October 20th, 2010 at 9:45:10 AM
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Does anyone know the average number of points hit by a shooter before they 7 out?

We all know the shooter can't keep hitting points forever. Sooner or later he eventually 7's out.

But, what do you guys think the average number of points a shooter is able to hit before rolling the seven and passing the dice?

We all know the shooter can't keep hitting points forever. Sooner or later he eventually 7's out.

But, what do you guys think the average number of points a shooter is able to hit before rolling the seven and passing the dice?

October 20th, 2010 at 10:20:44 AM
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Quote:Wizard

I believe that's rolls per shooter. Since rolls per passline decision is 3.376, that makes the average decisions per shooter about 2.5. Of those decisions, 27.1% are point winners, so the average number of points made by the shooter is about 0.68.

In other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?

Yes, you really do only have a bit better than 1 in 4 chance of establishing and making a point. A big chunk of winning passline bets happens from natural 7/11 rolls: 22.2% to be precise. When you add them in you get 49.3%, the overall chance of winning a line bet.

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563

October 20th, 2010 at 10:26:50 AM
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I was gonna say the same thing.

The Wiz linked to a page where he shows the math to come up with the 8.53 number.

Scroll down about 1/3 the page - just below the average rolls per hour charts.

It's 8.53 THROWS per shooter.

The Wiz linked to a page where he shows the math to come up with the 8.53 number.

Scroll down about 1/3 the page - just below the average rolls per hour charts.

It's 8.53 THROWS per shooter.

I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

October 20th, 2010 at 10:26:51 AM
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Quote:JimmyMacDoes anyone know the average number of points hit by a shooter before they 7 out?

A better question would be the math for:

The probability of a shooter making 0 points,

1 point,2 points, 3 points etc.

October 20th, 2010 at 10:36:58 AM
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You're right, I answered the wrong question.

Let x be the answer. The probability of eventually making the next point is (6/24)*(1/3)+(8/24)*(2/3)+(10/24)*(5/11) = 201/495 = 0.460606.

The expected number of points made is 0.460606/(1-0.460606) = 201/294 = 0.683673469.

Let x be the answer. The probability of eventually making the next point is (6/24)*(1/3)+(8/24)*(2/3)+(10/24)*(5/11) = 201/495 = 0.460606.

The expected number of points made is 0.460606/(1-0.460606) = 201/294 = 0.683673469.

It's not whether you win or lose; it's whether or not you had a good bet.

October 20th, 2010 at 10:58:18 AM
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Quote:guido111A better question would be the math for:

The probability of a shooter making 0 points,

1 point,2 points, 3 points etc.

I had run a sim of 8,522,945 million dice rolls. Here are those results:

1,000,000/shooters

682796/total points wins

count | points wins | % | or more | or less |
---|---|---|---|---|

594110 | 0 | 0.5941100000 | ||

241267 | 1 | 0.2412670000 | 0.4058900000 | 0.8353770000 |

97983 | 2 | 0.0979830000 | 0.1646230000 | 0.9333600000 |

39547 | 3 | 0.0395470000 | 0.0666400000 | 0.9729070000 |

16136 | 4 | 0.0161360000 | 0.0270930000 | 0.9890430000 |

6535 | 5 | 0.0065350000 | 0.0109570000 | 0.9955780000 |

2596 | 6 | 0.0025960000 | 0.0044220000 | 0.9981740000 |

1067 | 7 | 0.0010670000 | 0.0018260000 | 0.9992410000 |

425 | 8 | 0.0004250000 | 0.0007590000 | 0.9996660000 |

188 | 9 | 0.0001880000 | 0.0003340000 | 0.9998540000 |

91 | 10 | 0.0000910000 | 0.0001460000 | 0.9999450000 |

55 | 11+ | 0.0000550000 | 0.0000550000 | 1.0000000000 |

The math would be a challenge

7 winner chicken dinner!

October 20th, 2010 at 11:08:29 AM
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Quote:guido111A better question would be the math for:

The probability of a shooter making 0 points,

1 point,2 points, 3 points etc.

from the WoO site:

59.39% chance of not hitting a point

40.61% chance of hitting a point (or more)

edit:http://wizardofodds.com/askthewizard/craps-probability.html

October 20th, 2010 at 11:39:49 AM
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Quote:MathExtremistso the average number of points made by the shooter is about 0.68.

In other words, if you make even one point, you're having a better-than-average roll. Not very intuitive, is it?

Just goes to show that an "average" can be a useless number.

Knowing the standard deviation would give a better understanding to the "average", and that still may show a number that is still useless.

October 20th, 2010 at 12:09:28 PM
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Quote:WizardCan you give me the URL where I say that?

you do not exactly say that.

http://wizardofodds.com/askthewizard/craps-probability.html

" So the probability of making a point, given that a point was established is, (5/12)*(5/11)+(4/12)*(4/10)+(3/12)*(3/9) = 40.61%"

So,

59.39% chance of not hitting a point

40.61% chance of hitting a point (or more)