number of sevens, then why aren't these detectives privy to the conspiracy out there betting the don't pass and laying the odds?

- 2 or 12: 1,000
- 3 or 11: 600
- 4 or 10: 400
- 5 or 9: 300
- 6 or 8: 200

My question is what is average bonus win?

Click the following button for the answer.

Click the following button for the solution.

Let x be the answer. As long as the player doesn't roll a seven he can always expect future wins to be x, in addition to all previous wins. In other words, there is a memory-less property to throwing dice in that no matter how many rolls you have already thrown you are no closer to a seven than you were when you started.I won't go into the basics of dice probabilities but just say the probability of each total is as follows:

- 2: 1/36
- 3: 2/36
- 4: 3/36
- 5: 4/36
- 6: 5/36
- 7: 6/36
- 8: 5/36
- 9: 4/36
- 10: 3/36
- 11: 2/36
- 12: 1/36

Before considering the consolation prize, the value of x can be expressed as:

x = (1/36)*(1000 + x) + (2/36)*(600 + x) + (3/36)*(400 + x) + (4/36)*(300 + x) + (5/36)*(200 + x) + (5/36)*(200 + x) + (4/36)*(300 + x) + (3/36)*(400 + x) + (2/36)*(600 + x) + (1/36)*(1000 + x)Next, multiply both sides by 36:

36x = (1000 + x) + 2*(600 + x) + 3*(400 + x) + 4*(300 + x) + 5*(200 + x) + 5*(200 + x) + 4*(300 + x) + 3*(400 + x) + 2*(600 + x) + (1000 + x)36x = 11,200 + 30x

6x = 11,200

x = 11,200/6 = 1866.67.

Next, the value of the consolation prize is 700*(6/36) = 116.67.

Thus, the average win of the bonus is 1866.67 + 116.67 = 1983.33.

**Shooting/Rolling Dice - Dice Control**

**Rolling the dice in a game can be a contentious issue. A series of lucky throws may lead to resentment from other players who may believe, rightly or wrongly, that the dice are being held and thrown in a special way. Using these controlled dice throws is sometimes known as honest cheating. Make no mistake, there is nothing honest about it. **

**Casino Craps Shooting**

Brick and mortar casinos require a Craps player to throw or *shoot* the dice against a backboard so that they rebound before coming to rest (online casinos will simulate the rolls using software). This is supposed to ensure the roll is random but many crapshooters try to control the outcome and change the odds. The subject of controlled throws on casino Craps tables is a hot topic. It's sometimes called *rhythm rolling* and *dice setting* because players are said to develop a rhythm of holding and throwing the dice in a set way. If anyone could alter the outcome of the dice, even slightly, they would be rich beyond their wildest dreams and casinos *have* been scammed in the past. If you want to learn dice control and how to set the dice for a Craps game, try the links below.

Casinos have strict procedures for shooting dice that must be adhered to. For more details, see Casino Dice Procedure.

**Switching the Dice **

If a cheat is going to roll crooked dice in a game then the problem of switching them for the straight dice originally being used presents itself. An expert dice * 'mechanic'* can achieve a switch of the dice almost unnoticeably and with such smooth action that even someone who knows a switch is being made will have trouble spotting it. Of course, the crooked dice will have to be very similar in appearance to the straight ones, but this is always possible for a determined scammer.

A dice *mechanic* will substitute crooked or gaffed dice for the real ones and can quickly switch them back once they have gained their advantage. If you suspect a switch has taken place then you could examine the dice yourself. Remember, the dice you are playing with may not be the dice that started the game and cheats will switch the dice back and forth as it suits them. A cheat may even leave his crooked dice behind, just writing his loss off as part of his scam. .

The links below describe the professional techniques used in secretly exchanging dice, unnoticed.

**Techniques for Switching Dice...**

**A very discreet method for switching a pair of dice is... The Casino Thumb Palm Switch.****A very crafty method of switching a single die into a game is... The Mouth Switch.****An easy way to switch dice in a private game is... The Money Switch****An extremely difficult move to learn and perfect is... The Palm Switch****A move used in street games is The Two-Hand Palm Switch.**

**Hand Rolling Dice in the Private Game **

Without doubt it is possible to control throws in a private game and there are a number of techniques around. The best way to prevent a controlled throw is to insist that the dice are rattled and thrown against a backboard, although even this is no guarantee. The Greek shot is a technique of throwing a pair of dice so that they hit a backboard but so the bottom die does not turn. Only a real expert can pull this one off.

