TurboGenius Never bet on the "last" of something. (Informational)

Nice old system of mine.

Can it be improved according to the above? