The 5/39 Universe

AllenB

Member
IMAGINE
The Space Time Random Generator is the Sun at the Center of a Universe containing 39 Planets. Each day 5 Planets Crash into the Sun and are ejected to a height equal to the Number of Games it will take for it to Crash Again. If You go far enough Back in time it is possible to See the 5 Planets that Crashed and the position of the Other Planets. You can also Go Back in time to the First Unknown Ejection of a Planet (You do not know when it will Crash Again). That Happens a minimum of 40 Games Back; however, Repeats during those Games increases the Games Back to the Last Known Orbits for all Planets. As you move Forward in time Unknown Orbits Increase in number, until all Orbits are unknown after the Last Crash and Ejection. For the Next Draw it looks as if all of the Planets are equal distance from the Sun; but, We know they are Not. Visualize an Outer Ring at 40 Games until Crash. Each Planet appears to be on that Ring before the Next Crash and Ejection. At the moment of the Next Crash and Ejection the 5 Planets appear at 1 Day until Crash, are Grabbed by the Sun and are ejected to the Outer Ring. Is there a way to glimpse the locations of the other planets when the 5 Crashers appear? Probably Not; but, it deserves a look. Could there be a clue as to the force of the ejection of a Planet (that makes it stick to the outer Ring for a while before starting it’s decent)? Can we tell which of the planets are descending now?
You can Look at the unfolding Universe by taking the most recent 100-300 Games and arraigning them in Descending order in Excel. Start with the top of the list (the Oldest Game) and use the Match function to find out when the number will hit again. Do this in columns for all 39 Numbers (Planets). You then have a value for each planet that is equal to the days until it Crashes again. Set the Maximum Value to 40 so that the planets go to and stick on the outer Ring until the start to fall again. Use a circle chart to see the Universe. Set up the excel sheet so that you can set the game you want to look at by a single entry in a cell. Then you can cycle thru some draws and see how the universe behaved over time. Set the Chart in the window with the Game Key Cell, Start thru some Games and watch the Chart Change.
This will date me; but, I saw Big Brother and the Holding Company with Janis Joplin, the Paul Butterfield Blues Band and Cream with Eric Clapton at the Phillimore Auditorium in SF in 1966 or 1967. On the Walls there was this pulsing blob of light from the Light Show that accompanied the performances. The Charts pulsate just like those blobs of Light at the concert. The aroma in the Room was new to me then; but let’s say that that smell may have had something to do with the vision that prompted this post.
I am not saying that this is a way to figure anything out yet. It is just another way to look at Frequencies and Cycles and shows the Beauty and Randomness of the Ejections.
 

blitzed

Member
hiya AllenB /0 WILD!

So, a pseudo-planetary billiards game eh?

Makes me think of the Number Spiral site:
http://www.numberspiral.com/art/11.gif

CREAM wooooo \0/ Tales of Brave Ulysses! one of my top favs :)

cya,
blitzed:thumb:
 

Icewynd

Member
Wow! Janis Joplin and Eric Clapton in one concert. Awesome memory.

Your idea puts a visual to one of the ways that I do lotto analysis. What you will find is that some balls will come from low orbit (skips 0 to 4), some will come from a mid orbit (skips 5-39) and some will pop in from a high orbit (40+). And once in a while all the balls will come from the same band. The high orbit group is most likely to miss, mainly because there are few balls that inhabit the mesosphere.

The probabilities of each combination of low orbit and mid orbit balls are known from history. The difficulty is two-fold: 1) predict the proportion of the balls from each group; and 2)filter the balls in that orbit to eliminate those that will not hit.

