Craps Controlled Shooting

1. Craps Controlled Shooting Range
2. Craps Dice Throwing Tips
3. Craps Controlled Shooting
4. Craps Controlled Shooting Equipment
5. Craps Controlled Shooting Ball
6. Craps Controlled Shooting

Most craps players treat rolling the dice as a completely random occurrence. They may wear a lucky shirt or say their prayers beforehand, but they don’t do anything unique when rolling the dice.

However, a small subset of players believe that they can actually control the random craps’ outcomes. They claim that by holding the dice in a certain way and rolling them with a smooth motion, they occasionally manipulate the results.

Dubbed “controlled shooting,” this strategy promises long-term profits through craps. Unfortunately, controlled shooting (also known as dice control) is a complete lie.

Learn how to control dice from the men who mastered this technique. Our dice control video will teach you everything you need to know in order to beat the casino's. This DVD will have you learning the secrets of dice control right away. At a fraction of the cost of our seminars this is the perfect way to begin your dice control training. Dice control in Craps consists of eight physical elements in order to win. The Dominator's elements from this video will teach you where to stand, how to sca.

I’m going to discuss more on dice control along with why it doesn’t work. I’ll also cover what actually does work when trying to beat this game.

What Is Controlled Shooting?

The first step to controlled shooting begins with gripping the dice in a certain manner (setting). You want to hold the dice in a way that covers up undesirable combinations (usually anything that produces a seven) and exposes the combos that you do want.

Here’s an example on how this works with the “blanket roll” grip:

• Start by placing a pass line bet.
• Wait until a point number has been established.
• Hold the dice with the three’s in a V-shape (hides sevens).
• The goal is to toss the point number before a seven.
• You want a smooth, light roll so that the dice will lightly hit the wall and reduce randomness.

Craps Controlled Shooting Range

The second step is to ensure that you have some way to practice your toss. You can either rig a craps table in your home or even purchase a regulation-sized table (12×14 feet).

The latter option is expensive and takes up lots of space in a home. Assuming you rig a table that’s close to regulation size, then you can get the same effect.

Finally, you want to practice your toss for at least a couple of hours per day. The goal is to develop a smooth rolling motion that makes the dice kiss off the wall.

The lighter the dice ricochet off the wall, the less chance they have of breaking apart. By reducing the randomness, you supposedly have a better chance of producing desired combinations.

Controlled shooters measure their success through a concept known as Sevens: Rolls Ratio (RSR). This term alludes to the ratio of rolls that produce a seven versus those that don’t.

The odds of rolling a seven combination are one in six (6:1). A controlled shooter seeks to alter this ratio by rolling fewer sevens. You gain an advantage with an RSR of 6.3:1 (a seven once every 6.3 tosses) or higher.

Finally, you must put in the practice to become a better shooter. Experts recommend that you practice your toss at least two to three hours per day. You may need to keep this routine up for months before improving as a shooter.

The Alleged Promise of Dice Control

As with any advantage play method, the goal with dice control is to consistently win profits. Assuming you can toss the dice the same way each time, then you allegedly stand a better chance of winning.

Of course, you won’t successfully control every throw. Instead, you’re merely looking to change the odds once in a while.

Dice control proponents compare this process to an athlete honing their craft. A basketball player, for example, works tirelessly to improve their jump shot.

They’re never going to become so good that they make every single shot. But with more practice, they’ll shoot a higher percentage from the field.

Likewise, a wannabe craps pro doesn’t expect to avoid rolling a seven every time. They just need to roll other numbers consistently enough to overcome the odds and make profits.

This History Channel reenactment (which is completely ridiculous by the way) portrays what happens when a shooter gets hot. On their best nights, they may be able to roll desired numbers quite frequently.

Of course, just like an athlete, they’ll also have off-nights. Their goal is to experience enough hot nights to overcome poor performances and make long-term money.

Why Is Controlled Shooting a Fraud?

Dice control sounds amazing in theory. By practicing your toss, you can develop a skill that will bring guaranteed profits.

However, what sounds good in theory doesn’t always play out in practice. Casinos have designed craps to be perfectly random. And they’ve done a good job at accomplishing this goal too.

For starters, the table features pyramid-patterned wall. You’re expected to hit this wall, or at least make a reasonable attempt at doing so, on each toss.

The fact that you’re hurling two tiny plastic cubes across the table at a textured wall seems random enough. But controlled shooting proponents still claim that you can control tosses with some degree of success.

The problem, though, is that they can’t prove dice control works. Unlike with card counting, hole carding, or shuffle tracking, nobody has demonstrated that dice control is an undeniable success.

Instead, those saying that it works are the same people selling books and seminars on the subject. They’re not exactly the most impartial sources on the matter.

