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 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.Quote:that linkSkill 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.

BTW so far none of those Golden Touch folks seem to be chiming in here to help. Hmmm.

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.Quote:Gabes22I 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 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.

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]

Quote:Gabes22I 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).

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.Quote:TinManI’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).

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.

Quote:DeMangoTriple 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.