Crap Data
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.Quote: CasinoCrasherI 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.
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?
good luck looking! I would personally just roll the dice or look for actual dice roll files. Zumma comes to mind.Quote: CasinoCrasherI am looking for a formula or method to produce "theoretical" crap roll data.
Oh oh.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.
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
Quote: mustangsallyOh 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
this is about counting overlapping events that are not independent.Quote: SteenWhen 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.
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
Quote: mustangsallythis 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.
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.
Quote: CasinoCrasherI 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?
Quote: ThatDonGuyAt 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 . . .
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.
Quote: unJonThis 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.
Quote: ThatDonGuyConsecutive 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.
Quote: unJonThis 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.
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
Quote: CasinoCrasherI 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).
