Roulette Project
March 3rd, 2013 at 1:44:09 PM
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Hello Forum
I am doing a roulette project for a graduate statistics course where I am taking different systems and running them through a similator to analyze standard deviation, variance, when you go broke etc.
I came across a bunch of systems and I wanted to ask the following:
An author that I am reading says the following:
The probability that a black would show up after one red = 73.63%.
I believe that theP(black) = P(red) = 48.64%.
Each trial is independent so regardless of what the last spin is, it does not affect the next spin.
However, the conditional probability that red has already occurred, would change the calculation, is that correct?
So is it true that probability of any of the following orders would be = 73.63%?
Red, Black
Black, Red
Even, Odd
Odd, Even
I did some calculations, I believe that this problem is similar to flipping a coin. However, I think the author is wrong with the calculation.
How would you calculate the P(red,black)?
I am doing a roulette project for a graduate statistics course where I am taking different systems and running them through a similator to analyze standard deviation, variance, when you go broke etc.
I came across a bunch of systems and I wanted to ask the following:
An author that I am reading says the following:
The probability that a black would show up after one red = 73.63%.
I believe that theP(black) = P(red) = 48.64%.
Each trial is independent so regardless of what the last spin is, it does not affect the next spin.
However, the conditional probability that red has already occurred, would change the calculation, is that correct?
So is it true that probability of any of the following orders would be = 73.63%?
Red, Black
Black, Red
Even, Odd
Odd, Even
I did some calculations, I believe that this problem is similar to flipping a coin. However, I think the author is wrong with the calculation.
How would you calculate the P(red,black)?
There's always an edge, are you sharp enough to find it?
March 3rd, 2013 at 2:23:15 PM
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Since the ball does not have a memory the prob of black following a red that happened is (18/38)=.4736
If the question is at the start of the roll of getting red followed by a black, the odd is (18/38)(18/38)=.2244
same for even,odd =.2244
If the question is at the start of the roll of getting red followed by a black, the odd is (18/38)(18/38)=.2244
same for even,odd =.2244
Edward Snowden is not the criminal, the government is for violating the constitution!
March 3rd, 2013 at 9:32:59 PM
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I agree.
I'm dealing with european roulette also, but the idea is the same.
I got the caluculation from John Solitude's document.
www.freewebs.com/turbogenius/JSW.pdf
There are other calculations in the document that I believe are incorrect, but I'm not an expert by any means.
As I am testing these systems against a 10,000 spin simulator, this particular document contends that the wheel over time will undergo periods of balance/imbalance. After a particular number of spins, you expect the numbers to have the same distributions so if you see imbalance, then you should expect balancing to occur in areas that are not represented. I need to present the information to my class and not miss any critiques.
Can anyone comment on the validity of this theory?
I'm dealing with european roulette also, but the idea is the same.
I got the caluculation from John Solitude's document.
www.freewebs.com/turbogenius/JSW.pdf
There are other calculations in the document that I believe are incorrect, but I'm not an expert by any means.
As I am testing these systems against a 10,000 spin simulator, this particular document contends that the wheel over time will undergo periods of balance/imbalance. After a particular number of spins, you expect the numbers to have the same distributions so if you see imbalance, then you should expect balancing to occur in areas that are not represented. I need to present the information to my class and not miss any critiques.
Can anyone comment on the validity of this theory?
There's always an edge, are you sharp enough to find it?
March 3rd, 2013 at 9:57:53 PM
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I was waiting for this...
In the game of craps throwing a 12 has a 1/36 chance of being rolled, but if the dice are out of balance....
In the game of craps throwing a 12 has a 1/36 chance of being rolled, but if the dice are out of balance....
March 4th, 2013 at 11:41:32 AM
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Dice is a great example.
It sounds like you're saying the theory is bogus. However, for the sake of argument, if I said over 500 tosses I've seen approximately equal amounts of 11 particular numbers and not the final 12th number. Does it make sense that the final number should appear more than otherse to compensate (even out) the distribution? It sounds like the answer is no.It seems as if the author is basing his system on isn't mathematically correct. Which isn't a big problem for me, unless i assume his analysis to be true and submit it in my report.
But lets say this is a random number generator, instead of an actual roulette wheel. Since this isn't happening in nature, true randomness is only being simulated. Would we then expect the distribution to undergo "smoothing"?
It sounds like you're saying the theory is bogus. However, for the sake of argument, if I said over 500 tosses I've seen approximately equal amounts of 11 particular numbers and not the final 12th number. Does it make sense that the final number should appear more than otherse to compensate (even out) the distribution? It sounds like the answer is no.It seems as if the author is basing his system on isn't mathematically correct. Which isn't a big problem for me, unless i assume his analysis to be true and submit it in my report.
But lets say this is a random number generator, instead of an actual roulette wheel. Since this isn't happening in nature, true randomness is only being simulated. Would we then expect the distribution to undergo "smoothing"?
There's always an edge, are you sharp enough to find it?
March 4th, 2013 at 12:30:50 PM
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John Solitude's document.Quote: beenittyHello Forum
I am doing a roulette project for a graduate statistics course where I am taking different systems and running them through a similator to analyze standard deviation, variance, when you go broke etc.
I came across a bunch of systems and I wanted to ask the following:
An author that I am reading says the following:
The probability that a black would show up after one red = 73.63%.
Yep.
Lots of incorrect statements in that document.
That statement is on page 19.
But on other pages he talks about hitting at least one Black in X number of spins.
His math for that is right on. He is talking about wait times in those sentences.
The probability of seeing black (18/37) at least one time in 2 spins = 73.63%
the 3rd column below (Binomail Distribution of 2 spins)
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.2636961 0.0000000 1.0000000 0.2636961 0.7363039
1 0.4996348 0.2636961 0.7363039 0.7633309 0.2366691
2 0.2366691 0.7633309 0.2366691 1.0000000 0.0000000
Long hand calculation
That comes from the probability of seeing a Black on the first spin OR
the second spin given it did not occur on the first spin
48.6486% (18/37)
24.9817% (19/37 * 18/37)
Add them up since it can be either spin
For example
99.9988% chance of seeing at least 1 Black in 17 spins is a correct statement
99.9988% chance of seeing Black after 16 Reds is an incorrect statement as you know
He should have done a better job of proofing his document.
winsome johnny (not Win some johnny)
March 4th, 2013 at 12:34:16 PM
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You are talking about the "gambler's fallacy" that basically theorizes that a number that hasn't hit is "due." Because each spin is independent and the ball has no memory, you are right in stating that the odds for red/black are the same for each individual spin. Predicting the odds of a series of results for a series of spins at the point of the first spin is much more complicated and the calculations are only good at the time of the first spin. The numbers would have to be recalculated after each spin until the last one which would be the same as an individual spin. The results of a series of spins are what would follow a distribution curve with standard deviations, etc. It is the combination of a set of individual events, each of which is completely independent from every other individual spin that makes up the distribution curve. For a great explanation of this you should visit the WizardofOdds.com website and search gamblers fallacy. Good luck with your project.
March 4th, 2013 at 12:56:11 PM
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Quote: beenitty
But lets say this is a random number generator, instead of an actual roulette wheel. Since this isn't happening in nature, true randomness is only being simulated. Would we then expect the distribution to undergo "smoothing"?
A computer psuedo-random number generator should not add any 'smoothing' in. The chance of the next number coming up at any time should be equal everytime.
If it isn't, then it's not random.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
March 4th, 2013 at 12:59:07 PM
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Also the misunderstanding of the probability of a series of bets as they unfold.Quote: skrbornevryminYou are talking about the "gambler's fallacy" that basically theorizes that a number that hasn't hit is "due."
The author of the document uses "triggers" (going from my memory of reading that pdf) to tell when to start making a bet.
that if, for example,
black has not shown in the last 10 spins,
that is a great trigger to start betting for the next 10 spins because the probability of hitting at least 1 Black in the very next 10 spins is
99.8725%
(1 - (19/37)^10)
But he does not have to wait for 10 non-blacks in a row.
That is the fallacy.
the probability of hitting at least 1 Black in the very next 10 spins
regardless of what happened in the past is still 99.8725% starting from the first spin
winsome johnny (not Win some johnny)
March 4th, 2013 at 2:02:43 PM
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Wow, that John Solitude document was beyond absurd.
March 4th, 2013 at 2:24:51 PM
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Great information, thanks.
So it looks like this Solitude document is out because I cant present data based upon theory that is mathematically incorrect.
Can anyone suggest a system that is available, or at least the theory behind that system that would be interesting for analysis.
I have a list of the classic systems and hybrids but looking for something that makes sense in terms of probability mathematically regardless of the results.
(i.e. martingale, etc = out)
So it looks like this Solitude document is out because I cant present data based upon theory that is mathematically incorrect.
Can anyone suggest a system that is available, or at least the theory behind that system that would be interesting for analysis.
I have a list of the classic systems and hybrids but looking for something that makes sense in terms of probability mathematically regardless of the results.
(i.e. martingale, etc = out)
There's always an edge, are you sharp enough to find it?
March 4th, 2013 at 2:50:10 PM
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Quote: beenittyGreat information, thanks.
