November 3rd, 2013 at 6:30:44 AM
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First, thanks to everyone who maintains this website, and also thanks to those who participate and try to help.

I'm not new to the game, but I am newer at studying it. Thirty years ago I was taught the Pass w/ Odds method, and the Don't etc. I haven't really played much because I haven't lived near a casino until recently, and my preferred vacation is to the beach and my boat versus elsewhere.

Probability Table...this table is published as an answer to the question, "what is the probability of a shooter lasting x rolls in craps?"

Given that the average shooter lasts between 8-9 rolls, is this table saying that somewhere between rolls 1 and 9 the average shooter will roll a seven following an unresolved point? A possible scenario then might be: (1) rolls 1 and 2 are 5's, rolls 3 and 4 are 6's, rolls 5 and 6 are 8's, roll 7 is a 7, roll 8 is a 10, and roll 9 is a 7. In other words, the table is not saying that the average shooter will roll 8 times before he rolls his first seven, rather it's saying the average shooter will roll 8 times before disqualifying his turn. Is this a correct interpretation of the table?

I'm not new to the game, but I am newer at studying it. Thirty years ago I was taught the Pass w/ Odds method, and the Don't etc. I haven't really played much because I haven't lived near a casino until recently, and my preferred vacation is to the beach and my boat versus elsewhere.

Probability Table...this table is published as an answer to the question, "what is the probability of a shooter lasting x rolls in craps?"

Given that the average shooter lasts between 8-9 rolls, is this table saying that somewhere between rolls 1 and 9 the average shooter will roll a seven following an unresolved point? A possible scenario then might be: (1) rolls 1 and 2 are 5's, rolls 3 and 4 are 6's, rolls 5 and 6 are 8's, roll 7 is a 7, roll 8 is a 10, and roll 9 is a 7. In other words, the table is not saying that the average shooter will roll 8 times before he rolls his first seven, rather it's saying the average shooter will roll 8 times before disqualifying his turn. Is this a correct interpretation of the table?

...in the long run, you'll lose money.

November 3rd, 2013 at 7:01:01 AM
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Quote:nowakezoneIs this a correct interpretation of the table?

Yes. The expected rolls until any seven is 6.

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

November 3rd, 2013 at 7:22:48 AM
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Thanks Wizard.

I'm going to apologize in advance for starting my next thread because I know you've answered it here and on your other website before, but it still isn't making complete sense to me. The topic is betting the do and don't on the come out and taking the odds. If you'd prefer to zap my keyboard and ban me now I'll understand...lol!!! I just really like that bet so maybe hear me out, then zap me.

I'm going to apologize in advance for starting my next thread because I know you've answered it here and on your other website before, but it still isn't making complete sense to me. The topic is betting the do and don't on the come out and taking the odds. If you'd prefer to zap my keyboard and ban me now I'll understand...lol!!! I just really like that bet so maybe hear me out, then zap me.

...in the long run, you'll lose money.

November 3rd, 2013 at 8:00:27 AM
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I was shown a similar chart with very different figures.. I have to find it and post it

"I'm a DO'er and you my friend, are a Don'ter"
-Mark Walberg
pain and Gain

November 3rd, 2013 at 8:01:29 AM
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Quote:WizardYes. The expected rolls until any seven is 6.

Isn't it 3.8018 rolls?

(5/6)^x = .5

x * ln(5/6) = ln(.5)

x ~ 3.8018

November 3rd, 2013 at 9:22:10 AM
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You simply calculated the median (the 50/50 point)Quote:wudgedIsn't it 3.8018 rolls?

(5/6)^x = .5

x * ln(5/6) = ln(.5)

x ~ 3.8018

number of rolls to see a 7

rolls | Prob |
---|---|

1 | 16.6667% |

2 | 30.5556% |

3 | 42.1296% |

4 | 51.7747% |

5 | 59.8122% |

6 | 66.5102% |

7 | 72.0918% |

8 | 76.7432% |

9 | 80.6193% |

10 | 83.8494% |

11 | 86.5412% |

12 | 88.7843% |

13 | 90.6536% |

14 | 92.2113% |

15 | 93.5095% |

16 | 94.5912% |

17 | 95.4927% |

The length of a shooter's hand is a bit different.

