by PM I posed a question to the Wizard and he indicated if I posted it he would respond

the question has not been considered here (to my knowledge) or anywhere else that I know about

please don't attempt an answer until after the Wizard has posted

thank you

the question I posed was this:

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

but I don't know how to do the math to figure this out correctly

the Wizard answered my PM indicating that there is a way to figure this out although there may not be a handy formula

I'm looking forward to seeing his post

.

Quote:lilredrooster.

by PM I posed a question to the Wizard and he indicated if I posted it he would respond

the question has not been considered here (to my knowledge) or anywhere else that I know about

please don't attempt an answer until after the Wizard has posted

thank you

the question I posed was this:

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

but I don't know how to do the math to figure this out correctly

the Wizard answered my PM indicating that there is a way to figure this out although there may not be a handy formula

I'm looking forward to seeing his post

.

link to original post

For this one, you could either use the Beating Bonuses simulator or a binomial distribution calculator; the fact that these are flat bets at Even Money makes this quite easy to do if you are specifying a particular number of trials.

If you lose more trials than you win, then you are down money. The Binomial Distribution calculator can give you the probability of losing x or more where x is one more than half of all attempts.

You asked me not to attempt to do this until Wizard posts in this thread, so I won’t.

Quote:SOOPOOI was hoping for another roulette thread. I’m anxiously awaiting ‘but I make non random bets on the random outcomes, so I never lose…’. More sewage here….

link to original post

If we could ban EvenBob from just ONE roulette thread where he wasn't able to vomit his worthless masturbatory word salad of lies, we could actually have an interesting discussion about the game....

Quote:TigerWuQuote:SOOPOOI was hoping for another roulette thread. I’m anxiously awaiting ‘but I make non random bets on the random outcomes, so I never lose…’. More sewage here….

link to original post

If we could ban EvenBob from just ONE roulette thread where he wasn't able to vomit his worthless masturbatory word salad of lies, we could actually have an interesting discussion about the game....

link to original post

I disagree. There really can’t be anything interesting about roulette. Any bright high schooler can have it figured out from A to Z in an hour. Lilred’s question that started this thread can be answered with just a few minutes work by Mission, the Wiz, Mental, Ace2, and likely a host of others. Just looking at Lilred’s guesstimation he’s probably pretty close.

Quote:Mission146Quote:lilredrooster.

by PM I posed a question to the Wizard and he indicated if I posted it he would respond

the question has not been considered here (to my knowledge) or anywhere else that I know about

please don't attempt an answer until after the Wizard has posted

thank you

the question I posed was this:

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

but I don't know how to do the math to figure this out correctly

the Wizard answered my PM indicating that there is a way to figure this out although there may not be a handy formula

I'm looking forward to seeing his post

.

link to original post

For this one, you could either use the Beating Bonuses simulator or a binomial distribution calculator; the fact that these are flat bets at Even Money makes this quite easy to do if you are specifying a particular number of trials.

If you lose more trials than you win, then you are down money. The Binomial Distribution calculator can give you the probability of losing x or more where x is one more than half of all attempts.

You asked me not to attempt to do this until Wizard posts in this thread, so I won’t.

link to original post

I’d also like to add that what I have detailed above would work for literally ANY isolated Roulette bet, provided:

1. Running out of money is not possible.

-This is important because being able to bust out increases the bust rate substantially, though this effect would be diminished the greater the number of trials. The reason why is because, the greater the number of trials, the more likely you are to finish down anyway.

2. The bet amount is constant.

-The bet amount has to remain constant for obvious reasons.

The only added step would be to determine (based on the number of attempts) x, where x is the number of trials you’d have to lose in order to be down money. This would be a fairly trivial calculation that could easily be done with simple algebra solving for x where x is zero.***

This is possible for Roulette because amounts to be won (assuming a flat bet amount), unlike BJ and UTH (for two examples), are not variable.