Perhaps the biggest sign that controlled shooting is a hoax comes from casinos themselves. Gambling establishments make no effort to identify controlled shooters and ban them from the casino.

The casino only cares when controlled shooters take too long to set the dice. Some players take as long as 20 seconds to get their grips. This slows down the game and costs the casino profits.

Other than this situation, though, the house doesn’t care if you roll the dice in a certain manner and win while doing so. They fully realize that dice control is fake and won’t hurt them in the long run.

What You Can Do to Win in Craps Instead?

You’re not going to make regular profits with controlled dice today no matter how long you practice. It’s a fake advantage gambling technique that was made up to sell products.

However, you can improve your odds of beating craps through other means. Here are some tips for winning money with the game.

Make the Right Bets

Craps is one of the absolute best games with regard to the house edge. It features multiple regular bets that have low house advantages.

These wagers include:

• Pass line – 1.41% house edge
• Come – 1.41%
• Don’t pass line – 1.36%
• Don’t come – 1.36%

You can also make Place 6 and Place 8 wagers to enjoy solid odds as well. Both of those prop bets on a carry a 1.52% house edge.

Most casino games feature over a 2% house advantage or higher. Therefore, craps is definitely one of the best games when it comes to the odds.

Back Your Bets With Odds

Again, craps already features several excellent bets. But you can reduce the casino’s edge even further with odds.

A bet on the odds wager refers to bets that don’t have a house edge. Instead, they pay at your true odds of winning.

You must wait until a point number is established before placing one of these wagers. Here are the payouts depending upon the point number that your bet is on.

Pass Line

• 2:1 pay out on point numbers of 4 and 10.
• 3:2 pay out on point numbers of 5 and 9.
• 6:5 pay out on point numbers of 6 and 8.

Don’t Pass Line

• 1:2 pay out for point numbers of 4 and 10.
• 2:3 pay out for point numbers of 5 and 9.
• 5:6 pay out for point numbers of 6 and 8.

Odds are worth a multiple of your pass line or don’t pass line bet. You can take higher odds to lower the house edge further.

Here are the various odds bets along with how much they reduce the house advantage by:

• 1x odds: pass line 0.85%; don’t pass line 0.68%
• 2x odds: pass line 0.61%; don’t pass line 0.46%
• 3x odds: pass line 0.47%; don’t pass line 0.34%
• 5x odds: pass line 0.33%; don’t pass line 0.23%
• 10x odds: pass line 0.18%; don’t pass line 0.12%
• 20x odds: pass line 0.1%; don’t pass line 0.07%

Get Loyalty Rewards

You can collect rewards in online casinos right away after signing up and depositing. The software will automatically track your bets and reward you accordingly.

However, land-based casinos require that you sign up for a player’s club card before collecting loyalty benefits. You can register either online or at the player’s desk.

Comps in real money craps aren’t quite as lucrative as slots rewards, per say. But they do ensure that you pick up freebies on your play, such as cashback, meals, drinks, and more.

Look for Online Craps Bonuses

In addition to offering loyalty rewards, online casinos also feature deposit bonuses. These deals match your deposit by a certain percentage.

Here’s an example:

• A casino offers a 100% match bonus worth up to $500.
• You deposit $200 and qualify for a $200 bonus.
• You can cash out the $200 after meeting terms and conditions.

The terms and conditions are very important in this matter. You need to wager a certain multiple of your bonus (the playthrough) and follow other rules before cashing out.

Furthermore, craps wagers don’t count towards the bonus at every single casino. But as long as craps does qualify, then you should take advantage and earn free money on the side of your play.

Conclusion

There’s no shortcut to beating craps. You can’t just hold the dice in a certain manner and practice your toss to win riches.

Controlled shooting is a bogus strategy that was developed by charlatans who make money off the matter. Therefore, you should completely ignore this so-called advantage play method.

What you can do, though, is take basic steps towards beating craps. Pass line and don’t pass line bets, odds, rewards, and online bonuses all improve your chances of winning.

You may not swing the long-term edge in your favor with these steps. However, you can at least reduce the house advantage by quite a bit and pick up free money.

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

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.

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.

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%.

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.

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%.

Craps Dice Throwing Tips

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.

1. 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.

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.

3. 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.

4. 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.

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.

Craps Controlled Shooting

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?

The Hot Roll bonus round on slot machines awards the player the following number of coins according to the total of two dice. The player keeps collecting until he rolls a total of seven, which ends the bonus. If he rolls a seven on the first roll, he gets a consolation prize of 70 coins. Following are the prizes for all other totals besides seven:

• 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

Craps Controlled Shooting Equipment

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

Craps Controlled Shooting Ball

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.

Craps Controlled Shooting