Craps Dice Roll Probability
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*Craps Dice Roll Probability Dice
*Craps Dice Roll Probability
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*Craps Dice Roll Probability Rules
*Craps Dice Roll Probability Calculator
AnyDice is an advanced dice probability calculator, available online. It is created with roleplaying games in mind. A Brief History of Dice and Craps Dice are prehistory. The earliest discovered six-sided dice date around 3,000 B.C. (Schwartz, 2006). Modern dice have the same number pattern where the sum of opposing sides always equal 7 (e.g., 1/6, 2/5 and 3/4 are always opposite each other).
Therefore, the probability of rolling an 11 with two dice is 1/18. Every roll of two dice has 36 total outcomes. This table shows how to calculate the probability of rolling each number with two dice. You can use probability to figure out the odds of winning and losing in the popular casino dice game of craps. In the gambling game “craps,” a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2, 3, or 12. The casino game of Craps is played with a set of two perfectly balanced six-sided dice with each die side having white dots numbering one through six. The game is played by tossing the dice from one of the short ends of the table to the other, making sure that both dice hit the wall on the opposite side of the table.rickwolves91Hey Wizard,I’m a CJ/CIS major working in a Math department, and I’ve been asked to convert an open ended question test to multiple choice test. I found your site ’The Wizard of Odds’, and was directed here to ask my question. What is the probability of rolling a 2 given your roll only being even, when rolling a 20 sided die? Thank you for your help.-RMathExtremist
Hey Wizard,I’m a CJ/CIS major working in a Math department, and I’ve been asked to convert an open ended question test to multiple choice test. I found your site ’The Wizard of Odds’, and was directed here to ask my question. What is the probability of rolling a 2 given your roll only being even, when rolling a 20 sided die? Thank you for your help.-RI know what you probably meant, but you actually need to specify what’s on the 20-sided die. If it’s 10 through 200 in increments of 10, the probability is zero. Under the assumption that the d20 contains integers 1..20, here are the answers I’d list on a four-choice pick:a) 1/2b) 1/5c) 1/10d) 1/20Edit: What’s a ’CJ’ major?’In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice.’ -- Girolamo Cardano, 1563beachbumbabsAdministrator
I know what you probably meant, but you actually need to specify what’s on the 20-sided die. If it’s 10 through 200 in increments of 10, the probability is zero. Under the assumption that the d20 contains integers 1..20, here are the answers I’d list on a four-choice pick:a) 1/2b) 1/5c) 1/10d) 1/20Edit: What’s a ’CJ’ major?Funny, I read it differently for the same reason. I ’decided’ the question stipulated the 20 sides were numbered 2 thru 20, evens only, with 2 identical sides per number, before I read ME’s answer. Be interested to see the clarification.I’m guessing ’Criminal Justice/Crime Scene Investigation’ for the major.If the House lost every hand, they wouldn’t deal the game.TerribleTomI get confused at ’given your roll only being even’.I played some Dungeons & Dragons as a kid, so I’ve seen a d20 or two.With a d20 (integers 1 through 20) the odds of rolling a 2 would be (drum roll please)... 1 in 20.If you’re saying ’I’ve got a d20. The next roll will be an even number. What are the odds of that number being a 2?’ Then I’d have to say the odds are 1 in 10. But how does one guarantee that no odd number will be rolled?I’m thinking CJ/CIS is Criminal Justice/Criminal Information Systems.AxiomOfChoice
If you’re saying ’I’ve got a d20. The next roll will be an even number. What are the odds of that number being a 2?’ Then I’d have to say the odds are 1 in 10. But how does one guarantee that no odd number will be rolled?It’s a conditional probability problem. It’s guaranteed because the question says it’s guaranteed. It’s a math question, not a physics question.rdw4potusDesigning multiple-choice tests is one of my favorite things to do. I see this question, and all I think is ’gosh, i wonder how they’re doing answer placement. Is it randomized? Rotated? both?’My HS world geography teacher was a sadistic guy. He put patterns into all his tests, but he also rotated the correct answers. So my 100% score could go a,b,c,d,a,b,c,d... while my friend could have the same answers but need to respond d,c,b,a,d,c,b,a... Made colluding and/or selling answers a real pain.’So as the clock ticked and the day passed, opportunity met preparation, and luck happened.’ - Maurice Clarettrickwolves91CJ - Criminal JusticeI’m sorry, fellas. The numbers on the dice are 1-20. My understanding of this questions is, what is the probability you will roll an even number and what is the probability that even number will be 2? The next question on the test is ’Find the probability of drawing a king given that you draw a spade.’ I’m pretty sure both questions require the same method to solve.AxiomOfChoice