***Let me explain that better. We would start with a fixed number of trials. At that point, we would determine how many of those trials lost would put us even using zero as the target result and x as the number of trials lost (multiplied by probability of losing) and overall number of trials, then —-number of trials - x —-multiplied by gain on wins as the number of trials won. Using zero on the right side of the equation would closely approximate x were x not exact. In an even money bet, x would be exact (unless the number of trials were an odd number).

I hope this explanation is phrased correctly. It would be easier just to do it, but I’m not allowed, per the OP. JK

Quote:SOOPOOQuote:TigerWuQuote:SOOPOOI was hoping for another roulette thread. I’m anxiously awaiting ‘but I make non random bets on the random outcomes, so I never lose…’. More sewage here….

link to original post

If we could ban EvenBob from just ONE roulette thread where he wasn't able to vomit his worthless masturbatory word salad of lies, we could actually have an interesting discussion about the game....

link to original post

I disagree. There really can’t be anything interesting about roulette. Any bright high schooler can have it figured out from A to Z in an hour. Lilred’s question that started this thread can be answered with just a few minutes work by Mission, the Wiz, Mental, Ace2, and likely a host of others. Just looking at Lilred’s guesstimation he’s probably pretty close.

link to original post

Just because I can solve for it doesn’t make it not interesting. If I can ever convince at least one person in my life not to play a-EV game straight up, then I have done the best job I can.

Quote:Mission146Quote:Mission146Quote:lilredrooster.

by PM I posed a question to the Wizard and he indicated if I posted it he would respond

the question has not been considered here (to my knowledge) or anywhere else that I know about

please don't attempt an answer until after the Wizard has posted

thank you

the question I posed was this:

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

but I don't know how to do the math to figure this out correctly

the Wizard answered my PM indicating that there is a way to figure this out although there may not be a handy formula

I'm looking forward to seeing his post

.

link to original post

For this one, you could either use the Beating Bonuses simulator or a binomial distribution calculator; the fact that these are flat bets at Even Money makes this quite easy to do if you are specifying a particular number of trials.

If you lose more trials than you win, then you are down money. The Binomial Distribution calculator can give you the probability of losing x or more where x is one more than half of all attempts.

You asked me not to attempt to do this until Wizard posts in this thread, so I won’t.

link to original post

I’d also like to add that what I have detailed above would work for literally ANY isolated Roulette bet, provided:

1. Running out of money is not possible.

-This is important because being able to bust out increases the bust rate substantially, though this effect would be diminished the greater the number of trials. The reason why is because, the greater the number of trials, the more likely you are to finish down anyway.

2. The bet amount is constant.

-The bet amount has to remain constant for obvious reasons.

The only added step would be to determine (based on the number of attempts) x, where x is the number of trials you’d have to lose in order to be down money. This would be a fairly trivial calculation that could easily be done with simple algebra solving for x where x is zero.***

This is possible for Roulette because amounts to be won (assuming a flat bet amount), unlike BJ and UTH (for two examples), are not variable.

***Let me explain that better. We would start with a fixed number of trials. At that point, we would determine how many of those trials lost would put us even using zero as the target result and x as the number of trials lost (multiplied by probability of losing) and overall number of trials, then —-number of trials - x —-multiplied by gain on wins as the number of trials won. Using zero on the right side of the equation would closely approximate x were x not exact. In an even money bet, x would be exact (unless the number of trials were an odd number).

I hope this explanation is phrased correctly. It would be easier just to do it, but I’m not allowed, per the OP. JK

link to original post

you forgot to factor in the the ability to make non-random bets, which somehow changes everything.

Probability of having up to #Win success in #Spin trials (a loss or breakeven).

Prob #Win #Spin

99.99% 10000 20000

99.67% 5000 10000

81.24% 500 1000

75.90% 300 600

64.47% 50 100

Just use BINOMDIST(num_successes, num_trials, prob_success, cumulative) on any spreadsheet

BINOMDIST(#win, #spin, 18/37, 1)

I agree with SOOPOO. There is nothing interesting or hard about the statistics of flat betting roulette.

