CasinoCrasher
Joined: Oct 4, 2015
• Posts: 36
December 28th, 2017 at 2:21:27 PM permalink
I am looking for a formula or method to produce "theoretical" crap roll data. In other words I want a formula or method that will create a set of theoretical outcomes from 2 die, not a random number generator. A formula that will give you exactly what pure probability would predict, 6/36 chance of rolling a 7, 2 consecutive 7's every 36 rolls, ect.
OnceDear
Joined: Jun 1, 2014
• Posts: 3211
December 28th, 2017 at 2:36:09 PM permalink
Quote: CasinoCrasher

I am looking for a formula or method to produce "theoretical" crap roll data. In other words I want a formula or method that will create a set of theoretical outcomes from 2 die, not a random number generator. A formula that will give you exactly what pure probability would predict, 6/36 chance of rolling a 7, 2 consecutive 7's every 36 rolls, ect.

You can have a method to generate rolls that comply exactly with the long term probability: I.e they simply present you the exact 36 possible outcomes once per set of 36 rolls, or you can get an rng to show you realistic outcomes.
It is 'not at all realistic to see all 36 equally probable outcomes in one set of 36 rolls.
So what do you really want?
If you are enjoying the game, you're already winning.
TomG
Joined: Sep 26, 2010
• Posts: 1378
December 28th, 2017 at 2:38:36 PM permalink
Here is something very similar to what I use: https://www.online-calculator.com/scientific-calculator/
mustangsally
Joined: Mar 29, 2011
• Posts: 2392
December 28th, 2017 at 3:10:01 PM permalink
Quote: CasinoCrasher

I am looking for a formula or method to produce "theoretical" crap roll data.

good luck looking! I would personally just roll the dice or look for actual dice roll files. Zumma comes to mind.
Quote: CasinoCrasher

<snip> A formula that will give you exactly what pure probability would predict, 6/36 chance of rolling a 7, 2 consecutive 7's every 36 rolls, ect.

Oh oh.
two 7s in a row requires (on average) 42 rolls.

where did U get 36 rolls?
I pay lots of good \$US cash for truth.

wait time formula
1/p + 1/p^2
= 6 + 36
p=1/6

and
believe or knots
actual dice rolls show 42 to be true (as an average of course)
back to basketball and Lay bet winning

Sally
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AxelWolf
Joined: Oct 10, 2012
• Posts: 14570
Thanks for this post from:
December 29th, 2017 at 3:08:18 PM permalink
I love the title of the thread.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
Steen
Joined: Apr 7, 2014
• Posts: 110
January 5th, 2018 at 3:11:17 AM permalink
Quote: mustangsally

Oh oh.
two 7s in a row requires (on average) 42 rolls.

where did U get 36 rolls?
I pay lots of good \$US cash for truth.

wait time formula
1/p + 1/p^2
= 6 + 36
p=1/6

Yes, you're correct but so was the OP. It depends on how you define back-to-back 7's.

For example, consider four 7's in a row: 7777

How many back-to-back 7's would you count here? Three? Two?

When the second 7 rolls, we have the first instance of back-to-back 7's. What about when the third 7 rolls? If you count this as another instance of back-to-back 7's then 7777 contains three back-to-back 7's and the correct average number of rolls per occurrence is 36.

`If initializing script Then   name cs1.lastroll as "Last roll" :   name cs3.#events as "Number of events" :   name cs5.avg.rolls as "Avg number of rolls per event"Else   If dice total = 7 And cs1.lastroll = 7 Then      add 1 to cs3.#events :      cs5.avg.rolls = # of rolls / cs3.#events   EndIf   cs1.lastroll = dice total :EndIf`

However, if you count the third 7 as the beginning of another instance of back-to-back 7's then 7777 contains two back-to-back 7's and the correct average number of rolls per occurrence is 42.

`If initializing script Then   name cs1.lastroll as "Last roll" :   name cs3.#events as "Number of events" :   name cs5.avg.rolls as "Avg number of rolls per event"Else   If dice total = 7 And cs1.lastroll = 7 Then      add 1 to cs3.#events :      cs5.avg.rolls = # of rolls / cs3.#events :      cs1.lastroll = 0   Else      cs1.lastroll = dice total :   EndIfEndIf`

Steen
mustangsally
Joined: Mar 29, 2011
• Posts: 2392
January 5th, 2018 at 7:32:16 AM permalink
Quote: Steen

When the second 7 rolls, we have the first instance of back-to-back 7's. What about when the third 7 rolls? If you count this as another instance of back-to-back 7's then 7777 contains three back-to-back 7's and the correct average number of rolls per occurrence is 36.

this is about counting overlapping events that are not independent.
nothing basic here.
example dice roll sequence (2d6)
7,7
the 2nd 7 starts a new run of 7s length = 1
and a run=1

this confuses already confused gamblers (and some math folks too) imo
and brings up more questions than it answers(imo)

nice point to point 2
of course
no point was established with any 7 rolled
7
77
777
7777
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CasinoCrasher
Joined: Oct 4, 2015
• Posts: 36
August 14th, 2018 at 5:01:49 PM permalink
Quote: mustangsally

this is about counting overlapping events that are not independent.
nothing basic here.
example dice roll sequence (2d6)
7,7
the 2nd 7 starts a new run of 7s length = 1
and a run=1

this confuses already confused gamblers (and some math folks too) imo
and brings up more questions than it answers(imo)

nice point to point 2
of course
no point was established with any 7 rolled
7
77
777
7777

Sorry, what I meant by 2 consecutive was rolling two 7s back to back without any rolls in between, with two fair die. The chance of rolling a 7 at any point in time is 1/6 of course, so 2 7s back to back I was just saying is 1/6*1/6=1/36.
ChumpChange
Joined: Jun 15, 2018