CasinoCrasher
CasinoCrasher
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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
OnceDear
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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?
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
TomG
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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
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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.
ask hubby (hehe)

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
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odiousgambit
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
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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.
ask hubby (hehe)

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 :
EndIf
EndIf


Steen
mustangsally
mustangsally
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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
we already have a run=2
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
CasinoCrasher
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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
we already have a run=2
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
ChumpChange
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August 15th, 2018 at 7:17:28 AM permalink
Nobody rolls four 7's in a row except WinCraps...and me.
CasinoCrasher
CasinoCrasher
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August 16th, 2018 at 10:32:14 AM permalink
I would really like to see both, a RNG and a method that could produce results that comply exactly with long term probability.
Wizard
Administrator
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August 16th, 2018 at 12:35:27 PM permalink
The probability of rolling a total of t, n times in a row is [(6-abs(7-t))/36]^n.

For example, the probability of rolling 18 yo's in a row is [(6-abs(7-11))/36]^18 = 1 in 39,346,408,075,296,600,000,000.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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August 16th, 2018 at 5:50:42 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.


At what "level" do you want "exactly what pure probability would predict" - i.e. in how many rolls?

Let's say, as an example, 600,000. In rolls 1-600,000, there have to be exactly 100,000 7s - correct?
However, in order for it to be "exactly what pure probability would predict," rolls 2-600,001 would also have to be exactly 100,000 7s, as would rolls 3-600,002, rolls 4-600,003, and so on. This is possible only if roll N and roll (N + 600,000) are either both sevens or both not sevens. You will end up with 7s coming up in a cycle - why, yes, just like pretty much every math-based random number generator; how about that?
unJon
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August 16th, 2018 at 9:26:14 PM permalink
Quote: ThatDonGuy

At what "level" do you want "exactly what pure probability would predict" - i.e. in how many rolls?

Let's say, as an example, 600,000. In rolls 1-600,000, there have to be exactly 100,000 7s - correct?
However, in order for it to be "exactly what pure probability would predict," rolls 2-600,001 would also have to be exactly 100,000 7s, as would rolls 3-600,002, rolls 4-600,003, and so on. This is possible only if roll N and roll (N + 600,000) are either both sevens or both not sevens. You will end up with 7s coming up in a cycle - why, yes, just like pretty much every math-based random number generator; how about that?



This leads into the most trivial solution to the OPs question: just have the formula produce in an infinite cycle: 2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12 . . .
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Aaronians
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August 17th, 2018 at 3:21:19 AM permalink
Quote:

"This leads into the most trivial solution to the OPs question: just have the formula produce in an infinite cycle: 2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12 . . ."



I'm going to lean on this as well. Seems like a sensible solution.
ThatDonGuy
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August 17th, 2018 at 8:58:41 AM permalink
Quote: unJon

This leads into the most trivial solution to the OPs question: just have the formula produce in an infinite cycle: 2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12 . . .


Consecutive 7s appear too often in this cycle, don't they? One of the OP's "conditions" was, two consecutive 7s occur once every 36 rolls.
unJon
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August 17th, 2018 at 9:47:16 AM permalink
Quote: ThatDonGuy

Consecutive 7s appear too often in this cycle, don't they? One of the OP's "conditions" was, two consecutive 7s occur once every 36 rolls.


In that case you could super cycle each permutation of the above sequence and after the super cycle you would have a perfect sequence in accordance with the probabilities.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
CasinoCrasher
CasinoCrasher
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August 18th, 2018 at 3:53:39 PM permalink
Quote: unJon

This leads into the most trivial solution to the OPs question: just have the formula produce in an infinite cycle: 2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12 . . .



I agree that we are on the same page for sure. What I am looking for precisely is a formula that I can use to produce a purely theoretical crap data set or roll outcomes. What I mean is a random number generator that let’s say computer craps programs use in online gambling will produce “theoretical data” over an essentially infinite amount of rolls. I am looking for a formula that will produce dice roll outcome that would be purely theoretical over a finite number of rolls, let’s say a few thousand. There are 36 possible combinations with two common die, I would like it to produce every combination in 36 rolls and so on and so forth. Then also see all your theoretical doublets, triplets, etc at the exact number of rolls they would theoretically expected to be seen. See say your 2 consecutive 7s (7,7) in 36 rolls (6/36*6/36) and 3 consecutive 7s (7,7,7) every 216 rolls, and the same with all the other numbers.
Dalex64
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August 18th, 2018 at 4:51:57 PM permalink
You can do that with a computer program, but it will take a while.

Here's some python3

import itertools

for permutation in itertools.permutations( range(1, 37) ):
print(permutation)


That will print, in sets of 36, every permutation of a list containing the numbers of 1 to 36
ThatDonGuy
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August 18th, 2018 at 6:05:16 PM permalink
Quote: CasinoCrasher

I agree that we are on the same page for sure. What I am looking for precisely is a formula that I can use to produce a purely theoretical crap data set or roll outcomes. What I mean is a random number generator that let’s say computer craps programs use in online gambling will produce “theoretical data” over an essentially infinite amount of rolls. I am looking for a formula that will produce dice roll outcome that would be purely theoretical over a finite number of rolls, let’s say a few thousand. There are 36 possible combinations with two common die, I would like it to produce every combination in 36 rolls and so on and so forth. Then also see all your theoretical doublets, triplets, etc at the exact number of rolls they would theoretically expected to be seen. See say your 2 consecutive 7s (7,7) in 36 rolls (6/36*6/36) and 3 consecutive 7s (7,7,7) every 216 rolls, and the same with all the other numbers.


When you say "consecutive 7s every 36 rolls," do you mean, "consecutive 7s occurring 1/36 of the time"? You can't have consecutive 7s in every set of 36 consecutive rolls and one 7 in every set of 6 rolls simultaneously.

I'll look into this, but my gut feeling is, you can't generate a set of rolls that will meet every condition, unless you want to set a cap on how many times in a row a number "has to" show up. Also, if every specified sequence "has to" show up a certain number of times in every set of N consecutive rolls, then the pattern will cycle so that roll N + 1 has to be the same as roll 1 (otherwise at least one of the values will fail for the block of N consecutive rolls starting with roll 2).
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