Slot Machine Binomial Distribution
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Exam Numerical Ability Question Solution - The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Slot Machines The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Multiple-Job Holders According to the government 5.3% of those employed are multiple-job holders.
*Slot Machine Binomial Distribution Method
*Slot Machine Binomial Distribution Calculator
*Slot Machine Binomial Distribution Formula
QSCI 482/Dr. ConquestTAs Kennedy, Malinick
HW1--DUE *THIS* FRIDAY, OCTOBER 4, 2002. ALSO, IF YOU DID NOT FILL OUT A
CLASS SURVEY SHEET ON DAY 1, PLEASE DO SO BY THE END OF THE WEEK.
1. Attach a copy of a recent photo of yourself; it will help me learn
everyone’s names. I will return it if you wish. You don’t have to be
smiling [’mug shots’ are OK] but photo must be in good taste [no pictures
of your last skinny-dipping party, please. No ’Full Monty’s’.]. Thanks!
2. In a 1995 the Puget Sound Gilnetters Association conducted a field test
to test the effects of different kinds of fishing gear on alcid bird
entanglement (diving seabirds like rhinocerous auklet and common murre,
covered under the Migratory Bird Treaty Act).First, a test was done to
see if the total bird entanglements followed a Poisson distribution with
mean mu=.09, since this is what was found from previous field seasons.
(That is, the overall entanglement rate is about .09 birds per net set.)
Out of 449 net sets (a fishing net set in the water and fishing for the
same specified period of time), 418 of them had 0 birds caught, 30 sets
had 1 bird caught, and 1 of the sets experienced 2-or-more birds caught.
At the .05 level of significance, do a goodness-of-fit test to test
whether the observed entanglements follow a Poisson distribution with mean
mu=.09, against the alternative hypothesis that they do not follow such a
distribution. What do you conclude?[NOTE. The Poisson probability for a
particular count, X, is on p. 571 in Zar (in the 3rd ed. it’s p. 569) and
is: exp(-mu)*(mu^X)/X!.]
H0: bird entanglement follows a Poisson distribution with mean 0.09
Ha: bird entanglement follows some other distribution
Assumptions: Data are a random sample and independent, none of the expected values is less than 1 and no more than 20% are less than 5
Critical value: Chisq0.05,2 = 5.991
Test Statistic:Slot Machine Binomial Distribution Method
num.entangled
observed.countp(x)expected.countresidualChisq04180.914410.3557.6450.1424241300.08236.932-6.9321.301097210.0041.713-0.7130.296731
Sums44914491.740252
Test statistic = 1.740
Decision: 1.740<5.991, therefore we fail to reject H0.There is insufficient evidence to distinguish this data set from a Pois(0.09).
3. This exercise involves computing probabilities for the binomial
distribution (in Zar, the chapter titled ’More on Dichotomous Variables’,
or look under ’binomial distribution’ in any elementary statistics book).
According to a mathematician who works there, The Anchor Gaming Company is
the largest supplier of slot machines in the world. Let’s consider a
simple slot machine and compute some EXPECTED VALUES OF OUTCOMES FOR A
BINOMIAL PROBABILITY DISTIRIBUTION.Suppose a slot machine has 4 slots,
each of which shows a picture of either a cherry or a pear, and the slots
work independently. For a given slot, the probability that a cherry shows
up is 0.4, and the probability that a pear shows up is 0.6. So for a given
’pull’ of the machine, one can observe either: 4 cherries; 3 cherries and
1 pear; 2 cherries and 2 pears; 1 cherry and 3 pears; or 4 pears. Suppose
a given machine is ’pulled’ 10,000 times. How many times would we expect
to see:
a. 4 cherries
b. 3 cherries and 1 pear
c. 2 cherries and 2 pears
d. 1 cherry and 3 pears
e. 4 pears
a. Pr{4 cherries} = (.4)^4 = .0256; x 10,000 = 256 times.
b. Pr{3 cherries,1 pear} = 4 x (.4)^3 x .6 = .1536; x 10,000 = 1536
times.
c. Pr{2 cherries,2 pears} = 6 x (.4)^2 x (.6)^2 = .3456; x 10,000 = 3456
times.
d. Pr{1 cherry,3 pears} = 4 x .4 x (.6)^3 = .3456; x 10,000 = 3456
times.
e. Pr{4 pears} = (.6)^4 = .1296; x 10,000 = 1296 times.
