How can you tell if you are influencing the dice or not?
Skinny – 08:35am Mar 12, 2011 PT
Certified Instructor, GTC Primer/Refresher/TuneUp/Advanced/GTB Grad, 40-Roll/50-Roll/60-Roll Club, Post of Distinction Double Black Chip/Post of the Month Gold Chip
The best answer to this question is to record your practice throws, log them into Smart Craps and let it tell you if you are able to influence the dice or not. Smart Craps can prove, with statistical certainty, that you are influencing the dice outcomes. It provides you with a lot of other information as well such as:
the optimal dice sets and bets given your unique dice control skill.
your edge over the casino: how much money you can make playing craps.
Risk of Ruin (ROR) calculators and simulation.
I do not want you to think this post can preempt the capabilities of Smart Craps because that software is much more comprehensive than what I am about to give you.
But Alamo Tx asked me a good question a couple of days ago and I have been doing some thinking about how best to answer his post. Here is his question:
The Question is: Over 5000 rolls, for example, if Smart Craps tells you you’re hitting the 6 and 8 more often than random would indicate, is there a % that tells you something meaningful? Statistically, I know I hit more sixes and eights than random would dictate, but I can’t tell how significant it is. Statistically, I hit more box numbers as well…but that seems to be what would happen with a controlled throw. Of course, I don’t use Smart Craps for axis and double pitch statistics on anything but HW roll sets. I use Smart Craps on V3 and V2 only to calculate SRR.
I have come to the conclusion that the best way to answer this is to give him some statistical data to let him know how his results compare to what can be expected by that of a random roller. In order to do that I am first going to bore those of you who are not interested in math with some technical jargon.
If you do not like math your eyes are definitely going to glaze over in this next section. But I am providing it for completeness and to give those who are interested a background in what I am talking about. Bear with me through this technical treatise for a bit. I am taking some technical jargon from different sources on the web. At the end I will give you what I hope to be a simple method for figuring out how you are doing with your practice results that may have some meaning to you.
Technical crap begins now:
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Standard deviation is a widely used measurement of variability or diversity used in statistics and probability theory. It shows how much variation or "dispersion" there is from the "average" (mean, or expected/budgeted value).
Technically, the standard deviation of a statistical population, data set, or probability distribution is the square root of its variance. In addition to expressing the variability of a population, standard deviation is commonly used to measure confidence in statistical conclusions. For example, the margin of error in polling data is determined by calculating the expected standard deviation in the results if the same poll were to be conducted multiple times. The reported margin of error is typically about twice the standard deviation - the radius of a 95 percent confidence interval. In science, researchers commonly report the standard deviation of experimental data, and
only effects that fall far outside the range of standard deviation are considered statistically significant
– normal random error or variation in the measurements is in this way distinguished from causal variation.
In statistics, the 68-95-99.7 rule, or three-sigma rule, or empirical rule, states that for a normal distribution, nearly all values lie within 3 standard deviations of the mean.
About 68.26% of the values lie within 1 standard deviation of the mean. In statistical notation, this is represented as: μ ± 1σ.
About 95.44% of the values lie within 2 standard deviations of the mean. The statistical notation for this is: μ ± 2σ.
Nearly all (99.73%) of the values lie within 3 standard deviations of the mean. Statisticians use the following notation to represent this: μ ± 3σ.
The formula for the mean and standard deviation for a binomial distribution is: EXP = np and SD = the square root of (npq), where n is the total number of trials, p is the probability of success, and q is the probability of failure, EXP is the mean and SD is the standard deviation.
For example if I write EXP( 6/8 ) I will mean the expected number of sixes or eights that a random roller will throw for a specific number of rolls. That is also called the mean number of sixes or eights that one can expect to see on average for that specific number of rolls.
SD( 6/8 ) will mean the standard deviation for the sixes or eight. Thus for a random roller I can expect approximately:
68% of his sixes or eights to be within EXP( 6/8 ) +/- SD( 6/8 )
95% of his sixes or eights to be within EXP( 6/8 ) +/- [2 * SD( 6/8 )]
99.7% of his sixes or eights to be within EXP( 6/8 ) +/- [3 * SD( 6/8 )]
If you are able to throw more sixes than EXP( 6/8 ) + [3 * SD( 6/8 )] you are extremely likely to be influencing the dice.
The other thing you need to understand in the formula I am going to give you is what is meant by p and q where p is the probability of success, and q is the probability of failure. Let us stick with looking at how many sixes I expect a random shooter to throw. We know there are 5 ways out of 36 to throw a 6. I can throw a 1-5, 5-1, 2-3, 3-2 or 3-3 to make a six. Conversely there are 31 other combinations of the dice that do not add up to six.
The probability of a random roller throwing a six is 5/36.
The probability of a random roller not throwing a six is 31/36.
