December 28th, 2017 at 2:21:27 PM
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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.

December 28th, 2017 at 2:36:09 PM
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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?

December 28th, 2017 at 2:38:36 PM
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Here is something very similar to what I use: https://www.online-calculator.com/scientific-calculator/

December 28th, 2017 at 3:10:01 PM
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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

I Heart Vi Hart

December 29th, 2017 at 3:08:18 PM
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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♪♪

January 5th, 2018 at 3:11:17 AM
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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

January 5th, 2018 at 7:32:16 AM
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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

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