May 15th, 2018 at 8:20:34 AM
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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!

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!

May 15th, 2018 at 8:36:07 AM
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So craps side bet all small/tall/all?

https://wizardofodds.com/games/craps/appendix/5/

https://wizardofodds.com/games/craps/appendix/5/

Playing it correctly means you've already won.

May 15th, 2018 at 8:50:48 AM
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so all 21 combos B4 any 7Quote:DferrarioSo the probability of rolling all these combinations before a 7 comes up? Thanks!

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

I Heart Vi Hart

May 15th, 2018 at 9:02:21 AM
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1,6Quote:DferrarioWhat 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

2,5

3,4

are the combos for a 7

you do not want that right?

so 18 combos in all?

I Heart Vi Hart

May 15th, 2018 at 9:14:02 AM
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this was easier than I thought.Quote:DferrarioSo the probability of rolling all these combinations before a 7 comes up? Thanks!

> 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

I Heart Vi Hart

May 15th, 2018 at 9:26:57 AM
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Great point, yes, without the 7 combinations! Thanks for catching this!

May 15th, 2018 at 9:42:15 AM
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it is what you said at 1st.Quote:DferrarioGreat point, yes, without the 7 combinations! Thanks for catching this!

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

I Heart Vi Hart

May 15th, 2018 at 3:12:55 PM
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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? It’s amazing you are able to calculate this, I should have paid more attention in school!😄

May 15th, 2018 at 3:48:18 PM
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Hope you won your side bet!Quote:DferrarioThanks 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?

yes, closer to 1 in 7431

most everything can be calculated.Quote:DferrarioIt’s amazing you are able to calculate this, I should have paid more attention in school!😄

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

I Heart Vi Hart