(Please don't comment about how we can't make that assumption, etc, etc. I am asking you to make the assumption - just pretend for a second please.)

How many trials (dice rolls) would it take to show that the system is in fact a winning system?

Said another way - if I thought I had a winning system, and I simulated it 1000 times and it wins (more bank roll than starting bank roll) - would that be enough rolls to prove it? How about 2000 rolls? 5000?

I guess another way to say it, is given that the house edge with optimal strategy is 1.36 - 1.41% - how many rolls would it take with a 99% probability that bank roll is less than starting? If we get that number, could one use that number to then say - if I can show a positive bankroll after that number of rolls, there is only a 1% chance it isn't a winning system?

but it's not easy to calculate unless you flat-bet

If the "system" is "always bet 1", then, if the bet is red on a single-zero roulette wheel and your starting bankroll is 100, the probability of being ahead at any point before losing the entire 100 is about 90%.

In fact, if the probability p of winning a single bet < 1/2, then, as your initial bankroll increases, the probability of being ahead B bets before losing the entire bankroll approaches ( (p / (1 - p) )

^{B}. In the single-zero roulette case, p = 9/19, so 1 - p = 10/19, p / (1 - p) = 9/10, and the probability of being ahead by 1 bet before busting approaches 9/10 as your initial bankroll increases, being ahead by 2 bets before busting approaches 81/100, being ahead by 3 bets before busting approaches 729/1000, and so on.

If I am crunching my numbers correctly, for the single-zero roulette even-money flat bet system, if you try it 100 times, you are expected to win 90 times, and the standard deviation is 3 (I am not sure if there is an "easy" way to calculate this; in 100 trials, assuming the probability of being ahead at some point before being behind by 100 is 0.9, then the probability of n out of the 100 trials being wins is (100)C(n) x 0.9

^{n}x 0.1

^{100-n}; some Excel number crunching shows the mean is 90 and the standard deviation is 3), so, in this case, a 99% success rate would be +3 standard deviations.

Could you do the same for craps.

Assuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?

Quote:slackyhackyThanks.

Could you do the same for craps.

Assuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?

What bets are you making? Every bet has a different EV and SD, as well as if you're doing a combination of bets.

I'd determine what the N0 is for a game. N0 (in a -EV game) is the number of rolls/hands/slot spins/etc. played such that 1 standard deviation = expected value. After playing 1 N0 rounds, you'll be within 1 SD 68.2% of the time.

Just figure out how many rounds you gotta play such that EV = 3 SD's. Or just skip straight to it and figure it out for 4 or 5 SD's.

Quote:slackyhackyAssuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?

Do you mean ahead at any point, or just ahead at the end of the set of rolls?

If you mean being ahead at the end of the set of rolls, I get:

100: 40.47%

200 39.34%

500: 35.94%

1000: 31.64%

Yes, the chance of being ahead decreases with more rolls; this is because you are exposing more bets to the house edge.

The probability of being ahead at any point before reaching N rolls is going to get closer and closer to p / (1 - p) as N increases.

Wait a minute...that's actually the probability of being ahead before being N behind. What you want to know is a different problem.

However, after some number crunching, I get a probability of 90.6835% of being ahead at some point in 100 come-outs.

Bets | Probability |
---|---|

100 | 90.6835% |

200 | 92.943% |

300 | 93.9362% |

400 | 94.5228% |

500 | 94.9194% |

600 | 95.2093% |

700 | 95.4324% |

800 | 95.6105% |

900 | 95.7566% |

1000 | 95.8789% |

1100 | 95.9832% |

1200 | 96.0732% |

1300 | 96.1518% |

1400 | 96.2212% |

1500 | 96.2829% |

1600 | 96.3382% |

1700 | 96.388% |

1800 | 96.4331% |

1900 | 96.4743% |

2000 | 96.5119% |

Note that p = 1/2 - HE/200; in this case, the house edge is 1.41%, so p = 1/2 - 1.41/200 = 0.49295, and p / (1 - p) = about 97.22%.

Note: do not PM me with requests

I don't mean at any point - I mean at the end of the x number of rolls.

Thanks for the numbers.

I'm having a hard time following the last post.

The spreadsheet was for the probability of being ahead at any point right?

Your whole premise is flawed though. Money management just rearranges variance. You might be able to create a system that wins 99% of the time, but that 1% is going to ream you. The hard work has already been done: you can easily walk into a casino and do a roulette martingale tomorrow with less than a 1% chance of losing--- but God help you if it does.

Quote:slackyhackyI don't mean at any point - I mean at the end of the x number of rolls.

Thanks for the numbers.

Note that I calculated based on 100 (for example) bets - not 100 rolls. The 100 rolls problem is harder as you have to account for different numbers of rolls per bet.

Quote:slackyhackyI'm having a hard time following the last post.

The spreadsheet was for the probability of being ahead at any point right?

Correct - at first, I listed the formula for being ahead at any point before being a particular number behind, before noticing that you wanted to know the probability of being ahead at any point before a certain number of bets (which is what is on the spreadsheet).

Quote:ThatDonGuy

If you mean being ahead at the end of the set of rolls, I get:

100: 40.47%

200 39.34%

500: 35.94%

1000: 31.64%

Yes, the chance of being ahead decreases with more rolls; this is because you are exposing more bets to the house edge.

What is the equation you used to figure this out?

Quote:ThatDonGuy

Note that p = 1/2 - HE/200; in this case, the house edge is 1.41%, so p = 1/2 - 1.41/200 = 0.49295, and p / (1 - p) = about 97.22%.

What does this math represent? What does 49% mean? What does 97% mean?

Quote:slackyhackyWhat is the equation you used to figure this out?

If p is the probability of winning a particular even-money bet, then the probability of being ahead after N bets is:

(N)C(N) x p

^{N}

+ (N)C(N-1) x p

^{N-1}x (1 - p)

+ (N)C(N-2) x p

^{N-2}x (1 - p)

^{2}

+ (N)C(N-3) x p

^{N-3}x (1 - p)

^{3}

+ ...

+ (N)C(K) x p

^{K}x (1 - p)

^{N-K}

where (A)C(B) is the number of combinations of A things taken B at a time (this is also written as C(A,B), or Combin(A,B))

p is the probability of winning a single bet

K is the smallest integer > N/2 (e.g. if N = 100, K = 51; if N = 101, K is also 51)

There's no way to "simplify" this that I know of; you need to use a spreadsheet to calculate each value and then add them up.

Quote:slackyhackyWhat does this math represent? What does 49% mean? What does 97% mean?Quote:ThatDonGuyNote that p = 1/2 - HE/200; in this case, the house edge is 1.41%, so p = 1/2 - 1.41/200 = 0.49295, and p / (1 - p) = about 97.22%.

49.295% is the probability of winning the bet; if you know the house edge on a "win-or-lose" bet, the probability of winning = 1/2 - HE/2.

I shouldn't have used percent with 0.9722; the value is p / (1 - p), and is used in the formula I posted earlier in the thread about calculating the probability of being ahead at any point before losing your entire bankroll.

Quote:ThatDonGuyIf p is the probability of winning a particular even-money bet, then the probability of being ahead after N bets is:

(N)C(N) x p^{N}

+ (N)C(N-1) x p^{N-1}x (1 - p)

+ (N)C(N-2) x p^{N-2}x (1 - p)^{2}

+ (N)C(N-3) x p^{N-3}x (1 - p)^{3}

+ ...

+ (N)C(K) x p^{K}x (1 - p)^{N-K}

where (A)C(B) is the number of combinations of A things taken B at a time (this is also written as C(A,B), or Combin(A,B))

p is the probability of winning a single bet

K is the smallest integer > N/2 (e.g. if N = 100, K = 51; if N = 101, K is also 51)

There's no way to "simplify" this that I know of; you need to use a spreadsheet to calculate each value and then add them up.

49.295% is the probability of winning the bet; if you know the house edge on a "win-or-lose" bet, the probability of winning = 1/2 - HE/2.

I shouldn't have used percent with 0.9722; the value is p / (1 - p), and is used in the formula I posted earlier in the thread about calculating the probability of being ahead at any point before losing your entire bankroll.

Holy crap. I guess this is why you didn't want me to PM you.

Quote:bbvk05If you invented a money management system that defied the fundamental math of a game I'm not sure any number of trials would prove it, but I'm sure you'd be very famous with all the attention you'd attract in the mathematics world with your discovery. But if you had billions of quality simulations that showed something notable then you'd have my attention.