CJ - Criminal JusticeThe numbers on the 20 sided dice are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. If this reply sounds smartalically, I appologize. I don’t really understand what you mean needing to ’specify what’s on the 20-sided die.’ Anyhow, which is the answer; a, b, c, or d?There are 10 even numbers. If you know for sure that it is even, then the answer is 1/10.You can also solve this by going through the formula for conditional probability. To be honest I always have to look it up (check wikipedia for conditional probability), but you will end up dividing 1/20 by 10/20 and get 1/10.rickwolves91Thank you.rickwolves91CIS - Computer Information Systems (Know thy enemy)’But how does one guarantee that no odd number will be rolled?’ Probable probability.Thank you.I was wondering how to alter dice in the game of craps so it hits 7 or 11 every time can you help. thank you.
Alter then so that one die has a six on every side, and the other one has all ones and fives.Do you believe that ’wishful thinking’ on behalf of the players can affect the outcome of a game. Note that I’m not concerned with the SIZE of the effect, just your philosophical opinion. Also, do you think that the manner in which a player tosses the dice in craps can cause a bias (good or bad) in the outcome. As always, your site is AWESOME.
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.Just wondering about your opinion of altering the frequency table in craps by pre-setting the dice.
I’m very skeptical of it. I go over some of the experiments on the topic in my craps appendix 3.I recently learned some information about dicesetting strategies in craps. Some believe that you can set the dice a certain way before the throw, and by keeping the roll of the dice to just one axis of rotation, you can have fewer possible sevens with certain dice sets. I wanted to know if there is any truth to this or is it just a fallacy.
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.Are dice truly unbiased? It seems like the sides with the larger numbers which have more holes would be lighter than the sides with the smaller numbers and less holes. This seems to suggest that the heavier sides would more likely land face down with the larger numbers more likely landing face up. I can imagine a craps system that could try to exploit this principle, but I wonder if it would really work. What do you think?
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.Do you believe the toss of the dice at a Casino Craps table is truly random as a RNG would be, or are there good shooters and bad shooters either thru dice ’mechanics’ or plain sloppy throwing (short throws as an example), if real world Casino Craps is not truly random , how would I take advantage of this?
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.In the October issue of Casino Player magazine, Frank Scoblete wrote an article on controlled dice shooting where you state you lost $1800 to Stanford Wong when he rolled only 74 sevens in 500 rolls. Why did you bet on such a small sample (500)? A person who claims to be able to control the dice should be willing to demonstrate their skill with a least 50,000 rolls. Am I wrong in thinking that 500 rolls is such a small sample that just about anything could happen?
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.I know you’re skeptical of dice control. I have been practicing dice setting and controlled shooting for 3 months. What is the probability of throwing 78 sevens over 655 throws randomly? Thanks for the help :)
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%.Three years ago, in an Ask The Wizard column, you wrote: ’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.In the news today, a woman in Atlantic City rolled 154 times consecutively before sevening out at the Borgata. That means she rolled two dice 154 times, with no sevens. So I took (30/36)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 DataDice TotalObservations2631241451862375083693710271114127Total244
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 ResultsCraps Dice Roll Probability DiceDice TotalObservationsExpectedChi-Squared266.7777780.08925331213.5555560.17850641420.3333331.97267851827.1111113.06193162333.8888893.49872575040.6666672.14207783633.8888890.13151293727.1111113.607013102720.3333332.185792111413.5555560.0145721276.7777780.007286Total24424416.889344
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
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*Craps Dice Roll Probability Dice
*Craps Dice Roll Probability
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AnyDice is an advanced dice probability calculator, available online. It is created with roleplaying games in mind. A Brief History of Dice and Craps Dice are prehistory. The earliest discovered six-sided dice date around 3,000 B.C. (Schwartz, 2006). Modern dice have the same number pattern where the sum of opposing sides always equal 7 (e.g., 1/6, 2/5 and 3/4 are always opposite each other).