Quote:lilredrooster.

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

link to original post

The first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

Quote:EvenBobQuote:lilredrooster.

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

link to original post

The first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

link to original post[/q

I told you.The difference between fiction and reality is that fiction is supposed to make sense.

several posters seemed to indicate it is a simple question to answer

but that's not what I got from the Wizard's PM back to me

he wrote this:

"My solution would be brute force combinations in Excel. I don't know that there is a handy single formula for that."

.

Quote:TigerWuQuote:SOOPOO

link to original post

If we could ban EvenBob from just ONE roulette thread where he wasn't able to vomit his worthless masturbatory word salad of lies, we could actually have an interesting discussion about the game....

link to original post

That reaches personal insult.

3 days.

Quote:lilredrooster.

several posters seemed to indicate it is a simple question to answer

but that's not what I got from the Wizard's PM

link to original post

They think it's simple because they don't understand what's going on. They think the house edge has an effect on every bet you make in roulette even in the short term. That pesky HE makes the game impossible to beat. This is absolutely true if you're talking long-term betting randomly you're screwed. Put in the extreme short-term, where you make the next bet and where we all live, there are no rules there is only short-term variance and you can get an edge over the game for one or two spins if you know what you're doing.

Quote:EvenBobQuote:lilredrooster.

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

link to original post

The first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

link to original post

Dangit, I really hate to agree with EvenBob, but this is almost exactly what I was going to contribute to the "interesting discussion".

If you've got a voodoo piñata (or whatever) that predicts spins accurately, I think you've got a chance.

Otherwise, you're just blowing in the wind.

Quote:lilredrooster

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

.

link to original post

Here are the exact answers, to six decimal places.

Spins | Net Win | Push | Net Loss |
---|---|---|---|

100 | 0.265023 | 0.069282 | 0.665695 |

200 | 0.207117 | 0.042698 | 0.750185 |

300 | 0.165841 | 0.030361 | 0.803798 |

400 | 0.134792 | 0.022893 | 0.842315 |

500 | 0.110664 | 0.017826 | 0.871510 |

600 | 0.091518 | 0.014167 | 0.894315 |

700 | 0.076106 | 0.011418 | 0.912476 |

800 | 0.063567 | 0.009298 | 0.927135 |

900 | 0.053283 | 0.007631 | 0.939086 |

1000 | 0.044796 | 0.006302 | 0.948902 |

This was with the binomial distribution by brute force in Excel. I am not aware of a simple formula to get these figures, although the Gaussian curve approximation would get you pretty close.

Isaac Newton discovered the binomial theorem about 1665 and later stated, in 1676, without proof, the general form of the theorem (for any real number n), and a proof by John Colson was published in 1736. So, the exact formula for the answer has been known for three centuries.Quote:lilredrooster.

several posters seemed to indicate it is a simple question to answer

but that's not what I got from the Wizard's PM back to me

he wrote this:

"My solution would be brute force combinations in Excel. I don't know that there is a handy single formula for that."

link to original post

The binomial expansion is a simple formula when the number of spins, N, is small. It is a polynomial with N+1 terms. So Wiz is right that it is a PITA to calculate it by calculating the N+1 terms in the polynomial and summing them up when N is large. And there is no simple formula that gets an exact answer because the exact answer has N+1 terms and has already been simplified. Wiz correctly suggests that you can use a gaussian approximation for large N.

At first, I thought it odd that Wiz said plugging 3 numbers into Excel is 'brute force'. However, the algorithm that Excel does behind the scenes involves a very large number of floating point operations. Using it is easy peasy.

Quote:EvenBobThe first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

link to original post

Hijacking warning. Let's not have discussion of your secret method to beat roulette ruin another thread.

Quote:WizardQuote:EvenBobThe first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

link to original post

Hijacking warning. Let's not have discussion of your secret method to beat roulette ruin another thread.

link to original post

So how do I answer statements that are directed at me like this one.

https://wizardofvegas.com/forum/gambling/other-games/38716-house-edge-effect-on-roulette/#post908816

You be a man of your word and stop talking about Roullet. You can answer that you don't talk about it anymore here.Quote:EvenBobQuote:WizardQuote:EvenBobThe first question is, how would he be choosing where to bet. Is he going to be doing the equivalent of standing at the end of the layout and just throwing his chips and betting wherever they land or does he have a specific non-random way of playing. This is the most important question in the whole thing and you conveniently left it out.

link to original post

Hijacking warning. Let's not have discussion of your secret method to beat roulette ruin another thread.

link to original post

So how do I answer statements that are directed at me like this one.

https://wizardofvegas.com/forum/gambling/other-games/38716-house-edge-effect-on-roulette/#post908816

link to original post

Quote:EvenBobQuote:lilredrooster.

several posters seemed to indicate it is a simple question to answer

but that's not what I got from the Wizard's PM

link to original post

They think it's simple because they don't understand what's going on. They think the house edge has an effect on every bet you make in roulette even in the short term. That pesky HE makes the game impossible to beat. This is absolutely true if you're talking long-term betting randomly you're screwed. Put in the extreme short-term, where you make the next bet and where we all live, there are no rules there is only short-term variance and you can get an edge over the game for one or two spins if you know what you're doing.

link to original post

Do you think before you post or do your fingers simply open up their sphincter?

I think I'll stick to trying to doubling my session money by winning to +7 bets within 50 spins. I might run over on the number of spins. It'd be -13 bets or +7 bets with some error to be built in from bet increase/decrease creep.

Quote:billryanDo you think before you post or do your fingers simply open up their sphincter?

link to original post

Personal insult. I see you've already had three suspensions since September, so let's make it seven days this time.

Quote:MentalAt first, I thought it odd that Wiz said plugging 3 numbers into Excel is 'brute force'. However, the algorithm that Excel does behind the scenes involves a very large number of floating point operations. Using it is easy peasy.

link to original post

You're right. I should have done it the easy way. Here is how it works:

= BINOMDIST(Number of successes,Total trials, Probability of success, cumulative?)

The last term is 0 for exactly that many number of success and 1 or that many or less.

In the case of 100 bets:

51 or more wins = 49 or less losses = BINOMDIST(49,100,20/38,1)

Exactly 50 wins = BINOMDIST(50,100,18/38,0)

49 or less wins = BINOMDIST(49,100,18/38,1)

Quote:EvenBob{snip for clarity}

[...]They think the house edge has an effect on every bet you make in roulette even in the short term. That pesky HE makes the game impossible to beat. This is absolutely true if you're talking long-term betting randomly you're screwed. Put in the extreme short-term, where you make the next bet and where we all live, there are no rules there is only short-term variance and you can get an edge over the game for one or two spins if you know what you're doing.

link to original post

This assertion is false. The House Edge manifests itself in every bet because every casino short-pays winning bets, never paying true odds. Flat-betting red? The odds of the ball landing on red is 18/38 for a double-zero wheel, thus, the correct payout should be $1.05556 per dollar bet. The casino only gives you one dollar, shorting you five and a half cents. That's the House Edge.

Yes, variance is real--it is at the heart of the Gambler's Fallacy that a wheel (or dice, or cards, or RNG) is 'hot' or 'cold'. But the house will always take a bite out what it should really should be paying you. In many games, screwing with the pay tables is the only way the house can assert an edge, as well as collecting it. The only bet that I know of where the house will ever pay true odds is the 'free odds' bet in craps.

If EB has a beef with this, or wants to reply, I invite him to head over to https://wizardofvegas.com/forum/gambling/other-games/38710-roulette-of-spins/3/#post908822, where I offer some more details.