So, 2 easy steps. What could go wrong? :)
 

AllenB

Member
I am taking a look at a few balls this morning. Charts of a Balls orbits (ejection heights) show patterns that may be useful. I have a setup that runs the Games from the First unknown ejection to the present (around 25 games). I forced a fixed value on the ejection height for each ball in the Last 25 Games, so that I got a 5 out of 5 for the 2nd and 4th to the Last Game and 4 of 5 from the last game and the 3rd to Last Game. The inventory (all combinations that match the Fixed Value) was 26,334 for each Game. The forced Values did not perform well in the 22 Games before the Last 4. There were 3 games that Hit 4 Numbers from 4368, 126 and 252 Combinations. There were 10 Games with 3 of 5 and Inventories of 21, 462, 252, 792, 792, 1287, 3033, 6188 and 15,504.
I am Running a Calculated Value that I would Like Someone to Analyze. Lets Say that I have 3 Hits in an Inventory of 12. Using the combin function there are 792 Combinations. So If there are 3 Hits how many of those 792 combinations will have 3 hits. I an using this Formula =Combin(12-3,2)=36, Meaning that of 792 Combinations, 36 will have 3 hits. Can anyone confirm that this is correct. I am running an approximate win-loss analysis along side each game. I would rather look at orbit histories than wheel numbers at this point, so I would appreciate any help.
I will get back with what I find in an analysis of a few Balls later.
Thanks Ice, I would email you my worksheet; but, you usually don't answer?
Later
 

Icewynd

Member
I came at it in a slightly different way, but same answer.
12 balls can be combined in 792 different groups of 5. 12 balls can be combined in 220 different groups of 3. Therefore, if 3 of your 12 balls hit 792/220=36 will have the correct group of 3.

Fire away with the e-mail. I will check! :thumb:
 

AllenB

Member
Thanks for verifying the calculation.
As I Look further into this, I know that I am just seeing Skips or Outs in a different way. Instead of the Linear Graphs which I still use for Each Ball, I can now wee them all in a Circle. I feel like I get a better picture of What is going on if I "Plug" the ejection distance in when the Number hits rather than send it to the outer ring and watch it. Looking back in Ca5's Last 325 games only 2 Balls were ejected too or past 40 and Started to Fall withing 1-5 Games. I am experimenting with an idea (Formula) for estimating the Ejection heights. It starts with what We know. For the Next Game Here is what We Know for Each Ball.
1. The Current Skip or Out.
2. The Average Skip over the Game Block and for all Games (a value that rarely hits)
3. The Median (another Skip value that rarely hits).
4. The Mode (the Skip Value that hits most often).
Ejection Values are Grouped into Orbits as follows:
Very Low (VL) 1-3
Low (L) 4-6
Low Medium (LM) 7-9
High Medium (HM) 10-12
Low High (LH) 13-15
High (H) 16-Above
In a separate Table the Orbit Group Replaces the Skip Value. From this We see and count Orbit Groupings, and We can attempt to Estimate the Ejection Value by Group. Assigning the Groups a Value from 1-6, Creates a Line Chart from which to "Guess" the next Orbit Group Value for Each Ball.
I am currently trying to predicts the Orbit Value by looking at how many times a Ball is ejected to each Orbit Group in the Last 30 Draws. I am counting how many times a ball is ejected (hits) in the Last 10 and 30 Draws. For example #3 hit 2 Times in the Last 10 Draws and 7 Times in the Last 30 Draws. This tells me that in the Last 10 Draws the Ball was ejected 1 time to one of the 3 Lowest Groups and in 30 Games it was ejected 6 times to one of the 3 Lowest Group Orbits. It is the Last ejection that we do not know. I am looking to see if there is any formula relationship between The Last Ejection and the Previous 30 Draws. I am looking at each Ball for a formula that will get within 3 orbit values of it’s actual value. When The formula value minus the current Skip is within +/- 3 that ball should hit (well that’s the idea to be proven). Values within +/- 2 are the goal. Inventories containing 3 or 4 hits have a Limit. For 3 hitters the Limit is 10 Balls, and for 4 Hitters is 15 Balls. Those limits are based on the California Fantasy 5 Last 20 Game payouts of $13 and $356.