The following problems (4-6, on the back side of this sheet) fall under the heading, ’algebra review’. If you have not done algebra for awhile, this will give you needed practice. If this stuff comes easily to you, give a classmate some help--we’re all in this together.
4. You may recall that the formula for the standard error for a sample
mean X-bar is: Standard Error of X-bar = sqrt(sample variance/n), where n
is the sample size, and ’sqrt’ stands for ’square root of’. You are
reading a journal paper and trying to find out what the original sample
size was (which the authors failed to state, and the editors did not catch
the omission).In a table, the authors state that ’the standard error for
the data was 10.3 kg.’ Elsewhere in the paper, you find that the sample
variance was 1,697.44 kg^2. Now, solve for the original sample size, n.
10.3=sqrt(1697.44/n)
10.32 = 1697.44/n
n=1697.44/10.32
n=16
5. We have a random sample of n = 5 weights (kg) of a particular kind of
animal: 3.1, 3.4, 3.6, 3.7, 4.0. The sample mean, X-bar, is 3.56 kg.
Now compute the sample variance of these data, using the standard
’machine formula’ calculation for the sample variance, s^2:
s^2 = {Sum([Xi^2]) - [(Sum[Xi])^2]/n}/(n-1)
Sum([Xi2]) = 63.82;[Sum(Xi)]2 = 17.82 = 316.84; n = 5
s2 = (63.82-316.84/5)/4 = 0.113
6. The following data (X) are known to come from a lognormal distribution;
that is, the natural logarithms of the data, Y = ln(X), will follow a
normal (bell-shaped) distribution. Here are the data in the original
units:3.67, 4.01, 3.85, 3.92, 3.71, 3.88, 3.74, 3.82 ml.
a. Compute the sample mean of the log-transformed data.
Let yi = ln(xi)
ybar = sum(yi)/8 = 10.729/8 = 1.34
b. Compute the sample variance of the log-transformed data. Take the sqrt
to get the sample standard deviation.
s2y = (sum(yi2)-[sum(yi)]2/n)/(n-1)
= [14.396-10.729^2/8]/7 = 0.00089Slot Machine Binomial Distribution Calculator
sy = sqrt(0.00089) = 0.0298
c. It turns out that the 95% confidence interval (CI) for the mean of the
log-transformed data is xbar +- 2.365*(std. deviation/sqrt(n)). Compute the 95%
CI for the log-transformed data; then transform each of the endpoints back
to get the 95% CI in the original units [ml]. Comment upon the symmetry of
the CI for the log-transformed units [ln(ml)], and the back-transformed
units [ml].
1.34+/-2.365*0.0289/sqrt(8) = 1.34+/-0.0242
1.32<=mu(y)<=1.36
e1.32<=mu(x)<=e1.36
3.74<=mu(x)<=3.90
xbar=3.825
The CI in the transformed units is inherently symmetric about the mean because we added and subtracted the same value from the mean to obtain the end points.To assess symmetry in the original units, let’s measure the distance from the original mean and the lower and upper end points:
3.825-3.74 = 0.085
3.9-3.825 = 0.075
It appears that there is asymmetry in the CI for the original units, with the mean closer to the upper limit.This is not surprising because we utilized a non-linear transformation.That means that the relationship between the mean and the endpoints is not preserved when the data are back-transformed (it’s not simply a difference in scale).
Many people accounted for the difference as rounding error, and in fact different results were seen due to rounding.Points were subtracted if symmetry was commented on without actually any calculation to assess the symmetry (you can’t just say something is so, you have to show it is).