So it looks like this Solitude document is out because I cant present data based upon theory that is mathematically incorrect.
Can anyone suggest a system that is available, or at least the theory behind that system that would be interesting for analysis.
I have a list of the classic systems and hybrids but looking for something that makes sense in terms of probability mathematically regardless of the results.
(i.e. martingale, etc = out)
There are several in the "betting systems" section of this website. Ultimately, however, the analyses will all conclude the same thing: that expected loss (in the long term) is equal to the dollar amount wagered multiplied by the house edge of each seperate bet, without regard to the combination of such bets. In the short term, results can be temporarily skewed under some circumstances, but the long term results will always approximate the house edge multiplied by the amount wagered. For example, the martingale "works" until you loose the largest bet you are willing to make. How soon that will happen (the length of the "short term") depends only on a person's risk aversity (or lack of it) and luck.
On the other hand, you could use a system like the one you described and re-do the math correctly to show under which circumstances (if any) it would work in the short run, and why it would not work in the long run.
March 4th, 2013 at 4:01:41 PM
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I agree.
Bettor will always go broke.
To further the conversation above, Im curious about the following.
For anyone who's read Solitude's document, you know what Im talking about when he theorizes that the wheel should have balance.
Technically, thats another way of falling into gambler's fallacy.
So, lets say you could actually track in real time 10,000 spins. As you approach a large sample size, does his argument of balancing breaks down because you can never know when the wheel will exhibit "balancing"? Lets say you do see a section of the wheel that isn't seeing as many hits as other sections. After a particular number of spins (i.e. 1000 spins), if you saw this phenomenon, you could start betting that section but you don't know when the distribution will "balance" and it doesnt in theory have to do it all at the same time.
Is this correct analysis?
Bettor will always go broke.
To further the conversation above, Im curious about the following.
For anyone who's read Solitude's document, you know what Im talking about when he theorizes that the wheel should have balance.
Technically, thats another way of falling into gambler's fallacy.
So, lets say you could actually track in real time 10,000 spins. As you approach a large sample size, does his argument of balancing breaks down because you can never know when the wheel will exhibit "balancing"? Lets say you do see a section of the wheel that isn't seeing as many hits as other sections. After a particular number of spins (i.e. 1000 spins), if you saw this phenomenon, you could start betting that section but you don't know when the distribution will "balance" and it doesnt in theory have to do it all at the same time.
Is this correct analysis?
There's always an edge, are you sharp enough to find it?
March 4th, 2013 at 4:32:42 PM
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John Solitude apparently isn't familiar with the gambler's fallacy. What he has written is silly bunk.
October 30th, 2013 at 4:22:07 PM
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Super late. I never thanked everyone for their help in the forum. My project got the highest grade in the class, even with a standing ovation after, pretty cool.
THANKS EVERYONE!!
THANKS EVERYONE!!
There's always an edge, are you sharp enough to find it?
November 2nd, 2013 at 12:03:05 AM
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It seems to me this is exactly why they display the numbers that have hit - some bettors see a streak of black or red / odd or even and make a bet thinking "It's gotta hit!".
And they're right. It's gotta hit eventually. Will they still have money left when it does?
And they're right. It's gotta hit eventually. Will they still have money left when it does?
November 2nd, 2013 at 3:39:57 PM
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Great point.
Bankroll for a negative expectation game is a big deal. I've been playing alot at empire casino. They give you the last 42 numbers and the number of occurrences for each number in the last 300 spins, in addition to the hottest numbers and coldest numbers over the last (I think) 100 spins. Some of that data is useless but for a quant based bettor it does help you get an idea of what to bet. The average roulette player (from what I've seen) has their numbers that they like an play those. Our project confirmed what many ppl in this forum already know. Betting your lucky numbers is no better than randomly picking numbers and betting them. Some strategies are have less loss than others but they're all still constrained by the negative expectation. Having a larger bankroll does help you absorb loosing streaks but each spin is a losing proposition for sure.
Bankroll for a negative expectation game is a big deal. I've been playing alot at empire casino. They give you the last 42 numbers and the number of occurrences for each number in the last 300 spins, in addition to the hottest numbers and coldest numbers over the last (I think) 100 spins. Some of that data is useless but for a quant based bettor it does help you get an idea of what to bet. The average roulette player (from what I've seen) has their numbers that they like an play those. Our project confirmed what many ppl in this forum already know. Betting your lucky numbers is no better than randomly picking numbers and betting them. Some strategies are have less loss than others but they're all still constrained by the negative expectation. Having a larger bankroll does help you absorb loosing streaks but each spin is a losing proposition for sure.
There's always an edge, are you sharp enough to find it?
