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 diceQuote:odiousgambityep, 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/

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.Quote:Ace2If 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

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.

Quote:Ace2If 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 diceQuote:odiousgambityep, 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/

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.

Quote:DeMangoAmazing 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.

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.

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%

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!

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