To be profitable you must hit 3 with a 10 or Lower Ball inventory using $13 (Ca5 20 game average payout). 4 Hitters must be from 15 or Lower Ball inventories to be profitable using $356 (Ca5 20 Game average payout). The Chart below shows Combination Inventories and the if hit Nets.
Balls Inv 3 Hits ($13) Net 4 Hits ($356) Net
6 6 3 $39 $33 2 $712 $706
7 21 6 $78 $57 3 $1,068 $1,047
8 56 10 $130 $74 4 $1,424 $1,368
9 126 15 $195 $69 5 $1,781 $1,655
10 252 21 $273 $21 6 $2,137 $1,885
11 462 28 $364 ($98) 7 $2,493 $2,031
12 792 36 $468 ($324) 8 $2,849 $2,057
13 1287 45 $585 ($702) 9 $3,205 $1,918
14 2002 55 $715 ($1,287) 10 $3,561 $1,559
15 3003 66 $858 ($2,145) 11 $3,917 $914
16 4368 78 $1,014 ($3,354) 12 $4,273 ($95)
17 6188 91 $1,183 ($5,005) 13 $4,629 ($1,559)
18 8568 105 $1,365 ($7,203) 14 $4,985 ($3,583)
19 11628 120 $1,560 ($10,068) 15 $5,342 ($6,287)
20 15504 136 $1,768 ($13,736) 16 $5,698 ($9,806)
21 20349 153 $1,989 ($18,360) 17 $6,054 ($14,295)

Back to Formulas, Many ideas come to mind, but it takes time to setup the test on a system that is evolving and has some lapses in accuracy from my poor typing skills and late night ideas that don’t pan out.
Still I gotta say I LOVE THIS STUFF!!! It is an Excel Junkies ideal data to mine.
Ice, if you add the words “and respond if worthy thereof” to the end of your email statement, I might have some cool images for you to look at that can’t be posted here. I want to develop this a little more before putting up for others.
I have only been at this a few years, So all you folks that have “Been There” can watch and chuckle as I plod along; But, You never know, I Might kick something up or Maybe get Lucky.
Happy New Year Every One
And by the Way there is a total Eclipse of the Sun Coming on August 21, 2017. I think I will take a trip to somewhere on the center of the path Like Southern Idaho to See it.
 

Icewynd

Member
I experienced a total eclipse in the 70's -- eerie to see the world go dark in the middle of the day. I would recommend making the trip. One thing I remember was roosters crowing as some science type had brought a bunch of roosters to see if they responded to the changes in light or to some internal clock for their morning crowing.

With regard to lotto balls, the ones that return from low orbits several times in a row are what we call "hot" numbers, i.e. numbers with a high frequency count in recent games. Once we identify a "hot" ball we can expect it to hit on low skips (until it cools off, of course).

Its good to know what would be profitable, but would be better to know how to whittle down 39 balls to 10 while keeping 3 or more of the winners. I tend to go with low skip values (less than 10), followed by frequency (1 to 4 hits in last 16 games).
 

AllenB

Member
Beware of Black Holes, I fell into one yesterday and thought I Found something. I even Sent Ice an Email with an outlandish claim that I could eliminate 17 Numbers each Game and the Draws would come from the remaining 22. I apologize Ice. No Response expected or Deserved.
So anyway, I have this Crazy Idea that there is some kind of a background Shuffle that takes Place after every Draw. I was trying an Idea that might mimic that shuffle. This Entire Idea is ludicrous, but, I am still Stuck in The Black hole so here is the Delusion.
The idea is that Each Number has a Twin in an alternate universe. I tried to mimic this by using a "Dial" that Changed the Twin Pairings after Each Draw. I then Ran all of the Games and Counted Twin Hits. You guessed it, When I fixed an Error in the Excel Worksheet there if was. Random Again. The Twins hit with the identical frequency of the Balls even though they change partners after every Game. So here I am Looking and Feeling Stupid. I can't find the Winners When they Stay in one Place. I sure as hell won't be able to find them if they move, Right.
Excel Malfunctions have temporarily shut down power to my Rockets. I hope to Restart them today and Blast Out of this Black Hole so I can Go Find the Next one and go through this all over again.
Not Giving Up, But I have to be sure, before I make any more outlandish Claims and then Eat them in Pie.
 

blitzed

Member
hi y'all, yes a solar eclipse is wild...temp drops, birds are fooled and fly back into their nests for the night *LOL*

cya,
blitzed:thumb:
 

Icewynd

Member
Allen, check your e-mail.

Black holes! I'm sure we can all identify with spending too much time chasing down some promising lead only to find that it is all based on a false premise. Love your Star Trek/sci fi language about dealing with it.

Re: eclipse -- be sure to get your welders glass or safe viewing goggles before you leave for Idaho. Prices there will probably be through the roof.
 

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