Your whole premise is flawed though. Money management just rearranges variance. You might be able to create a system that wins 99% of the time, but that 1% is going to ream you. The hard work has already been done: you can easily walk into a casino and do a roulette martingale tomorrow with less than a 1% chance of losing--- but God help you if it does.

Maybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

Quote:slackyhackyMaybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

Then the system loses more often. You can't escape the house edge by makeing more bets that have a house edge.

Your roulette thing is rank superstition. That can be fun but it has no value. You're just following the "it's due" theory.

Quote:slackyhackyMaybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

Then the system loses more often. You can't escape the house edge by makeing more bets that have a house edge.

Your roulette thing is rank superstition. That can be fun but it has no value. You're just following the "it's due" theory.

So I couldn't figure out your formula -

Instead, I took your 4 points, plotted them in excel, did an equation fit - then used that formula to get some more data. It isn't exact, but I think it is close.

at 16000 roles, there is a 0.3% chance you will be winning.

at 32000 roles, there is a .003% chance you will be winning.

So - if I came up with system that was ahead after 32000 roles, the likelyhood that it is a loosing system is 0.003% chance.

That's pretty cool. Now I just need to come up with a system that will do that.

I still think it is better described by gamblers fallacy - or the fallacy of the maturity of chance.

What do you mean it has no value? I'm not sure if your point of saying that. Did you think that I believe I could sell the idea - this meaning the idea has value?

Of course it has no value. Let me give you some hints to my statement that should have given you the clue that I know it doesn't have value - I used the word "faulty". I also acknowledged that my thinking is prey to falllacy.

If you meant that - yes, you agree with me that my method of play has no value - ok then.

But you seemed to be trying to point out to me as if I didn't recognize it already - which in that case, your statement about value has no value.

Quote:slackyhackyI couldn't find a definition of "rank superstition" anywhere on the internet so I have no idea what that means.

I still think it is better described by gamblers fallacy - or the fallacy of the maturity of chance.

What do you mean it has no value? I'm not sure if your point of saying that. Did you think that I believe I could sell the idea - this meaning the idea has value?

Of course it has no value. Let me give you some hints to my statement that should have given you the clue that I know it doesn't have value - I used the word "faulty". I also acknowledged that my thinking is prey to falllacy.

If you meant that - yes, you agree with me that my method of play has no value - ok then.

But you seemed to be trying to point out to me as if I didn't recognize it already - which in that case, your statement about value has no value.

Acknowledging something is faulty then doing it anyway doesn't get you much credit in my book. It's one of the several things people call the gamblers fallacy.

Here's the relevant definition of "rank" that google provided:

Rank (especially of something bad or deficient): complete and utter (used for emphasis).

"rank stupidity" synonyms: downright, utter, outright, out-and-out, absolute, complete, sheer, arrant, thoroughgoing, unqualified, unmitigated, positive, perfect, patent, pure, total; archaicarrant

"rank stupidity"

Quote:slackyhackyOkay,

So I couldn't figure out your formula -

Instead, I took your 4 points, plotted them in excel, did an equation fit - then used that formula to get some more data. It isn't exact, but I think it is close.

at 16000 roles, there is a 0.3% chance you will be winning.

at 32000 roles, there is a .003% chance you will be winning.

So - if I came up with system that was ahead after 32000 roles, the likelyhood that it is a loosing system is 0.003% chance.

That's pretty cool. Now I just need to come up with a system that will do that.

I couldn't calculate the numbers directly after 1050 - it seems that 0.49295

^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

If you always live your life without doing things that you know are faulty - then you are an amazing person. You should write a book about your travels and choices - that would be an inspiring and wonderful read.

Wow. My equation wasn't even close. Although not all is lost. I learned how to fit data points to a line and get an equation from that fit. It's a pretty cool excel function. I tried a 4th order polynomial but exponential seemed to fit the best.

Quote:slackyhackyBbvk05,

If you always live your life without doing things that you know are faulty - then you are an amazing person. You should write a book about your travels and choices - that would be an inspiring and wonderful read.

I don't think that's what I said. I said it doesn't get you much credit. Most importantly, the faulty thing here is a superstitious belief. I'm sure I have superstitious beliefs, and I eradicate them when I find them.

