Quote:Zcore13Yeah, because nobody would notice you winning $270 multiple times a day all year long.

ZCore13

Where do you work and what is your position title?

Quote:WatchMeWinWhere do you work and what is your position title?

Lol. Did you just sat that in your best Arnold Schwarzenegger voice?

What is you name and who is your daddy?

ZCore13

Quote:Zcore13Lol. Did you just sat that in your best Arnold Schwarzenegger voice?

What is you name and who is your daddy?

ZCore13

Well... Care to answer? Now get to the Chopper!

Quote:WatchMeWinWell... Care to answer? Now get to the Chopper!

I've been a Dealer, Pit Boss, Shift Manager, Table Games Director and been a panelist at two National Conventions on Table Games. Ive also consulted on a handful of new table games and for a dealer school.

ZCore13

Quote:Zcore13I've been a Dealer, Pit Boss, Shift Manager, Table Games Director and been a panelist at two National Conventions on Table Games. Ive also consulted on a handful of new table games and for a dealer school.

ZCore13

Thanks for the info. Credible n much respect. What city are you located in? I may need some of your help down the road.

As I understand it, the question is what is the probability of success of turning $1,000 into $1,265, with the alternative being losing the full $1,000. If we can ignore the thin house edge by making only don't bets and laying full odds, the probability of success per session is 1000/1265 = 79.05%.

I'm told there is a question of the probability of 8 or more successes if this experiment is repeated 10 times. Again, ignoring the house edge, the following table answers that question.

Wins | Probability | Cummulative |
---|---|---|

10 | 0.095300 | 0.095300 |

9 | 0.252546 | 0.347846 |

8 | 0.301161 | 0.649008 |

7 | 0.212821 | 0.861828 |

6 | 0.098696 | 0.960524 |

5 | 0.031385 | 0.991909 |

4 | 0.006931 | 0.998840 |

3 | 0.001050 | 0.999889 |

2 | 0.000104 | 0.999994 |

1 | 0.000006 | 1.000000 |

0 | 0.000000 | 1.000000 |

Total | 1.000000 |

The cumulative column shows the probability of 8 to 10 successes is 64.90%. Of course, it will be a little less due to the house edge. I'd have to run simulation to determine that answer, which is going beyond the call of duty of what I'll do for free. However, if forced, with a very careful strategy to minimize total action, I'd put it at about 60%.

I hope this information is helpful.

I want somebody to show me a news story where that has happened.

Just one. Please

Quote:WizardI've been asked to join this thread. Forgive me if I don't read every post. Betting system threads tend to not be of much interest to me.

As I understand it, the question is what is the probability of success of turning $1,000 into $1,265, with the alternative being losing the full $1,000. If we can ignore the thin house edge by making only don't bets and laying full odds, the probability of success per session is 1000/1265 = 79.05%.

I'm told there is a question of the probability of 8 or more successes if this experiment is repeated 10 times. Again, ignoring the house edge, the following table answers that question.

Wins Probability Cummulative 10 0.095300 0.095300 9 0.252546 0.347846 8 0.301161 0.649008 7 0.212821 0.861828 6 0.098696 0.960524 5 0.031385 0.991909 4 0.006931 0.998840 3 0.001050 0.999889 2 0.000104 0.999994 1 0.000006 1.000000 0 0.000000 1.000000 Total 1.000000

The cumulative column shows the probability of 8 to 10 successes is 64.90%. Of course, it will be a little less due to the house edge. I'd have to run simulation to determine that answer, which is going beyond the call of duty of what I'll do for free. However, if forced, with a very careful strategy to minimize total action, I'd put it at about 60%.

I hope this information is helpful.

Thank you for your input, Wizard. So, what odds would you give to someone trying to achieve this task?

Quote:WizardI've been asked to join this thread. Forgive me if I don't read every post. Betting system threads tend to not be of much interest to me.

As I understand it, the question is what is the probability of success of turning $1,000 into $1,265, with the alternative being losing the full $1,000. If we can ignore the thin house edge by making only don't bets and laying full odds, the probability of success per session is 1000/1265 = 79.05%.

I'm told there is a question of the probability of 8 or more successes if this experiment is repeated 10 times. Again, ignoring the house edge, the following table answers that question.

Wins Probability Cummulative 10 0.095300 0.095300 9 0.252546 0.347846 8 0.301161 0.649008 7 0.212821 0.861828 6 0.098696 0.960524 5 0.031385 0.991909 4 0.006931 0.998840 3 0.001050 0.999889 2 0.000104 0.999994 1 0.000006 1.000000 0 0.000000 1.000000 Total 1.000000

The cumulative column shows the probability of 8 to 10 successes is 64.90%. Of course, it will be a little less due to the house edge. I'd have to run simulation to determine that answer, which is going beyond the call of duty of what I'll do for free. However, if forced, with a very careful strategy to minimize total action, I'd put it at about 60%.

I hope this information is helpful.

Thanks Wizard, I already showed the same. We concur pretty much exactly.

https://wizardofvegas.com/forum/gambling/craps/29954-winning-is-my-drug/7/#post620340

~64% probability of hitting the '8 or more' goal.

WMW asking for odds on the wager is absurd.

If, however he's willing to offer 5 to 1 that someone else could not do it, he's a bigger [insert appropriate derogatory description] than I ever imagined.