Online slot machines are truly random, and a casino knows what the payout percentages via a set of certifications. Each slot and random number generator part of the slot need to pass rigorous in-house testing as well as a suite of statistical tests created to detect bias. One of many standard tests for is each slot is the measure of its RTP, each slot has a fixed theoretical value of its return to player used as a baseline value. Auditors consider the probabilities of the subset and compare events part of the actual gameplay against the expected events. The primary tool uses for this type of analyses are called the chi-squared statistics and the binomial distribution. The auditor/s also analyze the events in successive rounds and the RNG needs to meet the standards of the industry, therefore the RNG implementation is to safeguard outcomes, while also ensuring values produced are uniformly distributed over the required interval.
Before a new online casino game can be made available to players the game must be free of operational flaws that allow that the house edge could be changed by the casino or the player. The most cited cause of dissatisfaction amongst online players is the belief that games can be modified to yield the desired outcomes of an individual player. Online casinos are tremendously cautious about players believing that mathematical tools, bots or system are used to beat their spins or games, this is why auditors main function is to ensure that each slot are accurate and stable in each round and obligated to express his opinion supported by facts to manipulate false believes that the outcome of a spin or round can be influenced. Auditors track the long-term RTP plus several other statistics over months and years and therefore picks up any change and when he does he raises a red flag that the slot is not operating correctly. Auditors focus on statistical stability and audits take place monthly and only once the auditor is satisfied with all outcomes he will certify the game as fair and RNG.Slot Machine Binomial Distribution FormulaOnline slots are audited and tested for software integrity by several world-renowned auditing and testing companies such as Certified Fair Gaming an industry leader in fairness and auditing found in 2003. GLI or Gaming Laboratories International is another global leader in RNG auditing and perform regular audits and test on software fairness and RNG. TST or Technical Systems Testing does independent testing and auditing and verification that the software offers 100% true RNG and fair results and that it has not been manipulated at all. Online casinos certified by eCOGRA are viewed by experienced players as the most trustworthy, eCOGRA is based in London and is an internationally approved testing laboratory accredited certified body and known for focussing on player protection. eCOGRA stands for eCommerce Online Gaming Regulations and Assurance and the majority of major online operators proudly display their eCOGRA stamp of approval on their main page.
Register here: http://gg.gg/ny3b3
https://diarynote-jp.indered.space
Exam Numerical Ability Question Solution - The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Slot Machines The probability of winning on a slot machine is 5%. If a person plays the machine 500 times, find the probability of winning 30 times. Use the normal approximation to the binomial distribution. Multiple-Job Holders According to the government 5.3% of those employed are multiple-job holders.
*Slot Machine Binomial Distribution Method
*Slot Machine Binomial Distribution Calculator
*Slot Machine Binomial Distribution Formula
QSCI 482/Dr. ConquestTAs Kennedy, Malinick
HW1--DUE *THIS* FRIDAY, OCTOBER 4, 2002. ALSO, IF YOU DID NOT FILL OUT A
CLASS SURVEY SHEET ON DAY 1, PLEASE DO SO BY THE END OF THE WEEK.
1. Attach a copy of a recent photo of yourself; it will help me learn
everyone’s names. I will return it if you wish. You don’t have to be
smiling [’mug shots’ are OK] but photo must be in good taste [no pictures
of your last skinny-dipping party, please. No ’Full Monty’s’.]. Thanks!
2. In a 1995 the Puget Sound Gilnetters Association conducted a field test
to test the effects of different kinds of fishing gear on alcid bird
entanglement (diving seabirds like rhinocerous auklet and common murre,
covered under the Migratory Bird Treaty Act).First, a test was done to
see if the total bird entanglements followed a Poisson distribution with
mean mu=.09, since this is what was found from previous field seasons.
(That is, the overall entanglement rate is about .09 birds per net set.)
Out of 449 net sets (a fishing net set in the water and fishing for the
same specified period of time), 418 of them had 0 birds caught, 30 sets
had 1 bird caught, and 1 of the sets experienced 2-or-more birds caught.
At the .05 level of significance, do a goodness-of-fit test to test
whether the observed entanglements follow a Poisson distribution with mean
mu=.09, against the alternative hypothesis that they do not follow such a
distribution. What do you conclude?[NOTE. The Poisson probability for a
particular count, X, is on p. 571 in Zar (in the 3rd ed. it’s p. 569) and
is: exp(-mu)*(mu^X)/X!.]