Quote:bbvk05I'm sure I have superstitious beliefs, and I eradicate them when I find them.

Superstitious believe is a great weapon at a craps table. Turning bets off because of a tray lizard or stick change or .......

Your math skill are very impressive. Thanks for that. What is your background?

I have such a blind spot for stats.

Quote:slackyhackyThatDonGuy -

Your math skill are very impressive. Thanks for that. What is your background?

I have such a blind spot for stats.

"Mathhead" pretty much from birth

Bachelor's degree in Computer Science from Cal-Berkeley, 1984

To be fair, I am not as much of a statistics expert as it appears here - sites like Wolfram Mathworld are my (and your) friend.

Quote:ThatDonGuyI couldn't calculate the numbers directly after 1050 - it seems that 0.49295

^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

So ya saying there is a chance?????

Quote:billryanQuote:ThatDonGuyI couldn't calculate the numbers directly after 1050 - it seems that 0.49295

^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

So ya saying there is a chance?????

Yes. There's also a "chance" that you will win 16,000 bets in a row. Not a very likely one, mind you, but it's greater than zero.

Quote:billryanQuote:ThatDonGuyI couldn't calculate the numbers directly after 1050 - it seems that 0.49295

^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

So ya saying there is a chance?????

Haha. Love it.

Quote:ThatDonGuyQuote:billryanQuote:ThatDonGuy^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

So ya saying there is a chance?????

Yes. There's also a "chance" that you will win 16,000 bets in a row. Not a very likely one, mind you, but it's greater than zero.

!!!Amazing!!!

Quote:ThatDonGuyQuote:billryanQuote:ThatDonGuy^{1051}exceeds the smallest possible positive value for a floating-point number in Visual Studio - but I did do a simulation of 250,000 sets of 32,000 bets, and got about a 3.65% chance of being ahead after 16,000 bets and 0.58% after 32,000.

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

So ya saying there is a chance?????

Yes. There's also a "chance" that you will win 16,000 bets in a row. Not a very likely one, mind you, but it's greater than zero.

Think about this.

52 weeks a year times 5 days a week= 260 days.

Lets go with 250.

Lets say 400 rolls per day.

250days x 400rolls/day = 100,000 rolls per year.

One Billion rolls (1,000,000,000) divided by

One hundred thousand rolls per year (100,000)= Ten Thousand Years

Could someone post One Billion craps dice rolls (1,000,000,000)

Formatted to fit the above example. or

Ten Thousand Folders each with two hundred fifty files

containing 400 craps dice rolls each.

Just read my joining email Posted below in wrong section. re-posting here.

mikegaurdwizard

mikegaurdwizard Nov 18, 2017

I have always liked a challenge. My first gambling challenge was to beat the field on the craps game. While using some useless betting strategies on binary options. Some sunlight got... anyway. The (Unbeatable) field has been beaten. My goal was to beat the field, I did. Now my goal is to sell both of my betting strategies. Which beat the field. Just have no idea how, Or what they would be worth. Looking for advice. Would have been nice to have figured this out in the 70's or 80's, or even the 90's. If not for Binary options, I am sure I would never had revisited gambling systems. Turned $250 into $28,000 with just 4000 Binary option trades 88% payout.

FYI my only interest in Gambling came in the late 70's early 80's when I heard The field could not be beaten. I took after a good friend, who would never allow himself to lose a challenge. If not for Binary Options the field would have won. During my quest to make money with Binary Options. I developed several systems most of which were based on useless gambling systems. I used the field to test my Binary Options systems. Then one day. I thought of something. I had never seen. Could not find it on the internet.

My Binary Options system worked on the field. Then I thought of a second system It also worked on the field.

I used to agree with #4. below. But the below concept is no longer true.

4. This forum generally respects free speech. However, if you are here to discuss your "winning" betting system, please know that all betting systems are equally worthless and none of them have ever, or can ever, beat the house in the long-run. In fact, they can't even dent the house advantage. Betting system believers face a hostile audience in my forum and usually leave angry at how close-minded the other members are on the topic. If you still feel compelled to discuss the topic, may I suggest the forums of John Patrick or the Flat Earth Society However, if you still wish to face the ridicule, you must confine posts about betting systems to the Betting System section

My challenge is for any-one who cares. Post a Billion Craps dice rolls.