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odiousgambit
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November 26th, 2020 at 7:23:36 AM permalink
yep, dice setting again. 

but I want to check some assumptions that I am not sure are correct, and my math could be wrong too

the question does come up, how much would a shooter have to deviate from random in order to have an advantage? So I go to the Wizard's pages, see first link, and find that the ratio that is used is called rolls to sevens, and the first increment seems to be a ration of 6.04 versus 6 for random. It strikes me right away as odd, since the change is noted in the denominator, ie 1/6 versus 1/6.04 ............ This is one thing I'm trying to check, can someone confirm that I am interpreting that right?

However if we try to put such a small change in the numerator, I can see why this is done. We get 0.99337748344/6 ........... yeah, that's awkward. [my math needs to be checked] . I am assuming it is not meant to mean a ratio of 6.04 rolls of seven to 6.00 random for the reason the 7s would be increasing with that shooter [and why pick 6 as a measure anyway?]

In the meantime it must be noted that the chart in the link shows player advantage at 6.04. For one thousand rolls, 166.667 [1000/6]would be random and 6.04 [1000/6.04] gets you 165.5629 per thousand, or about 1 roll of seven less, per thousand rolls! That's not much for sure. 

So I am trying to see if I am interpreting it all correctly, or not. 

https://wizardofodds.com/games/craps/appendix/4/
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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November 27th, 2020 at 7:45:46 AM permalink
groan if you wish, but I am pursuing this for reasons unclear even to myself

Quote: that link

Skill Factor

The skill factor is defined as the percentage of double-pitch throws that the skillful shooter turns into zero-pitch throws. A skill factor of zero would apply to a random shooter, where the probability of both a zero-pitch and double-pitch throw are each (2/3) × (2/3) × (1/4) = 1/9 = 11.11%. A skill factor of 12%, for example, would move 12% of double-pitches into zero-pitches. In this case, the probability of a double-pitch would be 11.11% × 88% = 9.78%, and for a zero-pitch would be 11.11% × 112% = 12.44%. All other outcomes would be the same as that of a random shooter.

I think I have a stumbling block here, because the Wizard gets into all this zero and double pitch stuff and understands it well no doubt. However, I'm trying to avoid a plunge into that deep of a pool and just try to answer the question,  Did I get the right answer when I conclude 'about 1 roll of seven less, per thousand rolls' " will put a player into +EV? If I am wrong, it means I am not understanding what rolls to sevens ratio means [I'm pretty sure]. Yet it doesn't seem realistic that such a modest change could mean going from -EV to +EV even though clearly it's not much. 

BTW so far none of those Golden Touch folks seem to be chiming in here to help. Hmmm.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Gabes22
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November 27th, 2020 at 9:11:42 AM permalink
I think the burden of proof is much higher than say x amount of rolls per 7. I think to prove that dice setting works you not only have to be able to prove that you are able to reduce the # of 7s but also be able to do it from multiple positions at the table because different positions require different arcs, trajectories and velocities. Much like in basketball an 8 foot jumper is different than a free throw is different than a 3 pt shot, rolling the dice is different from the hook than it is next to the stickman
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odiousgambit
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November 27th, 2020 at 10:01:06 AM permalink
Quote: Gabes22

I think the burden of proof is much higher than say x amount of rolls per 7. I think to prove that dice setting works you not only have to be able to prove that you are able to reduce the # of 7s but also be able to do it from multiple positions at the table because different positions require different arcs, trajectories and velocities. Much like in basketball an 8 foot jumper is different than a free throw is different than a 3 pt shot, rolling the dice is different from the hook than it is next to the stickman

for the most part I don't even try to be serious about it unless I am just to the right of the stickman. That's not to say I won't set the dice, but for example throwing it from one of the ends they have to go so far in most tables it seems quite ridiculous.

I have gotten away from setting the dice much, but I think I will get back to it. I don't believe in it! I do have fun with it when I do set them.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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November 28th, 2020 at 6:23:02 AM permalink
Remember you can always block a thread, ha ha, I'm going to keep going with this. Planning on doing some gambling for the first time since The Plague hit and I have to admit it's got me going a bit.


I'm at the point of being satisfied that my assumptions have been correct, rereading the link about rolls to sevens and also noting 1 divided by 6.04 and 0.99337748344 divided by 6 both equal 0.1655629139, so that has to be right. I also have to say the bit about 166.667 versus 165.5629 per thousand rolls is right, can't see how it would be wrong.


I'm still befuddled a bit by 'skill factor', not so much the idea but the ability to apply the math, so ... Well, anyway, one thing I note in the Wizard's pages on this is that somehow there is no mention of the 'flying V' dice set. If you can stay on axis with that set, it was shown to me*, it rules out a roll of 7 using 5+2 or 6+1. If staying on axis, a big 'if' ha ha, the only possible rolls are 6,5,7,8... 9,8,10,11... 7,6,8,9... or 4,3,5,6 , which is 2 sevens out of 16, or 1 out of 8 versus random's 1 out of 6**. You can check this out yourself with a pair of dice.[Edit: I'm taking this off as I am not sure I am including all possibilites, sorry] I wish we had a chart, then, with the effect of using even a random roll in the comeout and then following up with the flying V after a number is marked to be resolved. This has been my practice, but the Wizard's chart uses hardways sets for some reason.


Lol Craps dot com image for the flying V
Actually any dice that are set as identical pairs, but then have one die turned 180 degrees like that, might produce the 'less 7s' effect, as far as I can tell; it can certainly be flying 2s instead of flying 3s. [edit]


Including that link again........... https://wizardofodds.com/games/craps/appendix/4/


*by some member here, and thanks very much, but sorry I forget who
** see further downthread for possible error [posting soon]
Last edited by: odiousgambit on Nov 28, 2020
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
TinMan
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November 28th, 2020 at 6:40:11 AM permalink
Quote: Gabes22

I think the burden of proof is much higher than say x amount of rolls per 7. I think to prove that dice setting works you not only have to be able to prove that you are able to reduce the # of 7s but also be able to do it from multiple positions at the table because different positions require different arcs, trajectories and velocities. Much like in basketball an 8 foot jumper is different than a free throw is different than a 3 pt shot, rolling the dice is different from the hook than it is next to the stickman



I’m skeptical that a person could influence the dice but I wouldn’t require someone to demonstrate the ability to do it from different positions. It’s like asking someone to hole card from every spot at a BJ table. So long as you can do it from one spot, that’s plenty. You’d only play when space was open at your preferred spot (I’d guess stick right/left 1 is the ideal spot but I could be wrong).
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odiousgambit
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November 28th, 2020 at 6:53:39 AM permalink
Quote: TinMan

I’m skeptical that a person could influence the dice but I wouldn’t require someone to demonstrate the ability to do it from different positions. It’s like asking someone to hole card from every spot at a BJ table. So long as you can do it from one spot, that’s plenty. You’d only play when space was open at your preferred spot (I’d guess stick right/left 1 is the ideal spot but I could be wrong).

I've never seen if demonstrated at all, which may seem like a strange comment. Looking for slow-mo demonstrations on youtube, what is posted for the most part are the worst examples of throws that might have any chance at all of staying on axis. Without that the whole subject seems ridiculous to me. At least I can say I can do better than showing a throw that hits that alligator surface, good grief.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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November 28th, 2020 at 7:04:44 AM permalink
Quote:

If staying on axis, a big 'if' ha ha, the only possible rolls are 6,5,7,8... 9,8,10,11... 7,6,8,9... or 4,3,5,6 , which is 2 sevens out of 16, or 1 out of 8 versus random's 1 out of 6



Double checking this, I now get as the only possible rolls as 6,8,8,6... 9,8,10,11... 7,6,8,9... or 4,3,5,6 , which is only one 7 out of 16. I seem to remember it is two sevens but now I can't find my error if there is one.

it does seem to be a fact that there are only 16 possibilities, right? which way you have the V pointing matters, you get different numbers, but it doesn't really matter I think. Hooboy. I've managed to convince myself I shouldn't be so sure, and have made an edit upthread.
Last edited by: odiousgambit on Nov 28, 2020
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
DeMango
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November 28th, 2020 at 8:09:45 AM permalink
Triple check please. There will be 2 7's, double pitch 4/3's, on 16 on axis tosses.
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odiousgambit
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November 28th, 2020 at 1:45:49 PM permalink
Quote: DeMango

Triple check please. There will be 2 7's, double pitch 4/3's, on 16 on axis tosses.


Thanks. We agree that 4 times 4 is 16, good. 


I did say I needed help!


I was thrown for a bit by realizing you can get different sets of numbers, 'depending'. It is easy to intend to start one die rotating and the next thing you know you've got the other one rotating and get lost. 


OK, so with a die with 2 and 5 as the axis, giving the axis a right and left side by facing the 6, say, the 2 is on the left and the 5 is on the right, call it the 2-5 die. For the other die with 6 and 1 as the axis, arbitrarily facing the 5, the 6 is on the left and the 1 is on the right. The 6-1 die call it. 


With the 2-5 die unchanged after throw, the four numbers we get with the spin of the 6-1 die are 8, 7, 5, 6, the last through double-pitch.
2-5 die making a rotation on axis that 6 comes up, making the four possibles 11, 10, 8, 9 same way. 
2-5 die making another rotation the 4 comes, and the four possilbes are 9, 8, 2, 7 ...
2-5 die making last rotation, the 1 comes up and the four we get are 6, 5, 3, 4


we encountered two 7s out of 16


However, there can be a 5-2 die if we choose to face the 1 on the die, that is, 5 on the left, 2 on the right.
Going through the same way with the 6-1 die, 8, 7, 5, 6.... 6,5,3,4... 9,8,6,7... 11,10,8,9............. two 7s


There can be a 1-6 axis die if you choose to face the 2. Against a 2-5 die that does the rotation on axis:
1-6 die unchanged, 9,7,4,6 
1-6 die has the 2 come up, 8,6,3,5
1-6 die has the 4 come up, 10,8,5,7
1-6 die has the 5 come up, 11,9,6,8.................... process also has two 7s


1-6 die now needs the numbers from a 5-2 die. Somebody may tell me these are not the only possibles but I hope this is it


4,7,9,6... 3,6,8,5... 5,8,10,7... 6,9,11,8            ....................... two 7s


So there are 16 possibilities, but only for each of four scenarios, making 64, which has nearly killed me. Note that a player does not need to decide anything but to have the flying V, or even flying 2s, all of which is quickly set.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
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November 28th, 2020 at 2:36:35 PM permalink
Quote: odiousgambit

yep, dice setting again. 

but I want to check some assumptions that I am not sure are correct, and my math could be wrong too

the question does come up, how much would a shooter have to deviate from random in order to have an advantage? So I go to the Wizard's pages, see first link, and find that the ratio that is used is called rolls to sevens, and the first increment seems to be a ration of 6.04 versus 6 for random. It strikes me right away as odd, since the change is noted in the denominator, ie 1/6 versus 1/6.04 ............ This is one thing I'm trying to check, can someone confirm that I am interpreting that right?

However if we try to put such a small change in the numerator, I can see why this is done. We get 0.99337748344/6 ........... yeah, that's awkward. [my math needs to be checked] . I am assuming it is not meant to mean a ratio of 6.04 rolls of seven to 6.00 random for the reason the 7s would be increasing with that shooter [and why pick 6 as a measure anyway?]

In the meantime it must be noted that the chart in the link shows player advantage at 6.04. For one thousand rolls, 166.667 [1000/6]would be random and 6.04 [1000/6.04] gets you 165.5629 per thousand, or about 1 roll of seven less, per thousand rolls! That's not much for sure. 

So I am trying to see if I am interpreting it all correctly, or not. 

https://wizardofodds.com/games/craps/appendix/4/

If you get less than 120 sevens in a thousand rolls, that would be somewhat convincing. 4 SDs south of expectations, so about 1 in 30,000 chance without influencing the dice
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odiousgambit
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November 29th, 2020 at 4:52:16 AM permalink
Quote: Ace2

If you get less than 120 sevens in a thousand rolls, that would be somewhat convincing.  4 SDs south of expectations, so about 1 in 30,000 chance without influencing the dice

I have to say I'd be convinced alright. You'd never be able to pull me out of the casino, then one day I'd be thrown out. 


