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January 27th, 2018 at 7:35:17 PM
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I am interested in knowing how often a come out roll will occur in craps (once every x rolls). According to the Wizard of Odds, the probability of a DECISION on the pass line is once in every 3.38 rolls. That would include rolling a natural on the come out roll. I'm not necessarily looking for pass line decisions, but rather the frequency of come out rolls in craps.

January 27th, 2018 at 9:53:08 PM
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the probability of a roll being a comeout roll

should be 165/557

that would be 1 in 3.375376 rolls on average

Here

35,097 Zumma actual dice rolls

10,635 come out rolls

and my super 1 million actual dice roll collection

296,192 come out rolls

rolls/come out rolls (the program tracks this for us)

Zumma = 1 in 3.30 rolls

super 1 million = 1 in 3.3761884 rolls

hope this is clear

Zumma

super 1 million

Sally

should be 165/557

that would be 1 in 3.375376 rolls on average

Here

35,097 Zumma actual dice rolls

10,635 come out rolls

and my super 1 million actual dice roll collection

296,192 come out rolls

rolls/come out rolls (the program tracks this for us)

Zumma = 1 in 3.30 rolls

super 1 million = 1 in 3.3761884 rolls

hope this is clear

Zumma

super 1 million

Sally

I Heart Vi Hart

January 29th, 2018 at 1:30:16 PM
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Thanks so much Sally!

May 2nd, 2019 at 1:18:53 PM
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Hi Sally:

Can you explain the 165/557?? I am racking my brain to figure out the 3.38 rolls .

thankx very much!!

barry

Can you explain the 165/557?? I am racking my brain to figure out the 3.38 rolls .

thankx very much!!

barry

May 2nd, 2019 at 3:20:26 PM
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you are not alone. this article shows how this is calculated. Donald Catlin been gone now for some time but his articles and math lives on.Quote:BarryHi Sally:

Can you explain the 165/557?? I am racking my brain to figure out the 3.38 rolls .

http://catlin.casinocitytimes.com/article/how-long-is-a-craps-roll-1240

"First let us calculate the expected number of rolls for a Pass Line decision. To do this we will calculate the expected number of rolls for each of the possible things that will lead to a Pass Line decision and weight each one by the probability that it occurs."

"Since 1 = 165/165, the total is (165 + 392)/165 or 557/165.

This fraction is approximately 3.376.

In other words, if you played many, many Crap games (each decision is a game), kept track of the total number of rolls, and divided the number of rolls by the number of games, the result would be approximately 3.376.

According to Stewart Ethier from the University of Utah as quoted in Peter Griffin's book (Extra Stuff: Gambling Ramblings, Huntington Press, Las Vegas, 1991, p. 168), this result was first published in the American Mathematical Monthly in 1909."

took me sometime to actually have the data table sink in, then I was free.

good luck

added: the distribution (up to 30 rolls)

roll N | ends on Nth roll | ends by Nth roll |
---|---|---|

0 | 0 | |

1 | 0.333333333 | 0.333333333 |

2 | 0.188271605 | 0.521604938 |

3 | 0.134773663 | 0.656378601 |

4 | 0.096567311 | 0.752945911 |

5 | 0.0692571 | 0.822203011 |

6 | 0.049717715 | 0.871920727 |

7 | 0.035725128 | 0.907645855 |

8 | 0.025695361 | 0.933341216 |

9 | 0.018499325 | 0.95184054 |

10 | 0.013331487 | 0.965172027 |

11 | 0.009616645 | 0.974788673 |

12 | 0.006943702 | 0.981732374 |

13 | 0.005018575 | 0.98675095 |

14 | 0.003630703 | 0.990381653 |

15 | 0.002629179 | 0.993010832 |

16 | 0.001905753 | 0.994916585 |

17 | 0.001382697 | 0.996299282 |

18 | 0.001004149 | 0.997303432 |

19 | 0.000729922 | 0.998033354 |

20 | 0.000531076 | 0.99856443 |

21 | 0.000386754 | 0.998951184 |

22 | 0.000281906 | 0.99923309 |

23 | 0.000205665 | 0.999438755 |

24 | 0.000150175 | 0.99958893 |

25 | 0.000109751 | 0.99969868 |

Last edited by: 7craps on May 2, 2019

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