Make em all craps bet
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November 4th, 2017 at 4:59:07 PM
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Hello folks.
Here to ask advice about the maths behind the make 'em all craps side bet.
For those who don't know, the bet is a wager that a shooter will roll every number 2-6 and 8-12 before rolling a 7.
I have seen the wizard's answers to the house edge on his site and it lines up with the brute force calculations I have seen simulated previously. However I was wondering if there was another way to do it? If anyone could point me in the right direction of where to look to start finding answers I would really appreciate it.
Similarly, if this has already been brought up somewhere else please point me towards it and I will be happy to read that!
Thanks in advance for any hints or pointers anyone can offer!
Here to ask advice about the maths behind the make 'em all craps side bet.
For those who don't know, the bet is a wager that a shooter will roll every number 2-6 and 8-12 before rolling a 7.
I have seen the wizard's answers to the house edge on his site and it lines up with the brute force calculations I have seen simulated previously. However I was wondering if there was another way to do it? If anyone could point me in the right direction of where to look to start finding answers I would really appreciate it.
Similarly, if this has already been brought up somewhere else please point me towards it and I will be happy to read that!
Thanks in advance for any hints or pointers anyone can offer!
November 5th, 2017 at 12:22:25 AM
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Up to 21% edge these lower payout days, correct?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
November 5th, 2017 at 1:52:29 AM
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Here is my All Tall Small spreadsheet. Feel free to ask any questions.
“Man Babes” #AxelFabulous
November 5th, 2017 at 4:59:44 AM
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Thanks Miplet! That really helpful! I started doing something similar and then thought I might be trying too hard so stopped haha.
But to clarify, you have 2^10 states as a result of having either made or not made each number correct? I'm not enough of an excel wiz to understand everything that's going on but most of it makes intuitive sense. I am curious as to how using the mod(floor ... gives you the 1024 different possible orders like it does? Is there any chance you could dumb that down for me a tiny bit? haha
Also, the reason I stopped trying to make a sheet like that, is I thought maybe I could find a formula to solve the whole thing?
For example, P(9)= 4/36 P(7or9)1=10/36 -> P(7|7or9) = 1/9 / 5/18 = 1/9 * 18/5 = 2/5 which is obv why the odds bet pays 3/2.
Anyway, if i can do that, can I not do a formula for odds of a 5 and a 9 before a seven? I don't honestly even know if its possible just seems like to me that it should be! If you can give me any advice as well regarding that I would be very grateful!
Cheers for your time, man.
But to clarify, you have 2^10 states as a result of having either made or not made each number correct? I'm not enough of an excel wiz to understand everything that's going on but most of it makes intuitive sense. I am curious as to how using the mod(floor ... gives you the 1024 different possible orders like it does? Is there any chance you could dumb that down for me a tiny bit? haha
Also, the reason I stopped trying to make a sheet like that, is I thought maybe I could find a formula to solve the whole thing?
For example, P(9)= 4/36 P(7or9)1=10/36 -> P(7|7or9) = 1/9 / 5/18 = 1/9 * 18/5 = 2/5 which is obv why the odds bet pays 3/2.
Anyway, if i can do that, can I not do a formula for odds of a 5 and a 9 before a seven? I don't honestly even know if its possible just seems like to me that it should be! If you can give me any advice as well regarding that I would be very grateful!
Cheers for your time, man.
November 5th, 2017 at 6:50:26 AM
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This is not related to the math but you might enjoy the story. We where playing craps one night and my wife got the dice. She proceeded to roll the All Tall Small bet hitting every number. She rolled a few numbers afterward and sevened out. We went to dinner and came back to the table. She gets the dice again and roll the ATS again! Twice I one night on back to back rolls!! She then rolled every number again except for one small number and one tall number. She was two numbers away from hitting it twice in one roll. Never had a roll close to that again but it was fun when it happened!!
November 5th, 2017 at 7:50:35 AM
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Quote: DeMangoUp to 21% edge these lower payout days, correct?
That's easy peasy for you DI people to beat, right??
DUHHIIIIIIIII HEARD THAT!
November 5th, 2017 at 10:11:31 AM
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Just another brick in the wall.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
November 5th, 2017 at 10:12:13 AM
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How about the new payouts, 30-1 and 150-1? Reply to Miplet
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
November 5th, 2017 at 10:55:57 AM
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Yes the new payouts suck. 18.3% and 20.6% if those are "to 1" pays.Quote: DeMangoHow about the new payouts, 30-1 and 150-1? Reply to Miplet
“Man Babes” #AxelFabulous
November 5th, 2017 at 11:21:39 AM
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Quote: FalsettoThanks Miplet! That really helpful! I started doing something similar and then thought I might be trying too hard so stopped haha.