This response is not for the Wiz, but for anyone who does not see how trivial it is to use the BINOMDIST() function in excel to answer probability questions even for very large numbers of binary trials.Quote:WizardQuote:MentalAt first, I thought it odd that Wiz said plugging 3 numbers into Excel is 'brute force'. However, the algorithm that Excel does behind the scenes involves a very large number of floating point operations. Using it is easy peasy.

link to original post

You're right. I should have done it the easy way. Here is how it works:

= BINOMDIST(Number of successes,Total trials, Probability of success, cumulative?)

The last term is 0 for exactly that many number of success and 1 or that many or less.

In the case of 100 bets:

51 or more wins = 49 or less losses = BINOMDIST(49,100,20/38,1)

Exactly 50 wins = BINOMDIST(50,100,18/38,0)

49 or less wins = BINOMDIST(49,100,18/38,1)

link to original post

You could get the same result as BINOMDIST(49,100,18/38,1) by calculating (sum over N=0..49 of BINOMDIST(N,100,18/38,0)). It is not that hard to calculate all of the terms and create the sum formula. The point is that the value 1 in the fourth argument tells excel to do the summation for you.

BINOMDIST(N,T,P,1) will give you the cumulative probability that you have between zero and N wins in T trials if the probability of every individual trial is the constant P. If you want to know how many trials will end with at least M wins but less than N wins, just subtract BINOMDIST(M-1,T,P,1) from BINOMDIST(N-1,T,P,1).

The answer to the question "What is the sum of 137 and 428?" has not been considered here (to my knowledge) or anywhere else that I know about. This does not mean it isn't easy to calculate the answer or that people with limited math skills cannot easily calculate it with a spreadsheet. The BINOMDIST() should be in every gambler's toolkit. It implements an accurate algorithm to answer an infinite number of questions similar to OP's question. I use BINOMDIST() frequently.

Quote:BillHasRetiredQuote:EvenBob{snip for clarity}

[...]They think the house edge has an effect on every bet you make in roulette even in the short term. That pesky HE makes the game impossible to beat. This is absolutely true if you're talking long-term betting randomly you're screwed. Put in the extreme short-term, where you make the next bet and where we all live, there are no rules there is only short-term variance and you can get an edge over the game for one or two spins if you know what you're doing.

link to original post

This assertion is false. The House Edge manifests itself in every bet because every casino short-pays winning bets, never paying true odds. Flat-betting red? The odds of the ball landing on red is 18/38 for a double-zero wheel, thus, the correct payout should be $1.05556 per dollar bet. The casino only gives you one dollar, shorting you five and a half cents. That's the House Edge.

Yes, variance is real--it is at the heart of the Gambler's Fallacy that a wheel (or dice, or cards, or RNG) is 'hot' or 'cold'. But the house will always take a bite out what it should really should be paying you. In many games, screwing with the pay tables is the only way the house can assert an edge, as well as collecting it. The only bet that I know of where the house will ever pay true odds is the 'free odds' bet in craps.

If EB has a beef with this, or wants to reply, I invite him to head over to https://wizardofvegas.com/forum/gambling/other-games/38710-roulette-of-spins/3/#post908822, where I offer some more details.

link to original post

I'm afraid I've been told I can only discuss my roulette method in the one thread I have going about roulette. Look what happened in this thread, my posts threw two totally innocent bystanders into complete delirium, I injured their delicate sensitivities to such an extent that they were hurling personal insults at me and they both got suspended. I am wracked with guilt over this and don't want to see that happen to you. Please address anything about my roulette method in the thread where I'm allowed to discuss it. The gamble thread.

Quote:WizardQuote:lilredrooster

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

.

link to original post

Here are the exact answers, to six decimal places.

Spins Net Win Push Net Loss 100 0.265023 0.069282 0.665695 200 0.207117 0.042698 0.750185 300 0.165841 0.030361 0.803798 400 0.134792 0.022893 0.842315 500 0.110664 0.017826 0.871510 600 0.091518 0.014167 0.894315 700 0.076106 0.011418 0.912476 800 0.063567 0.009298 0.927135 900 0.053283 0.007631 0.939086 1000 0.044796 0.006302 0.948902

This was with the binomial distribution by brute force in Excel. I am not aware of a simple formula to get these figures, although the Gaussian curve approximation would get you pretty close.

link to original post

I do not understand what these numbers mean. How do you get a push in roulette. Can someone break this down into layman's terms. I did not major in math I majored in American History and they did not cover this.