H0: bird entanglement follows a Poisson distribution with mean 0.09
Ha: bird entanglement follows some other distribution
Assumptions: Data are a random sample and independent, none of the expected values is less than 1 and no more than 20% are less than 5
Critical value: Chisq0.05,2 = 5.991
Test Statistic:Slot Machine Binomial Distribution Method
num.entangled
observed.countp(x)expected.countresidualChisq04180.914410.3557.6450.1424241300.08236.932-6.9321.301097210.0041.713-0.7130.296731
Sums44914491.740252
Test statistic = 1.740
Decision: 1.740<5.991, therefore we fail to reject H0.There is insufficient evidence to distinguish this data set from a Pois(0.09).
3. This exercise involves computing probabilities for the binomial
distribution (in Zar, the chapter titled ’More on Dichotomous Variables’,
or look under ’binomial distribution’ in any elementary statistics book).
According to a mathematician who works there, The Anchor Gaming Company is
the largest supplier of slot machines in the world. Let’s consider a
simple slot machine and compute some EXPECTED VALUES OF OUTCOMES FOR A
BINOMIAL PROBABILITY DISTIRIBUTION.Suppose a slot machine has 4 slots,
each of which shows a picture of either a cherry or a pear, and the slots
work independently. For a given slot, the probability that a cherry shows
up is 0.4, and the probability that a pear shows up is 0.6. So for a given
’pull’ of the machine, one can observe either: 4 cherries; 3 cherries and
1 pear; 2 cherries and 2 pears; 1 cherry and 3 pears; or 4 pears. Suppose
a given machine is ’pulled’ 10,000 times. How many times would we expect
to see:
a. 4 cherries
b. 3 cherries and 1 pear
c. 2 cherries and 2 pears
d. 1 cherry and 3 pears
e. 4 pears
a. Pr{4 cherries} = (.4)^4 = .0256; x 10,000 = 256 times.
b. Pr{3 cherries,1 pear} = 4 x (.4)^3 x .6 = .1536; x 10,000 = 1536
times.
c. Pr{2 cherries,2 pears} = 6 x (.4)^2 x (.6)^2 = .3456; x 10,000 = 3456
times.
d. Pr{1 cherry,3 pears} = 4 x .4 x (.6)^3 = .3456; x 10,000 = 3456
times.
e. Pr{4 pears} = (.6)^4 = .1296; x 10,000 = 1296 times.
The following problems (4-6, on the back side of this sheet) fall under the heading, ’algebra review’. If you have not done algebra for awhile, this will give you needed practice. If this stuff comes easily to you, give a classmate some help--we’re all in this together.
4. You may recall that the formula for the standard error for a sample
mean X-bar is: Standard Error of X-bar = sqrt(sample variance/n), where n
is the sample size, and ’sqrt’ stands for ’square root of’. You are
reading a journal paper and trying to find out what the original sample
size was (which the authors failed to state, and the editors did not catch
the omission).In a table, the authors state that ’the standard error for
the data was 10.3 kg.’ Elsewhere in the paper, you find that the sample
variance was 1,697.44 kg^2. Now, solve for the original sample size, n.
10.3=sqrt(1697.44/n)
10.32 = 1697.44/n
n=1697.44/10.32
n=16
5. We have a random sample of n = 5 weights (kg) of a particular kind of
animal: 3.1, 3.4, 3.6, 3.7, 4.0. The sample mean, X-bar, is 3.56 kg.
Now compute the sample variance of these data, using the standard
’machine formula’ calculation for the sample variance, s^2:
s^2 = {Sum([Xi^2]) - [(Sum[Xi])^2]/n}/(n-1)
Sum([Xi2]) = 63.82;[Sum(Xi)]2 = 17.82 = 316.84; n = 5
s2 = (63.82-316.84/5)/4 = 0.113
6. The following data (X) are known to come from a lognormal distribution;
that is, the natural logarithms of the data, Y = ln(X), will follow a
normal (bell-shaped) distribution. Here are the data in the original
units:3.67, 4.01, 3.85, 3.92, 3.71, 3.88, 3.74, 3.82 ml.
a. Compute the sample mean of the log-transformed data.