Thus the probability of success (p) is 5/36 and the probability of failure (q) is 31/36.
Technical crap ends here.
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I said I would give you what I hope to be a simple method for figuring out how you are doing with your practice results that may have some meaning to you. So how do I intend to do that in light of all that technical gibberish above?
I am going to give you a formula to let you know where your results for box numbers fall on a normal distribution curve, typically known as a bell curve. If the number of box numbers you threw in practice is outside the 99.7% range then you are definitely able to throw more of that specific box number than can be expected by a random roller. When you are outside the 3 sigma (99.7%) range for any specific number, you are most definitely influencing the dice in a fashion that is highly unlikely to be merely random. At that range you are greater than 99.7% to be influencing the dice.
First I am going to show how this works for the 5,000 rolls Alamo Tx said he had documented.
I am going to be working with two general formulas:
EXP = np and SD = the square root of (npq)
If I want to look at sixes or eights the numbers are as follows:
n = 5,000
p = 5/36
q = 1 – 5/36 = 31/36
EXP( 6/8 ) = np = 5000*5/36 = 694.4
SD( 6/8 ) = square root (npq) = square root [5000*(5/36)*(31/36)] = 24.5
68% of the sixes or eights are between 694.4-24.5 and 694.4+24.5…or between 669.94 and 718.9.
95% of the sixes or eights are between 694.4-(2*24.5) and 694.4+(2*24.5)…or between 645.4 and 743.4.
99.7% of the sixes or eights are between 694.4-(3*24.5) and 694.4+(3*24.5)…or between 620.9 and 767.9.
What this boils down to is that for 5,000 practice rolls if you are able to roll 768 or more sixes you are most definitely influencing the dice in a fashion that is not likely to be random. The same is true if you roll 768 or more eights.
I will now do the calculation for you for the other box numbers.
If I want to look at fives or nines the numbers are as follows:
n = 5,000
p = 4/36
q = 1 – 4/36 = 32/36
EXP(5/9) = np = 5000*4/36 = 555.6
SD(5/9) = square root (npq) = square root [5000*(4/36)*(32/36)] = 22.2
68% of the fives or nines are between 555.6-22.2 and 555.6+22.2…or between 533.4 and 577.8.
95% of the fives or nines are between 555.5-(2*22.2) and 555.5+(2*22.2)…or between 511.2 and 600.
99.7% of the fives or nines are between 555.5-(3*22.2) and 555.5+(3*22.2)…or between 489 and 622.2.
What this boils down to is that for 5,000 practice rolls if you are able to roll 623 or more fives you are most definitely influencing the dice in a fashion that is not likely to be random. The same is true if you roll 623 or more nines.
If I want to look at fours or tens the numbers are as follows:
n = 5,000
p = 3/36
q = 1 – 3/36 = 33/36
EXP(4/10) = np = 5000*3/36 = 416.7
SD(4/10) = square root (npq) = square root [5000*(3/36)*(33/36)] = 19.5
68% of the fours or tens are between 416.7-19.5 and 416.7+19.5…or between 397.2 and 436.2.
95% of the fours or tens are between 416.7-(2*19.5) and 416.7+(2*19.5)…or between 377.7 and 455.7.
99.7% of the fours or tens are between 416.7-(3*19.5) and 416.7+(3*19.5)…or between 358.2 and 475.2.
What this boils down to is that for 5,000 practice rolls if you are able to roll 476 or more fours you are most definitely influencing the dice in a fashion that is not likely to be random. The same is true if you roll 476 or more tens.
OK, maybe I got a little more technical here at the end then I led you to believe I would. Sorry about that. But I wanted you to see how the formula works so you might be able to apply it to your own data for your practice rolls for a different sample size than 5,000.
First let me summarize the results for 5,000 rolls.
If you throw more than 768 sixes or eights in 5,000 rolls you are 99.7% certain you are influencing the dice.
If you throw more than 623 fives or nines in 5,000 rolls you are 99.7% certain you are influencing the dice.
If you throw more than 476 four or tens in 5,000 rolls you are 99.7% certain you are influencing the dice.
If you throw fewer box numbers than that you can tell where you fall on the binomial distribution from the data I gave you above. Alamo TX said, "I know I hit more sixes and eights than random would dictate, but I can’t tell how significant it is." I hope this helps you to understand how significant your numbers are in relation to random. Are you in the 68%, 95%, 99.7% or greater than 99.7% range?
Now if you have a different number of rolls than 5,000 I hope you can follow the formula I gave you above to get the numbers for those ranges. You only have to change n in the formula I gave you by making n equal to the number of rolls in your sample size. All the other numbers in the formula are the same.
Calculate the mean (EXP) and standard deviation (SD) for your number of rolls and then determine the 68-95-99.7 data as I did above.
Does this help answer the question and is anyone able to follow what I have done here?
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