Quote:EvenBob

I'm afraid I've been told I can only discuss my roulette method in the one thread I have going about roulette. Look what happened in this thread, my posts threw two totally innocent bystanders into complete delirium, I injured their delicate sensitivities to such an extent that they were hurling personal insults at me and they both got suspended. I am wracked with guilt over this and don't want to see that happen to you. Please address anything about my roulette method in the thread where I'm allowed to discuss it. The gamble thread.

link to original post

I have answered this cut-and-pasted 'reply' over on the "Roulette - # of spins" thread: https://wizardofvegas.com/forum/gambling/other-games/38710-roulette-of-spins/3/#post908851

If you have an intelligent response to any of my posts, then respond from your pied-a-terre and post the link here. Life is too short for me to search for some obscure thread to where you've been banished.

Quote:BillHasRetiredQuote:EvenBob

I'm afraid I've been told I can only discuss my roulette method in the one thread I have going about roulette. Look what happened in this thread, my posts threw two totally innocent bystanders into complete delirium, I injured their delicate sensitivities to such an extent that they were hurling personal insults at me and they both got suspended. I am wracked with guilt over this and don't want to see that happen to you. Please address anything about my roulette method in the thread where I'm allowed to discuss it. The gamble thread.

link to original post

I have answered this cut-and-pasted 'reply' over on the "Roulette - # of spins" thread: https://wizardofvegas.com/forum/gambling/other-games/38710-roulette-of-spins/3/#post908851

If you have an intelligent response to any of my posts, then respond from your pied-a-terre and post the link here. Life is too short for me to search for some obscure thread to where you've been banished.

link to original post

Hardly obscure it's the top of the list and it's the only place I'm allowed to discuss this.

https://wizardofvegas.com/forum/off-topic/general/3831-how-much-gamble-do-you-have/25/#post908858

Quote:EvenBobQuote:WizardQuote:lilredrooster

if a player flat bets any even chance (one unit on one even chance) - what is the probability he will be down after a certain number of spins in double zero Roulette

my estimate is:

after 1,000 spins - 98% probability the player will be down

after 600 spins - 92% probability the player will be down

after 100 spins - 80% probability the player will be down

.

link to original post

Here are the exact answers, to six decimal places.

Spins Net Win Push Net Loss 100 0.265023 0.069282 0.665695 200 0.207117 0.042698 0.750185 300 0.165841 0.030361 0.803798 400 0.134792 0.022893 0.842315 500 0.110664 0.017826 0.871510 600 0.091518 0.014167 0.894315 700 0.076106 0.011418 0.912476 800 0.063567 0.009298 0.927135 900 0.053283 0.007631 0.939086 1000 0.044796 0.006302 0.948902

This was with the binomial distribution by brute force in Excel. I am not aware of a simple formula to get these figures, although the Gaussian curve approximation would get you pretty close.

link to original post

I do not understand what these numbers mean. How do you get a push in roulette. Can someone break this down into layman's terms. I did not major in math I majored in American History and they did not cover this.

link to original post

EvenBob,

By "Push" the Wiz means you have broken even.

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

Quote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

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So if you bet 10,000 times those figures remain the same?

Quote:EvenBobQuote:BillHasRetiredQuote:EvenBob{snip for clarity}

[...]They think the house edge has an effect on every bet you make in roulette even in the short term. That pesky HE makes the game impossible to beat. This is absolutely true if you're talking long-term betting randomly you're screwed. Put in the extreme short-term, where you make the next bet and where we all live, there are no rules there is only short-term variance and you can get an edge over the game for one or two spins if you know what you're doing.

