Dferrario
Dferrario
Joined: May 15, 2018
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May 15th, 2018 at 8:20:34 AM permalink
What is the probability of rolling 2 dice and rolling every combination of dice before a 7 comes up? Not looking for the 36 combinations, looking for the probability without duplicates. (Example: Counting 1,2 and 2,1 as only one combination) I believe looking at it this way there are 21 combinations.

1,1. 1,2. 1,3. 1,4. 1,5. 1,6
2,2. 2,3. 2,4. 2,5. 2,6
3,3 3,4. 3,5. 3,6
4,4. 4,5. 4,6
5,5. 5,6
6,6

So the probability of rolling all these combinations before a 7 comes up? Thanks!
Romes
Romes
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May 15th, 2018 at 8:36:07 AM permalink
So craps side bet all small/tall/all?

https://wizardofodds.com/games/craps/appendix/5/
Playing it correctly means you've already won.
mustangsally
mustangsally
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May 15th, 2018 at 8:50:48 AM permalink
Quote: Dferrario

So the probability of rolling all these combinations before a 7 comes up? Thanks!

so all 21 combos B4 any 7

instead of the 10 sums that the All Bonus Craps bet is. that is 1 in 190

so I say very close to 0 (how close do you want?)
without calculating it.
too many states for a simple transition matrix solution and too small values for inclusion-exclusion

there might be one that could do this while having fun with it

edit
Sally
Last edited by: mustangsally on May 15, 2018
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mustangsally
mustangsally
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Thanks for this post from:
Dferrario
May 15th, 2018 at 9:02:21 AM permalink
Quote: Dferrario

What is the probability of rolling 2 dice and rolling every combination of dice before a 7 comes up?

1,1. 1,2. 1,3. 1,4. 1,5. 1,6
2,2. 2,3. 2,4. 2,5. 2,6
3,3 3,4. 3,5. 3,6
4,4. 4,5. 4,6
5,5. 5,6
6,6

1,6
2,5
3,4
are the combos for a 7
you do not want that right?
so 18 combos in all?
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mustangsally
mustangsally
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May 15th, 2018 at 9:14:02 AM permalink
Quote: Dferrario

So the probability of rolling all these combinations before a 7 comes up? Thanks!

this was easier than I thought.
> p
[1] 0.0001345821
> a=1/p
> a # 1 in
[1] 7430.409
> a-1 # odds against
[1] 7429.409


compare All Bonus bet too below

I used R
 ##################################################################
> # Probability of rolling a subset of numbers before a single number
> ##################################################################
> start.time <- Sys.time()
> options(scipen=999)
>
> numbers = c(2,3,4,5,6,8,9,10,11,12,7) # Last must occur only after all others in any order
>
> in_36 = ifelse(numbers <= 7, numbers-1, 13-numbers) # Ways to make each number
> i = length(in_36)
> p = 0
> for (j in 1:(i-1)) { # Last number before j numbers
+ terms = combn(in_36[1:(i-1)],j) # Matrix w/combos of j numbers in C(i-1,j) columns
+ for (k in 1:ncol(terms)) { # Sum each column, compute and add probabilities
+ p = p + (-1)^(j+1) * in_36/(in_36 + sum(terms[1:j,k]))
+ }
+ }
> end.time <- Sys.time()
> time <- end.time - start.time
> time
Time difference of 0.1562679 secs
> p = 1-p
> p
[1] 0.005257704
> a=1/p
> a # 1 in
[1] 190.1971
> a-1 # odds against
[1] 189.1971
> ##################################################################
> # Probability of rolling a subset of numbers before a single number
> ##################################################################
> start.time <- Sys.time()
> options(scipen=999)
>
> numbers = c(2,3,4,5,6,4,5,6,8,6,8,9,8,9,10,10,11,12,7) # Last must occur only after all others in any order
>
> in_36 = c(1,2,2,2,2,1,2,2,2,1,2,2,1,2,2,1,2,1,6) # Ways to make each number
> i = length(in_36)
> p = 0
> for (j in 1:(i-1)) { # Last number before j numbers
+ terms = combn(in_36[1:(i-1)],j) # Matrix w/combos of j numbers in C(i-1,j) columns
+ for (k in 1:ncol(terms)) { # Sum each column, compute and add probabilities
+ p = p + (-1)^(j+1) * in_36/(in_36 + sum(terms[1:j,k]))
+ }
+ }
> end.time <- Sys.time()
> time <- end.time - start.time
> time
Time difference of 1.200368 secs
> p=1-p
> p
[1] 0.0001345821
> a=1/p
> a # 1 in
[1] 7430.409
> a-1 # odds against
[1] 7429.409


Sally
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Dferrario
Dferrario
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May 15th, 2018 at 9:26:57 AM permalink
Great point, yes, without the 7 combinations! Thanks for catching this!
mustangsally
mustangsally
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May 15th, 2018 at 9:42:15 AM permalink
Quote: Dferrario

Great point, yes, without the 7 combinations! Thanks for catching this!

it is what you said at 1st.
the subset of combos can roll in any order (the order does not matter)
in the code I used

you can also use the code for smaller subsets of combos too

wondering why the question?
Sally
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Dferrario
Dferrario
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May 15th, 2018 at 3:12:55 PM permalink
Thanks Mustangsally. My dealer friend and I had a side bet. So to confirm it is roughly a 1-7,500 chance of rolling all 18 combinations before a seven comes up? Itís amazing you are able to calculate this, I should have paid more attention in school!😄
mustangsally
mustangsally
Joined: Mar 29, 2011
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Thanks for this post from:
Dferrario
May 15th, 2018 at 3:48:18 PM permalink
Quote: Dferrario

Thanks Mustangsally. My dealer friend and I had a side bet. So to confirm it is roughly a 1-7,500 chance of rolling all 18 combinations before a seven comes up?

Hope you won your side bet!
yes, closer to 1 in 7431
Quote: Dferrario

Itís amazing you are able to calculate this, I should have paid more attention in school!😄

most everything can be calculated.
credit is to BruceZ who did the original code.
I changed it to suit my interests.

The math behind it is more involved as it uses a concept in set counting named inclusion- exclusion.
the code looks to see the probability of NOT getting all the combos and we subtract that from 1 to get the answer.

a dice example. the chance to NOT get at least one 1 on 2 dice =
5 ways * 5 ways = 25 ways out of 6*6 = 36 total ways = 25/36.

so 1 - 25/36 = 11/36 = the probability of getting at least one 1 when rolling 2d6

or 1/6 + 1/6 = 2/6 or 12/36
so we overcounted when the dice rolls 1,1
that should only count as one 1.
so 12/36 - 1/36 = 11/36
(one can see this in a list of all the possible sums)
Permutations with repetition (n=6, r=2)
Using Items: 1,2,3,4,5,6

List has 36 entries.
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
3,1
3,2
3,3
3,4
3,5
3,6
4,1
4,2
4,3
4,4
4,5
4,6
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6

many ways to use the math to our advantage of finding the right answer.
Sally
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Dferrario
Dferrario
Joined: May 15, 2018
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May 15th, 2018 at 6:31:05 PM permalink
One more math question for you:
Do you know the odds of not rolling a 7, for 10, 20, 30, 40 and 50 rolls?

Looking for 1 in 6, 1 in 50 etc.

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