Therefore, the probability of rolling an 11 with two dice is 1/18. Every roll of two dice has 36 total outcomes. This table shows how to calculate the probability of rolling each number with two dice. You can use probability to figure out the odds of winning and losing in the popular casino dice game of craps. In the gambling game “craps,” a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2, 3, or 12. The casino game of Craps is played with a set of two perfectly balanced six-sided dice with each die side having white dots numbering one through six. The game is played by tossing the dice from one of the short ends of the table to the other, making sure that both dice hit the wall on the opposite side of the table.rickwolves91Hey Wizard,I’m a CJ/CIS major working in a Math department, and I’ve been asked to convert an open ended question test to multiple choice test. I found your site ’The Wizard of Odds’, and was directed here to ask my question. What is the probability of rolling a 2 given your roll only being even, when rolling a 20 sided die? Thank you for your help.-RMathExtremist
Hey Wizard,I’m a CJ/CIS major working in a Math department, and I’ve been asked to convert an open ended question test to multiple choice test. I found your site ’The Wizard of Odds’, and was directed here to ask my question. What is the probability of rolling a 2 given your roll only being even, when rolling a 20 sided die? Thank you for your help.-RI know what you probably meant, but you actually need to specify what’s on the 20-sided die. If it’s 10 through 200 in increments of 10, the probability is zero. Under the assumption that the d20 contains integers 1..20, here are the answers I’d list on a four-choice pick:a) 1/2b) 1/5c) 1/10d) 1/20Edit: What’s a ’CJ’ major?’In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice.’ -- Girolamo Cardano, 1563beachbumbabsAdministrator
I know what you probably meant, but you actually need to specify what’s on the 20-sided die. If it’s 10 through 200 in increments of 10, the probability is zero. Under the assumption that the d20 contains integers 1..20, here are the answers I’d list on a four-choice pick:a) 1/2b) 1/5c) 1/10d) 1/20Edit: What’s a ’CJ’ major?Funny, I read it differently for the same reason. I ’decided’ the question stipulated the 20 sides were numbered 2 thru 20, evens only, with 2 identical sides per number, before I read ME’s answer. Be interested to see the clarification.I’m guessing ’Criminal Justice/Crime Scene Investigation’ for the major.If the House lost every hand, they wouldn’t deal the game.TerribleTomI get confused at ’given your roll only being even’.I played some Dungeons & Dragons as a kid, so I’ve seen a d20 or two.With a d20 (integers 1 through 20) the odds of rolling a 2 would be (drum roll please)... 1 in 20.If you’re saying ’I’ve got a d20. The next roll will be an even number. What are the odds of that number being a 2?’ Then I’d have to say the odds are 1 in 10. But how does one guarantee that no odd number will be rolled?I’m thinking CJ/CIS is Criminal Justice/Criminal Information Systems.AxiomOfChoice
If you’re saying ’I’ve got a d20. The next roll will be an even number. What are the odds of that number being a 2?’ Then I’d have to say the odds are 1 in 10. But how does one guarantee that no odd number will be rolled?It’s a conditional probability problem. It’s guaranteed because the question says it’s guaranteed. It’s a math question, not a physics question.rdw4potusDesigning multiple-choice tests is one of my favorite things to do. I see this question, and all I think is ’gosh, i wonder how they’re doing answer placement. Is it randomized? Rotated? both?’My HS world geography teacher was a sadistic guy. He put patterns into all his tests, but he also rotated the correct answers. So my 100% score could go a,b,c,d,a,b,c,d... while my friend could have the same answers but need to respond d,c,b,a,d,c,b,a... Made colluding and/or selling answers a real pain.’So as the clock ticked and the day passed, opportunity met preparation, and luck happened.’ - Maurice Clarettrickwolves91CJ - Criminal JusticeI’m sorry, fellas. The numbers on the dice are 1-20. My understanding of this questions is, what is the probability you will roll an even number and what is the probability that even number will be 2? The next question on the test is ’Find the probability of drawing a king given that you draw a spade.’ I’m pretty sure both questions require the same method to solve.AxiomOfChoice