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This assertion is false. The House Edge manifests itself in every bet because every casino short-pays winning bets, never paying true odds. Flat-betting red? The odds of the ball landing on red is 18/38 for a double-zero wheel, thus, the correct payout should be $1.05556 per dollar bet. The casino only gives you one dollar, shorting you five and a half cents. That's the House Edge.

Yes, variance is real--it is at the heart of the Gambler's Fallacy that a wheel (or dice, or cards, or RNG) is 'hot' or 'cold'. But the house will always take a bite out what it should really should be paying you. In many games, screwing with the pay tables is the only way the house can assert an edge, as well as collecting it. The only bet that I know of where the house will ever pay true odds is the 'free odds' bet in craps.

If EB has a beef with this, or wants to reply, I invite him to head over to https://wizardofvegas.com/forum/gambling/other-games/38710-roulette-of-spins/3/#post908822, where I offer some more details.

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I'm afraid I've been told I can only discuss my roulette method in the one thread I have going about roulette. Look what happened in this thread, my posts threw two totally innocent bystanders into complete delirium, I injured their delicate sensitivities to such an extent that they were hurling personal insults at me and they both got suspended. I am wracked with guilt over this and don't want to see that happen to you. Please address anything about my roulette method in the thread where I'm allowed to discuss it. The gamble thread.

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No, your posts are just a complete and total waste of time to read and seem to pervade every thread even remotely about Roulette.

Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

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So if you bet 10,000 times those figures remain the same?

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My god no! All it means that out of the 10,000 bets you won 5,000 and lost 5,000. The likelihood of that exact event is extremely low. MUCH lower than the likelihood of winning exactly 50 out of 100. Which is of course much lower than the chance of winning exactly 1 out of 2….

Quote:DieterQuote:TigerWuQuote:SOOPOO

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If we could ban EvenBob from just ONE roulette thread where he wasn't able to vomit his worthless masturbatory word salad of lies, we could actually have an interesting discussion about the game....

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That reaches personal insult.

3 days.

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How does EvenBob not take a ban for derailing the thread? The math of Roulette does not depend on anything non-random in the context of the OP's post.

Fortunately, it seems that you guys have taken the action of restricting EB's Roulette methods to one thread. I'll have to seek that thread out in order to block it so it doesn't waste space on my recent posts list.

If there are no objections, will there be any further math-related Roulette questions, or has the ship sailed?

Quote:WizardQuote:EvenBob

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Hijacking warning. Let's not have discussion of your secret method to beat roulette ruin another thread.

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I missed that post; sorry about that.

Quote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

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So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

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So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

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Let me double down on Wizard's admonition: EvenBob should restrict his discussion of roulette to the one (other) thread -and not engage in roulette discussions in this thread.

I am reversing the guidance I issued above. EvenBob is indeed allowed to ask questions about roulette and discuss roulette on this thread, as long as he doesn't engage in a discussion about his 'roulette method' on this thread. I am a little skeptical that he can refrain from espousing his ideas, but I am giving him enough rope and will see what he does with it.Quote:gordonm888Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

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So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

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Let me double down on Wizard's admonition: EvenBob should restrict his discussion of roulette to the one (other) thread -and not engage in roulette discussions in this thread.

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Quote:WizardThis was with the binomial distribution by brute force in Excel. I am not aware of a simple formula to get these figures, although the Gaussian curve approximation would get you pretty close.

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I don't think there is a "simple formula" as it is the sum of n results, where n is the number of spins.

Take the 100 spins case as an example.

Let p(n) be the probability that n of the 100 bets won; the probability that it is a net loss = p(0) + p(1) + p(2) + ... + p(49).

For double-zero roulette, p(n) = C(100, n) (9/19)^n (10/19)^(100-n)

p(0) = (10/19)^100

For n > 0, p(n) = (101 - n) / n x 9 / 10 x p(n-1) (where "x" is the multiplication symbol)

However, I don't think there is a way to simplify the sum.

