Thread Rating:

markdcasino
markdcasino
Joined: Jan 27, 2018
  • Threads: 3
  • Posts: 7
January 27th, 2018 at 7:35:17 PM permalink
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.
mustangsally
mustangsally
Joined: Mar 29, 2011
  • Threads: 25
  • Posts: 2463
January 27th, 2018 at 9:53:08 PM permalink
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
I Heart Vi Hart
markdcasino
markdcasino
Joined: Jan 27, 2018
  • Threads: 3
  • Posts: 7
January 29th, 2018 at 1:30:16 PM permalink
Thanks so much Sally!
Barry
Barry
Joined: Oct 11, 2018
  • Threads: 1
  • Posts: 5
May 2nd, 2019 at 1:18:53 PM permalink
Hi Sally:

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

thankx very much!!

barry
7craps
7craps
Joined: Jan 23, 2010
  • Threads: 18
  • Posts: 1977
May 2nd, 2019 at 3:20:26 PM permalink
Quote: Barry

Hi Sally:

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

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.

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 Nends on Nth rollends by Nth roll
00
10.3333333330.333333333
20.1882716050.521604938
30.1347736630.656378601
40.0965673110.752945911
50.06925710.822203011
60.0497177150.871920727
70.0357251280.907645855
80.0256953610.933341216
90.0184993250.95184054
100.0133314870.965172027
110.0096166450.974788673
120.0069437020.981732374
130.0050185750.98675095
140.0036307030.990381653
150.0026291790.993010832
160.0019057530.994916585
170.0013826970.996299282
180.0010041490.997303432
190.0007299220.998033354
200.0005310760.99856443
210.0003867540.998951184
220.0002819060.99923309
230.0002056650.999438755
240.0001501750.99958893
250.0001097510.99969868
Last edited by: 7craps on May 2, 2019
winsome johnny (not Win some johnny)

  • Jump to: