# Due Theory

Discussion in 'General Craps Discussion' started by badddoin, Feb 25, 2018.

1. von duck, Mar 1, 2018

### von duck Member

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Your analogy with the coin flip stinks. Total bullshit. If something occurs once in about 1500 throws, then it will occur about 40 times in 60,000 throws, period. However I bet your numbers ARE correct for the average set of casino dice. WIN CRAPS, MY ASS. You are a "GUN" sent in by the industry, to put out fires.

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2. lone irish digit, Mar 1, 2018

### lone irish digit Member

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This is Barney,

Are you saying the greatest Mr Steen is the cowboy with the bad arm?

Thank you very much

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3. random_roller, Mar 1, 2018

### random_roller Member

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@von duck

If rolls 1 to 43 do *NOT* contain a 7 (roll #44 is a 7), do you consider there to have been one (1) streak of 40 rolls without a 7 or nine (9) such streaks?

1-40 (really 1-43 for craps purposes) or

1-40
1-41
1-42
1-43
2-41
2-42
2-43
3-42
3-43

For craps, there would be 1 such streak that really mattered based upon the above information. Rolls 1-43 are all part of the same (no 7) streak. The 7-less streak -- however long -- doesn't end until there is a 7 rolled. However, if the question is framed as how many intervals of 40 (or more) rolls without a 7 are there, then nine (9) could be an appropriate response (unless for evaluation purposes, rolls can't overlap/be double counted).

I took a quick look at what Steen posted. Thanks for sharing, Steen. At one time long ago I could have written some, uh, respectable code, but those days are long gone. That being said, it appears once the iterative search found a streak of 40 non-7 numbers, it would increment the counter totaling the # of such streaks. However, it would continue evaluating numbers without incrementing the counter even if the next 800 consecutive numbers were non-7. Only a 7 would "reset" the search for the next 40 consecutive non-7 numbers. So in my above example, it would find one (1) such streak of 40 non-7 numbers, not 9. Thus, if there are several streaks of considerably more than 40 rolls without a 7 in the data, that's going to impact the overall search results.

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Last edited: Mar 1, 2018
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4. ### Steen Member

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Stinks? Bullshit? My ass?

Wow, if I'd only known that you'd be using such powerful arguments as these, I'd have never bothered!

I'll be sure to call my bosses in "the industry" and tell them the jig is up. They can't duck the von duck any more because he really knows his stuff.

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5. random_roller, Mar 1, 2018

### random_roller Member

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@Steen

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6. von duck, Mar 1, 2018

### von duck Member

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The question was, how many rolls would be 40 OR MORE. Some of the 40 rolls would be more. If you want to make an analogy, then explain the bet, "4 rolls without a 7" and what the odds are on it. The same math that gives you 4 rolls without a 7 will lead you to 40 rolls without a 7. Don't try to blow smoke up my ass. I know a phony, and I know "canned" response when I see one.

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7. random_roller, Mar 1, 2018

### random_roller Member

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You're so far off base on this one, I thought it might be April 1st already.

Feel free to show your math. It is orders of magnitude easier to have 4 consecutive rolls without a 7 than it is 40 consecutive rolls without a 7. If that wasn't the case, craps would cease to exist, at least as we know it today.

Not everything is a conspiracy.

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8. von duck, Mar 1, 2018

### von duck Member

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Simplify it. Do the math to find 4 rolls without a 7. Then carry it out to 40 rolls without a 7. It's the same thing. Out of 100 try's, how many times would you do 4 rolls without a 7. 100 X .833333, that product X .83, that product X.83, that product X .83 = 48.225. Now go to the wizards site, and check the odds on "4 rolls no 7" bet. When between 4 and 40 does the math stop working. I believe, my math to be correct.

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9. TDVegas, Mar 1, 2018

### TDVegas Member

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10. von duck, Mar 1, 2018

### von duck Member

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Yeah TD, I notice how you showed us a lot of math, with this post. You, and steen both know where this line of discussion leads. Most of the members don't, but I do. We're going there! No, YOU probably don't know, unless Steen told you.

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11. von duck, Mar 1, 2018

### von duck Member

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Nice B.J. RR. How long have you two known each other?

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12. \$nakeEye\$, Mar 1, 2018

### \$nakeEye\$ Member

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I am surprised that you honestly feel that way -

Your chances of winning the lottery never change - for better or worse - regardless WHAT the prize amount -

Granted that MORE people play when the prize money soars - BUT - in the Final Analysis -

Your chances of winning remain constant !

\$...eE..\$

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13. random_roller, Mar 1, 2018

### random_roller Member

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No idea who Steen is. First post I've read from him (or her) was the one that was posted earlier to this thread. I don't use Wincraps. If I had, you probably would have seen a screenshot or 3 of that application in one of my posts by now.

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14. Onautopilot, Mar 1, 2018

### Onautopilot Member

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You are way off base here Duck! I guess being civil is not your strong suit, hu?

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15. random_roller, Mar 1, 2018

### random_roller Member

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There is no mandate hitting 40 consecutive non-7 rolls has to happen within Y rolls. There is only a theoretical calculation for X possible consecutive non-7 numbers (2-12, excluding 7) appearing within Y rolls. Every time a 7 is rolled, though, the reset button is hit and the count starts over. Think of Sisyphus pushing that boulder up the hill but never quite getting there. That happens a lot when attempting to reach 40 consecutive non-7 rolls. The reset button also gets tripped when attempting to reach 4 consecutive non-7 rolls, but at a considerably (exponentially) lesser clip. BTW, think of how (in)frequently you have witnessed a shooter having a 40 roll turn throwing the dice -- and that likely included a few come-out roll 7s scattered throughout.

If there are 60,000 rolls, @10K 7s (1/6; normal distribution) have to be scattered throughout. That's a lot of 7s to disperse. If it was a perfectly normal distribution of 60,000, there would literally be a 7 every 6 rolls. That virtually never happens, so there will be blocks of rolls of varying sizes between 7s. I just ran 60,000 rolls (6 sets of 10,000 rolls each) in Excel. The top 25 blocks appear below (Block Size = 0 means back-to-back 7s). A total of five (5) instances of 40 or more consecutive rolls without a 7 occurred. The largest block of consecutive non-7s was 72 (1 occurrence), followed by 52 (1 occurrence).

The above table has been truncated for formatting purposes.

The above table is the distribution of rolls.

Develop your own model using Excel if you have questions/concerns about Wincraps. Find out for yourself. In the end, you can believe what you choose to believe, even when you're wrong. That is your call.

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16. von duck, Mar 1, 2018

### von duck Member

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Assuming that everything is run honestly. Good to know I still have a 1/290,000,000 chance, of course, being struck by lightning, 290 times, while I'm waiting to win, is gonna take a lot of the fun out of it.

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17. \$nakeEye\$, Mar 1, 2018

### \$nakeEye\$ Member

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You CAN significantly reduce your odds of being struck by lightning 290 times while waiting to win the lottery -

Just don't play golf in Florida !

\$...eE..\$

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18. von duck, Mar 1, 2018

### von duck Member

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TDV "likes this post? He no more knows what this means than shit.
STEEN, is the guy that that wrote the "Win Craps" program, and therefore has an obvious, conflict of interest. My math education is limited to what I learned in high school, so it's not a "lock". I'm going to do some further research on this problem, but, apart from any tool that may have been corrupted by the industry. This guy came from somewhere, and for some reason. The timing is too convenient, for things to be otherwise. My question still stands, "when does the math that gives you the correct answer for 4 rolls no/7, stop working. Roll 6, roll 22, roll 31, when? Let's ask the question a different way. How many rolls out of 57,549, does 4rolls no/7'occur. So, 57.549 X .8333.....= 47,957.5 X .8333...=39964.6 X .83333.....= 33303.8 X .83333 = 27753.2./ 57,549 = .48225, so, when does the math quit working correctly?

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19. von duck, Mar 1, 2018

### von duck Member

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So, show me the math, not a computer simulation, the actual math that will give you the correct answer. This guy Steen, knows what the next question will be, do you?if so, ask it, otherwise stand back and watch.

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20. von duck, Mar 2, 2018

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