The exact fraction for 100 spins is

2,629,551

,088,564,363,934,099,207,801,030,879,996

,616,689,794,636,837,506,594,944,850,079

,351,175,800,000,000,000,000,000,000,000

,000,000,000,000,000,000,000,000,000,000

/ 3,950,085

,484,118,525,497,024,052,097,529,573,238

,897,049,443,942,607,252,543,835,470,215

,867,415,609,371,393,307,890,303,310,063

,934,009,000,876,624,970,111,349,493,579

for the record Are you declaring what Bob has indicated he does is in fact a method? Please clarify that statement. Because before I thought you said that everything you knew about his play indicated a system.Quote:WizardQuote:EvenBob

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Hijacking warning. Let's not have discussion of your secret method to beat roulette ruin another thread.

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Bob isn't supposed to be talking about roulette on this forum period. He made a deal with the mods and the forum. Mike should make him stick to his word.Quote:gordonm888I am reversing the guidance I issued above. EvenBob is indeed allowed to ask questions about roulette and discuss roulette on this thread, as long as he doesn't engage in a discussion about his 'roulette method' on this thread. I am a little skeptical that he can refrain from espousing his ideas, but I am giving him enough rope and will see what he does with it.Quote:gordonm888Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

link to original post

So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

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Let me double down on Wizard's admonition: EvenBob should restrict his discussion of roulette to the one (other) thread -and not engage in roulette discussions in this thread.

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What green, Bob ignors green as per his admission.Quote:ChumpChangeJust because I'm betting Black and 19 out of 25 spins comes up Red and there's only 1 Green, only the 1 Green counts as HA.

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Quote:AxelWolfBob isn't supposed to be talking about roulette on this forum period. He made a deal with the mods and the forum. Mike should make him stick to his word.Quote:gordonm888I am reversing the guidance I issued above. EvenBob is indeed allowed to ask questions about roulette and discuss roulette on this thread, as long as he doesn't engage in a discussion about his 'roulette method' on this thread. I am a little skeptical that he can refrain from espousing his ideas, but I am giving him enough rope and will see what he does with it.Quote:gordonm888Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

link to original post

So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

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Let me double down on Wizard's admonition: EvenBob should restrict his discussion of roulette to the one (other) thread -and not engage in roulette discussions in this thread.

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Axel, please link to the exact post you keep mentioning. I remember it in concept but would like to see it exactly. Thanks.

I also frequently get a PM question that is better posted to the group, and tell the questioner to post it.Quote:lilredroosterby PM I posed a question to the Wizard and he indicated if I posted it …

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After all, why should I spend time answering a PRIVATE question that others may benefit from?

Go look at the thread where he created a poll about about his roulette play and if wre wanted him to stop posting about roulette and close the thread. I don't know off hand what the name of it was because he had a few shut down at his request.Quote:SOOPOOQuote:AxelWolfBob isn't supposed to be talking about roulette on this forum period. He made a deal with the mods and the forum. Mike should make him stick to his word.Quote:gordonm888Quote:gordonm888Quote:EvenBobQuote:DogHand

Thus, if you play 100 spins of roulette betting 1 unit each time on Red (or any other 1:1 payout bet), about 26.5% of the time you'll finish ahead, about 7.0% of the time you'll finish even, and the other 66.5% of the time you'll finish behind.

Hope this helps!

Dog Hand

link to original post

So does this answer the OP's original question? I'm not understanding this and nobody is discussing it. So 1/3 of the time you'll be even or ahead, and 2/3 of the time you'll finish behind after 100 bets? Am I getting that right? How does it change after 1,000 bets.

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Let me double down on Wizard's admonition: EvenBob should restrict his discussion of roulette to the one (other) thread -and not engage in roulette discussions in this thread.

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Axel, please link to the exact post you keep mentioning. I remember it in concept but would like to see it exactly. Thanks.

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He has been avoiding the question and dismissing it because he knows he F'D UP.