But I see your point. Let's say someone demonstrates his stuff and does 155 sevens in a thousand ... if that was a real indication of his abilities, it'd be a shame, because no one would really believe it and for sure there would be no way to know other than to point out it is within probably 1 or 2 standard deviations. 


A guy getting obsessed with this sort of thing has plenty of company, I keep telling myself. Stanford Wong even wrote a book.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
unJon
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November 29th, 2020 at 6:35:36 AM permalink
Quote: Ace2

Quote: odiousgambit

yep, dice setting again. 

but I want to check some assumptions that I am not sure are correct, and my math could be wrong too

the question does come up, how much would a shooter have to deviate from random in order to have an advantage? So I go to the Wizard's pages, see first link, and find that the ratio that is used is called rolls to sevens, and the first increment seems to be a ration of 6.04 versus 6 for random. It strikes me right away as odd, since the change is noted in the denominator, ie 1/6 versus 1/6.04 ............ This is one thing I'm trying to check, can someone confirm that I am interpreting that right?

However if we try to put such a small change in the numerator, I can see why this is done. We get 0.99337748344/6 ........... yeah, that's awkward. [my math needs to be checked] . I am assuming it is not meant to mean a ratio of 6.04 rolls of seven to 6.00 random for the reason the 7s would be increasing with that shooter [and why pick 6 as a measure anyway?]

In the meantime it must be noted that the chart in the link shows player advantage at 6.04. For one thousand rolls, 166.667 [1000/6]would be random and 6.04 [1000/6.04] gets you 165.5629 per thousand, or about 1 roll of seven less, per thousand rolls! That's not much for sure. 

So I am trying to see if I am interpreting it all correctly, or not. 

https://wizardofodds.com/games/craps/appendix/4/

If you get less than 120 sevens in a thousand rolls, that would be somewhat convincing. 4 SDs south of expectations, so about 1 in 30,000 chance without influencing the dice



I think that’s expecting too much based on the “theory” of DI. Let’s assume that the shooter can keep the dice on axis 100% of the time but it’s otherwise random. That would leave 2 sevens out of 16 or 125 sevens expected out of 1,000.

Alternatively, let’s assume a shooter can “affect” one role out of 6 through spin correlation so zero sevens expected that role. That would be 138.89 sevens out of 1,000.

In other words, expect you’ll need a bigger sample size to test for a smaller effect.
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DeMango
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November 29th, 2020 at 7:12:56 AM permalink
Amazing that most "experts" assume the way to go is "On Axis" and the V3 is the only way to go and you must also avoid sevens on the come out.
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unJon
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November 29th, 2020 at 8:06:45 AM permalink
Quote: DeMango

Amazing that most "experts" assume the way to go is "On Axis" and the V3 is the only way to go and you must also avoid sevens on the come out.



No expert here. I set for fun. I used to set hard ways on coke out but the ATS bet made me stop. I’ll switch between V2, V3 and crossed 6s depending on what I’m hunting for.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
odiousgambit
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December 1st, 2020 at 10:57:15 AM permalink
I'm still going with this. 


I wanted to see just how many less 7s a player needs to roll before it's +EV in rightside Craps. So, I came up with this, and maybe you'll tell me it can't be done this way, but it seems to offer an insight unless I am very much mistaken. So I took the following, 

the familiar figures for each number that can be rolled:
2....1/36   rolled and  loses 1 unit
3.....2/36  rolled and loses 1
4.....3/36  rolled and wins 3/9 of the time  
10.....3/36  rolled and wins 3/9 of the time 
5...4/36  rolled and wins 4/10 of the time 
9...4/36  rolled and wins 4/10 of the time
6...5/36 rolled and wins 5/11 of the time 
8...5/36 rolled and wins 5/11 of the time
7    6/36     wins 1
11  2/36    wins 1
12  1/36     loses 1