Let yi = ln(xi)
ybar = sum(yi)/8 = 10.729/8 = 1.34
b. Compute the sample variance of the log-transformed data. Take the sqrt
to get the sample standard deviation.
s2y = (sum(yi2)-[sum(yi)]2/n)/(n-1)
= [14.396-10.729^2/8]/7 = 0.00089Slot Machine Binomial Distribution Calculator
sy = sqrt(0.00089) = 0.0298
c. It turns out that the 95% confidence interval (CI) for the mean of the
log-transformed data is xbar +- 2.365*(std. deviation/sqrt(n)). Compute the 95%
CI for the log-transformed data; then transform each of the endpoints back
to get the 95% CI in the original units [ml]. Comment upon the symmetry of
the CI for the log-transformed units [ln(ml)], and the back-transformed
units [ml].
1.34+/-2.365*0.0289/sqrt(8) = 1.34+/-0.0242
1.32<=mu(y)<=1.36
e1.32<=mu(x)<=e1.36
3.74<=mu(x)<=3.90
xbar=3.825
The CI in the transformed units is inherently symmetric about the mean because we added and subtracted the same value from the mean to obtain the end points.To assess symmetry in the original units, let’s measure the distance from the original mean and the lower and upper end points:
3.825-3.74 = 0.085
3.9-3.825 = 0.075
It appears that there is asymmetry in the CI for the original units, with the mean closer to the upper limit.This is not surprising because we utilized a non-linear transformation.That means that the relationship between the mean and the endpoints is not preserved when the data are back-transformed (it’s not simply a difference in scale).
Many people accounted for the difference as rounding error, and in fact different results were seen due to rounding.Points were subtracted if symmetry was commented on without actually any calculation to assess the symmetry (you can’t just say something is so, you have to show it is).
Online slot machines are truly random, and a casino knows what the payout percentages via a set of certifications. Each slot and random number generator part of the slot need to pass rigorous in-house testing as well as a suite of statistical tests created to detect bias. One of many standard tests for is each slot is the measure of its RTP, each slot has a fixed theoretical value of its return to player used as a baseline value. Auditors consider the probabilities of the subset and compare events part of the actual gameplay against the expected events. The primary tool uses for this type of analyses are called the chi-squared statistics and the binomial distribution. The auditor/s also analyze the events in successive rounds and the RNG needs to meet the standards of the industry, therefore the RNG implementation is to safeguard outcomes, while also ensuring values produced are uniformly distributed over the required interval.
Before a new online casino game can be made available to players the game must be free of operational flaws that allow that the house edge could be changed by the casino or the player. The most cited cause of dissatisfaction amongst online players is the belief that games can be modified to yield the desired outcomes of an individual player. Online casinos are tremendously cautious about players believing that mathematical tools, bots or system are used to beat their spins or games, this is why auditors main function is to ensure that each slot are accurate and stable in each round and obligated to express his opinion supported by facts to manipulate false believes that the outcome of a spin or round can be influenced. Auditors track the long-term RTP plus several other statistics over months and years and therefore picks up any change and when he does he raises a red flag that the slot is not operating correctly. Auditors focus on statistical stability and audits take place monthly and only once the auditor is satisfied with all outcomes he will certify the game as fair and RNG.Slot Machine Binomial Distribution FormulaOnline slots are audited and tested for software integrity by several world-renowned auditing and testing companies such as Certified Fair Gaming an industry leader in fairness and auditing found in 2003. GLI or Gaming Laboratories International is another global leader in RNG auditing and perform regular audits and test on software fairness and RNG. TST or Technical Systems Testing does independent testing and auditing and verification that the software offers 100% true RNG and fair results and that it has not been manipulated at all. Online casinos certified by eCOGRA are viewed by experienced players as the most trustworthy, eCOGRA is based in London and is an internationally approved testing laboratory accredited certified body and known for focussing on player protection. eCOGRA stands for eCommerce Online Gaming Regulations and Assurance and the majority of major online operators proudly display their eCOGRA stamp of approval on their main page.
Register here: http://gg.gg/ny3b3
https://diarynote-jp.indered.space
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