CJ - Criminal JusticeThe numbers on the 20 sided dice are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. If this reply sounds smartalically, I appologize. I don’t really understand what you mean needing to ’specify what’s on the 20-sided die.’ Anyhow, which is the answer; a, b, c, or d?There are 10 even numbers. If you know for sure that it is even, then the answer is 1/10.You can also solve this by going through the formula for conditional probability. To be honest I always have to look it up (check wikipedia for conditional probability), but you will end up dividing 1/20 by 10/20 and get 1/10.rickwolves91Thank you.rickwolves91CIS - Computer Information Systems (Know thy enemy)’But how does one guarantee that no odd number will be rolled?’ Probable probability.Thank you.I was wondering how to alter dice in the game of craps so it hits 7 or 11 every time can you help. thank you.
Alter then so that one die has a six on every side, and the other one has all ones and fives.Do you believe that ’wishful thinking’ on behalf of the players can affect the outcome of a game. Note that I’m not concerned with the SIZE of the effect, just your philosophical opinion. Also, do you think that the manner in which a player tosses the dice in craps can cause a bias (good or bad) in the outcome. As always, your site is AWESOME.
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.Just wondering about your opinion of altering the frequency table in craps by pre-setting the dice.
I’m very skeptical of it. I go over some of the experiments on the topic in my craps appendix 3.I recently learned some information about dicesetting strategies in craps. Some believe that you can set the dice a certain way before the throw, and by keeping the roll of the dice to just one axis of rotation, you can have fewer possible sevens with certain dice sets. I wanted to know if there is any truth to this or is it just a fallacy.
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.Are dice truly unbiased? It seems like the sides with the larger numbers which have more holes would be lighter than the sides with the smaller numbers and less holes. This seems to suggest that the heavier sides would more likely land face down with the larger numbers more likely landing face up. I can imagine a craps system that could try to exploit this principle, but I wonder if it would really work. What do you think?
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.Do you believe the toss of the dice at a Casino Craps table is truly random as a RNG would be, or are there good shooters and bad shooters either thru dice ’mechanics’ or plain sloppy throwing (short throws as an example), if real world Casino Craps is not truly random , how would I take advantage of this?
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.In the October issue of Casino Player magazine, Frank Scoblete wrote an article on controlled dice shooting where you state you lost $1800 to Stanford Wong when he rolled only 74 sevens in 500 rolls. Why did you bet on such a small sample (500)? A person who claims to be able to control the dice should be willing to demonstrate their skill with a least 50,000 rolls. Am I wrong in thinking that 500 rolls is such a small sample that just about anything could happen?
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.I know you’re skeptical of dice control. I have been practicing dice setting and controlled shooting for 3 months. What is the probability of throwing 78 sevens over 655 throws randomly? Thanks for the help :)
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%.Three years ago, in an Ask The Wizard column, you wrote: ’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.In the news today, a woman in Atlantic City rolled 154 times consecutively before sevening out at the Borgata. That means she rolled two dice 154 times, with no sevens. So I took (30/36)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 DataDice TotalObservations2631241451862375083693710271114127Total244
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 ResultsCraps Dice Roll Probability DiceDice TotalObservationsExpectedChi-Squared266.7777780.08925331213.5555560.17850641420.3333331.97267851827.1111113.06193162333.8888893.49872575040.6666672.14207783633.8888890.13151293727.1111113.607013102720.3333332.185792111413.5555560.0145721276.7777780.007286Total24424416.889344
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
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