I put it in a calculator as 
[1/12*1/3]+[1/12*1/3]+[1/9*2/5]+[1/9*2/5]+[5/36*5/11]+[5/36*5/11]+[1/6]+[1/18]
for the winning combinations, in order those are 4,10,5,9,6,8, when winning, then 7,11 in come out

for the losing combinations, 
-[1/12*2/3]-[1/12*2/3]-[1/9*3/5]-[1/9*3/5]-[5/36*6/11]-[5/36*6/11]-[1/36]-[1/36]-[1/18]
which is 4,10,5,9,6,8 when losing, then 2,12, 3 in come out

the sum of the winners is +0.4929292929292929
the sum of the losers is -0.5070707070707071
and the sum of both is the familiar -0.0141414141414142 house edge

So after pondering this quite a bit, decided what it would look like if the shooter rolled one less 7 in about one thousand rolls, and chose 972 rolls, a multiple of 36. I then postulated that if the shooter rolled one less 7, it would be when he set the dice for that in numbers to be resolved, while rolling without setting dice in the come outs. So in the equation, only the chances when resolving did I want to alter. To make the shooter slightly more unlucky, what is rolled instead of 7-out will be deemed to be a roll that does not resolve, so only the part of the equation that gives the wins is unchanged while the chances of a loss do indeed decrease slightly for each. 

So when rolling to resolve, instead of the ratio 162 per 972 rolls [1/6] such shooter enjoys a mere 161, while the chances, say, for rolling an 8 stay at 5/36 or 135 per 972 rolls. This makes 161 ways to lose and 135 ways to win, 296 total ways, and 161/296, the chances of losing, now to be 0.5439189189189189 instead of the 6/11 that comes with random results, or 0.5454545454545455, a difference of 0.0015356265356266 which seems to fit. So in the losing equation above, the same process for each takes place and I get:

for the 5s and 9s, 4/36, or 108/972, 161 ways to lose vs 108, total of 269, 161/269 checks as 0.5985130111524164 chances instead of the 0.6 of random's 3/5

for the 4s and 10s, 3/36 = 81/972, and 161+81 = 242 total ways, 161/242 = 0.6652892561983471 which is mighty close to 2/3 and 67%

-[1/12*161/242]-[1/12*161/242]-[1/9*3/5]-[1/9*161/269]-[5/36*161/296]-[5/36*161/296]-[1/36]-[1/36]-[1/18] = -0.5062493547496927 

that is reduced from -0.5070707070707071 however, 
+0.4929292929292929-0.5062493547496927 = -0.0133200618203998

I get a house edge of -1.33% for the 6.04 shooter versus -1.41% and that is definitely progress but the shooter has to be better it seems.

Now I do realize that this process is a little wonky, but to me it does not seem totally flawed. Or can somebody tell me that it is? If not totally flawed I am going to continue with it.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
ChumpChange
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December 1st, 2020 at 11:40:41 AM permalink
I set for 7's on the come-out in hopes that I'll hit 6 come-out winners in a row. If I make a point, I get another chance for another 6 come-out winners.
odiousgambit
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December 2nd, 2020 at 6:51:29 AM permalink
caught an error in the above for the shooter who gets one less 7 in 972 [with other stipulations]

it should have been:

-[1/12*161/242]-[1/12*161/242]-[1/9*161/269]-[1/9*161/269]-[5/36*161/296]-[5/36*161/296]-[1/36]-[1/36]-[1/18] = -0.5060841337666279

that is reduced from -0.5070707070707071 however, 
+0.4929292929292929--0.5060841337666279 = -0.013154840837335

house edge of -1.31% versus the skill-less 1.41%
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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December 2nd, 2020 at 6:59:55 AM permalink
Trial and error continues, to see at what point rolling less 7s in an unknown way gets a player to +EV. The next level of skill noted by the Wizard is a 6.08 rolls to sevens shooter which indeed means we go to about two missing 7s after 1000 in theory if continuing with the assumption this means one 7 per 6.08 rolls, 1/6.08 instead of 1/6. I'll continue with the above stipulations, that the come-out rolls are without setting the dice and random results, and that rolling less 7s helps avoid 7-outs but does not increase the number of wins. That latter stipulation will have to change when we go to higher rolls to sevens, and that is going to give me a headache it looks like. I'm not expecting much change but let's see. 