Why the Wizard refuses to call this, the length of a shooter's hand,

by what other gaming math writers call it, is beyond me.

like calling a ROSE a 'thorny stemmed pretty petal flower'

or maybe a 'Crap Out' when the shooter rolls a 7 with a point established?

close

How about this book

The Doctrine of Chances

Probabilistic Aspects of Gambling

Stewart N. Ethier (yes, he can easily be my newest boyfriend)

15.2 The Shooter’s Hand

"A shooter rolls the dice until he sevens out, that is, until he misses a point.

When this happens, all bets are resolved and a new shooter comes out. The

sequence of rolls by a shooter, from the initial come out to the seven out,

is known as the shooter’s hand.

so the expected length of the shooter’s hand follows easily...

Table 15.2 The distribution of the length L of the shooter’s hand."

This is easily (Sally says so) calculated in a spreadsheet using either a Markov chain

or a simple recursive formula.

I thought the Wizard showed how he did this, not the Markov chain, but that seems to be very difficult to find.

added:

I found how the Wizard did this the non- Markov chain way.

that coming soon.

Here is my transition matrix in Excel and

raised to the 153 power (using a simple udf - user defined function)

by summing the 4 cells in the cor row.

(green in the photo)

0.000000000178882426152234

1 in 5,590,264,071.83

of being in one of the 4 states

1) the come out roll

2) point of 4 or 10

3) point of 5 or 9

4) point of 6 or 8

The 7 out value is actually close but has rounding errors in Excel.

I found how the Wizard did this the non- Markov chain way.

that coming soon.

Here is my transition matrix in Excel and

raised to the 153 power (using a simple udf - user defined function)

by summing the 4 cells in the cor row.

(green in the photo)

0.000000000178882426152234

1 in 5,590,264,071.83

of being in one of the 4 states

1) the come out roll

2) point of 4 or 10

3) point of 5 or 9

4) point of 6 or 8

The 7 out value is actually close but has rounding errors in Excel.

Remember

23 May 2009, at the Borgata Hotel Casino & Spa in Atlantic City, Patricia DeMauro

finally sevening out at the 154th roll?

The media loves this kind of story.

The Wizard was quoted in the news about it.

Had only a simulation completed for the answer that was close.

from another math guy

http://forumserver.twoplustwo.com/25/probability/154-rolls-craps-494244/

"It is easy to solve this problem exactly, and you just need a program that can multiply 4x4 matrices. Excel can do this. I found the exact probability of rolling at least 154 times before missing a point to be about 1 in 5.6 billion.

The wizardofodds website is reporting the number 1 in 5.3 billion (not 3.5 billion as Mizzles has posted), but this is based on a monte carlo simulation rather than an exact calculation, and he admits that he did not run enough samples for accuracy at 154 rolls, and he is extrapolating his error.

This isn't the first time that the wizard has relied on simulation for something that can be computed exactly with relative ease, and then reported more digits than could be justified by the size of the simulation. This problem cries out for a state space solution as shown below."

Sally says not really exactly (in bold)

I think this says it better

"Results are rounded to X significant digits"

The wizardofodds website is silent on this event today. (searching Google)

must have got lost in the shuffle.

I do not play Craps these days.

Too many hands lasting at most 6 rolls. (not the average but the median. here they are not the same)

Takes the fun out of playing (for me) if you want to keep rolling the dice.

Sally

I Heart Vi Hart

November 3rd, 2013 at 9:28:28 AM
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These were the numbers I remember seeing

"I'm a DO'er and you my friend, are a Don'ter"
-Mark Walberg
pain and Gain

November 3rd, 2013 at 9:47:07 AM
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Quote:mustangsallyYou simply calculated the median (the 50/50 point)

number of rolls to see a 7

How is this different than "expected rolls until any 7" ?

November 3rd, 2013 at 9:53:31 AM
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average (mean) verses the medianQuote:wudgedHow is this different than "expected rolls until any 7" ?

The mean is just the average of the numbers.

It is easy to calculate: add up all the numbers, then divide by how many numbers there are

The Median is the "middle number" (in a sorted list of numbers).

http://www.mathsisfun.com/mean.html

http://www.mathsisfun.com/median.html

took me some extra time to understand the differences

Sally

I Heart Vi Hart