**Techniques for Controlled Throws by Hand... **

**The first move cheats learn is rattling the dice without rotation... The Lock Grip****The first, most common, and easiest controlled throw cheats learn is... The Blanket Roll****The next most common controlled throw is... The Whip Shot****A controlled throw with a backboard is a variation on the Whip Shot... The Greek Shot****A crooked throw for private games is... The Spin Shot****A variation on the Greek Shot that doesn't utilise a backboard is the****Reverse Greek Shot.**

** Conventions/Rules For Rolling Dice in the Private Game**

The dice should be vigorously shaken and cleanly rolled/thrown. Should a die not land flat on one side, but tipped at an angle, then it is declared *cocked* and all the dice must be re-thrown. Similarly, if a die is resting on top of another it is declared *stacked* and all the dice must be re-thrown.

**Dice Throwing Cups**

A cup to hold and roll the dice *can* help to overcome the problem of controlled throws and *can* help to ensure that the dice are shaken properly and fall at random. However, the use of throwing cups for dice does not always guarantee the dice fall at random. In fact, some cheats prefer them. Crooked dice can just as easily be thrown and even straight dice can be controlled with practice. Cheats develop techniques where the dice are placed in the cup and are then slid out of the horizontally held cup rather than rolled.

Both the Greeks and Romans used specially made throwing cups and some incorporated crossbars inside to prevent controlled rolls.

Today purpose made * ribbed dice cups* can be purchased with a rubber *ribbed* interior to specifically prevent controlled throws. *Tripped cups* have a lip around the rim to trip the dice as they come out. A *slick cup* is polished and smooth on the inside to facilitate controlled throws. Cheap dice cups are available in plastic while more expensive deluxe dice cups can be bought made from real leather. Dice cups are sometimes referred to as a 'box' by dice players.

The best safety check against controlled throws with a cup is to insist that the dice are vigorously rattled and that the dice cup is completely tipped upside down.

Dice cups are suited to games where more than three dice are being rolled because shaking and rolling more than three dice in a player's hand becomes awkward and difficult. They are also useful for games where the dice must be concealed from other players.

**Techniques for Controlled Throws with a Dice Cup... **

**A relatively simple crook's move with a dice cup is a... Dice Cup False Shake****A crooks move to control a single die of several with a dice cup is... The Dice Hold Up****A variation on the Hold Up is...****The Dice Hold Up & Slide**

**Dice Trays and Bowls**

Dice trays are used to prevent dice rolling off or outside the playing area. Usually wooden, they come in various shapes, often hexagonal, octagonal, rectangular, circular etc, and have a surrounding wall an inch or few high, often with a green felt-lined base. Sometimes a groove is included in the surround for holding out-of-play dice.

The French game of *Shut The Box* is usually played with a traditional wooden box which incorporates a throwing tray.

Chinese dice games (like *C**ockfighting*, *G**rasping Eight*, *Strung Flowers,**Heaven and Nine*, *Pursuing Sheep* and *Sic-Bo***)** are often traditionally played with a flat-bottomed porcelain bowl to throw the dice into. Highly decorative antique dice bowls from the Ming and other dynasties often fetch high prices at auction.

There are crescent-shaped wooden bowls available for storing out-of-play dice and for presenting them to players. You may come across them in casinos on their Craps tables holding a number of special casino dice for the shooter to make a selection from.

**Dice Rolling Devices**

Operators of the game *Chuck-A-Luck* often use a specially made *chuck cage*(an hourglass-shaped wire cage that rotates) to roll the dice. This is supposed to assure players that the outcome is random. However, crooked dice could be substituted for straight ones including *electric dice* with an electromagnet placed underneath the *chuck cage*.

The games of *Grand Hazard* and *Chuck-A-Luck* used to be played using a special *hazard chute*. This was a cone or horn shaped device that often had inclined trips inside to ensure random rolls of the dice. (For more on *hazard chutes* see The Wild West term 'Tinhorn'.)

Spring loaded transparent domes with dice inside have been used for children's dice and board games. They are operated by pushing down on the plastic dome and then releasing it. The base then snaps or springs up causing the dice to jump and bounce around inside.

## Craps Dice Rolling Techniques

Many patents exist for other dice rolling devices that haven't succeeded in gaining widespread use. Maybe some of these ingenious contraptions will be featured in the future on * dice-play*.

## Craps Rolling Techniques

**Beware** of crooked dice and controlled throws. Don't play with strangers but rather stick to a friendly game with people you know well and trust. Should you ever find that you are in a crooked dice game the best advice is to cut your losses and take no further part. Even if you are positive the game is rigged the situation could rapidly turn ugly if you were to accuse someone.