Sticking to the admittedly deviating 972 rolls, again instead of the ratio 162 per 972 rolls [1/6] such shooter enjoys a mere 160, while the chances for rolling a 6 or 8 stay at 5/36 or 135 per 972 rolls, total of 295 possibles for resolving. 

for the 5s and 9s, 4/36, or 108/972, 160 ways to lose vs 108, total of 268 possibles. 

for the 4s and 10s, 3/36 = 81/972, and 160+81 = 241 total ways. 


-[1/12*160/241]-[1/12*160/241]-[1/9*160/268]-[1/9*160/268]-[5/36*160/295]-[5/36*160/295]-[1/36]-[1/36]-[1/18] = -0.5050902973936376

-0.5050902973936376+0.4929292929292929 = -0.0121610044643447

-1.22% ; so we are getting somewhere but not to +EV

How do I solve the need to adjust the winning side of the equation if I go to, say, 6 less rolls than random of seven, per 972? Help accepted!
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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December 3rd, 2020 at 2:48:47 AM permalink
I certainly hope you have blocked this thread if it bothers you.

I apologize since I too can get irritated by the subject of dice control. The people who populate that universe are full of delusional types and clearly even some despicable operators preying on the gullible. Then there are those who just try to have fun with it, including me, since those bitten by the Craps bug are going to play anyway. After all, I have seen the proof that the casinos do worry about you if you set the dice. That alone adds a big element of fun to it. For me sometimes the fun even includes trying to crunch some numbers.

It would be nice to know just what point in a player's rolls per sevens he finally eclipses the house edge for a particular scenario, still working on that and I think I know where to go with this next.

I will only post again here with something of greater interest, and will use a blog post to record what I have so far. Anything else of more minor interest will also go in the blog.

Final note, as for myself, I have changed my throw completely. One prior problem was when the dice landed with my old throw, they landed together and just exploded apart nearly every time. This is fixed now. I have also made slight improvement in getting the dice to stay on axis until they hit something, at which point they may stay on axis if they hit perfectly flat to that surface. I actually think I was working on this in my subconscious as I haven't played Craps since last December, but when I picked up some dice I knew what to try right away.

My guess is that +EV will be unattainable even if I bought some equipment and practiced constantly*. The not staying on axis problem will probably not really get fixed. Ah, make that simply 'will not get fixed'. But I will have some fun trying.

* which I'm not going to do
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Gabes22
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December 3rd, 2020 at 4:57:08 AM permalink
In craps, not only do you have to minimize 7s but you have to prove you are able to make more points. If you can get the 7 up from 1 in 6 rolls to 1 in 6.25 rolls you have to equally prove you can hit the 4, 5, 6, 8, 9 and 10 at higher rates once that particular point is established
A flute with no holes is not a flute, a donut with no holes is a danish
ChumpChange
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December 3rd, 2020 at 9:24:05 AM permalink
If I set for the point of 5 and keep throwing 9's, I should put a progressive bet down on the 9.
odiousgambit
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December 3rd, 2020 at 10:37:47 AM permalink
Quote: Gabes22

In craps, not only do you have to minimize 7s but you have to prove you are able to make more points. If you can get the 7 up from 1 in 6 rolls to 1 in 6.25 rolls you have to equally prove you can hit the 4, 5, 6, 8, 9 and 10 at higher rates once that particular point is established

If you are talking about my number crunching, it is true that I am showing less 7-outs without showing increase in hitting numbers to be resolved. This is what I meant when I said it is a little wonky, and the house edge would actually be reduced more than I show.

So I don't want to continue with even higher rolls to sevens factors until I can fix that. The conundrum is once you stipulate rolling less 7-outs, you do have to figure it will mean a better rate for resolving in favor of the rightside bettors. If the shooter doesn't roll a 7-out, something else will be rolled, and a proportion of the time it will be a meaningless number and a proportion of the time it will 'hit' the number. I think this is a matter of just figuring out the proportions and I am working on that.

I don't want to switch to a special set that targets the number to resolve, though, but want to know the answer sticking to just trying to roll less 7-outs.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
DeMango
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December 3rd, 2020 at 1:08:16 PM permalink
Are we using BoneTracker to resolve all these issues? Excel program.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
odiousgambit
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December 3rd, 2020 at 3:14:29 PM permalink
I don't have the program and am not used to Excel

Anybody out there who can use this feel free
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Zcore13
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ChumpChangeRogerKint
December 4th, 2020 at 3:58:12 AM permalink
I am an employee of a Casino. Former Table Games Director,, current Pit Supervisor. All the personal opinions I post are my own and do not represent the opinions of the Casino or Tribe that I work for.
odiousgambit
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December 4th, 2020 at 4:41:24 AM permalink
As a matter of fact, I don't have a job.