But to clarify, you have 2^10 states as a result of having either made or not made each number correct? I'm not enough of an excel wiz to understand everything that's going on but most of it makes intuitive sense. I am curious as to how using the mod(floor ... gives you the 1024 different possible orders like it does? Is there any chance you could dumb that down for me a tiny bit? haha
Also, the reason I stopped trying to make a sheet like that, is I thought maybe I could find a formula to solve the whole thing?
For example, P(9)= 4/36 P(7or9)1=10/36 -> P(7|7or9) = 1/9 / 5/18 = 1/9 * 18/5 = 2/5 which is obv why the odds bet pays 3/2.
Anyway, if i can do that, can I not do a formula for odds of a 5 and a 9 before a seven? I don't honestly even know if its possible just seems like to me that it should be! If you can give me any advice as well regarding that I would be very grateful!
Cheers for your time, man.
The =mod(floor(X,Y,1),2) formula was what I came up with to get the individual bits from a number. X is the number and Y is the bit (always a power of 2). Let's do 5 for example
1s bit. mod(floor(5,1,1),2) 5/1=5 and the remainder of 5/2 is 1
2s bit. mod(floor(5,2,1),2) 5/2=2 when rounded down and the remainder of 2/2 is 0
4s bit. mod(floor(5,4,1),2) 5/4=1 when rounded down and the remainder of 1/2 is 1
8s bit. mod(floor(5,8,1),2) 5/8=0 when rounded down and the remainder of 0/2 is 0
16s bit and higher are all like 8s bit and end up as 0
“Man Babes” #AxelFabulous
November 5th, 2017 at 8:33:18 PM
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Took me a while but I get that now. And I would never have figured that out as a way to populate the matrix haha. Really nice job!
November 6th, 2017 at 11:06:54 PM
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someone did, but it is not like a+b-c = pQuote: FalsettoAlso, the reason I stopped trying to make a sheet like that, is I thought maybe I could find a formula to solve the whole thing?
it uses the inclusion/exclusion principle.
BruceZ(?) did this some time ago
and wrote some R code for it.
to read about it hereQuote: FalsettoAnyway, if i can do that,
can I not do a formula for odds of a 5 and a 9 before a seven?
I don't honestly even know if its possible just seems like to me that it should be!
https://forumserver.twoplustwo.com/25/probability/craps-probability-1208676/
here is the code I have
###################################################
# Odds against rolling a subset of numbers before a single number
###################################################
nums = c(5,9,7) # Last must occur only after all others in any order
in_36 = ifelse(nums <= 7, nums-1, 13-nums) # 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]))
}
}
p # prob of not rolling a subset of numbers before a single number
1-p # prob of rolling a subset of numbers before a single number
a=1/(1-p) # 1 in
a
a-1
Odds.to.1 = 1/(1-p) - 1
Odds.to.1
returns this
> p # prob of not rolling a subset of numbers before a single number
[1] 0.7714286
> 1-p # prob of rolling a subset of numbers before a single number
[1] 0.2285714
> a=1/(1-p) # 1 in
> a
[1] 4.375
> a-1
[1] 3.375
> Odds.to.1 = 1/(1-p) - 1
> Odds.to.1
[1] 3.375
>
Sally
I did this in Excel using a transition matrix.
this is way faster and easier once you know what is up
and one can run this R code online it you do not want R on your machine
http://aleph.sagemath.org/
(select R 1st)
[1] 0.7714286
[1] 0.2285714
[1] 4.375
[1] 3.375
[1] 3.375
here is for All
> ##################################################################
> # Odds against rolling a subset of numbers before a single number
> ##################################################################
> nums = 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(nums <= 7, nums-1, 13-nums) # 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]))
+ }
+ }
> p # prob of not rolling a subset of numbers before a single number
[1] 0.9947423
> 1-p # prob of rolling a subset of numbers before a single number
[1] 0.005257704
> a=1/(1-p) # 1 in
> a
[1] 190.1971
> a-1
[1] 189.1971
> Odds.to.1 = 1/(1-p) - 1
> Odds.to.1
[1] 189.1971
>
hope this can help out
Hi BruceZ!
I have yet to win that All bet!
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