And as a matter of fact, in this thread I have plenty of disclaimer about the prospects for dice setting, as much as the Wizard, I'd say, who did in fact put on his site a ton of information about it, along with that disclaimer. As for this thread, even the title is a disclaimer of sorts.

As for casino personnel, they should just admit they're paranoid about dice setting. I've seen them with their hair on fire about some dice setters who, as I watched them with 'their stuff', was able to conclude there was no possibility whatsoever they were getting anything but random results. You can only conclude these hair-enflamed paranoids might try to spend some time educating themselves about it instead of making themselves look like morons.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
odiousgambit
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December 12th, 2020 at 10:25:53 AM permalink
I was looking at the Wall Street Journal and saw an image that might explain why I sometimes don't do so good at craps. I might go into this more at my blog [but maybe not today]

Was the person who made the image a craps player having a laugh?

image from today's article "the-dicey-economics-of-investing-in-oil-during-covid-19"

the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Doc
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odiousgambit
December 12th, 2020 at 12:13:02 PM permalink
Yep, a very interesting die. With it having two faces with just a single pip each (and neither of them opposite the 6 face that we can see), I wonder how many and which of the other usual face values are missing. Might make for some advantageous place bet opportunities.
onebok
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December 15th, 2020 at 2:45:17 PM permalink
While there are "gurus" that swear by the dice being able to be thrown on-axis, the likelihood of getting the dice to perform
on-axis from start to finish of a Dice-Influenced or DI toss is very unlikely.
Assuming you have enough skill to toss the dice so that they hit the table surface, roll to the pyramid wall and react off of it and
come to rest of their own lost energy, rather than, along the way, hitting the side wall, chips, puck, the other die, etc., the dice are
still likely to be tumbled and rotated at the contact with the table top or contact with the pyramid rubber, and usually both.
High speed video shreds the idea of distinguishing on-axis vs. correlation for modes of executing a controlled toss.
Any DI-wannabee who can reduce the occurrence of sevens by 3 standard deviations will be more than happy to accept any
percentage of off-axis or on-axis tosses that simply aren't sevens.

IMO, Wizard data for SRR of 6.04 vs. 6.08, etc. is strictly statistics. From a practicable point of view, any DI with 6.08 or even 6.5
could spend their lifetime never knowing whether they were having a good Variance win or skill-based win. This goes toward your
desire to measure +EV while not knowing the bucket of good or bad numbers those non-sevens should be placed.

My guess as to how you might proceed in your "wonky" quest:
Place results of 2,3,4,10,11,12 in the garbage bucket(non-winner, non-wagered) and 5,6,8,9 in the winner bucket for all
post-comeout tosses...it is likely that most DI's will limit themselves to betting inside numbers, especially with $25 and $50
minimum bets becoming the norm around larger cities in these Covid times.
I believe that one can only hope to reduce sevens without an ability to target inside vs garbage buckets unless hugely familiar
with the table being used.
I would be very interested in what you might come up with. Without your results, I am leaning more towards the ballpark of
3 or 4 standard deviations before you can really have a +EV that is practicable and even measurable without requiring large
and troublesome sample size.
I do want to be wrong about that ballpark.
odiousgambit
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December 15th, 2020 at 5:45:31 PM permalink
Well, I'm the wrong guy in many ways for you. I can get obsessed with it for a while, but it does go away, seeing as how I have no real belief in it.

And as it turned out in this particular case, the particular table was all wrong for trying my new throw, as Covid concerns had them allow only three spots on the table, none of which suited me at all.

One thing that had me going was the realization that the apparent conclusion from the Wizard's work on it from years ago: it seems the slightest ability to deviate from 1 in 6 throws being a 7 will wipe out house edge. I was trying to get confirmation that I was reading that right and never got it, so there's that, but I think so.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
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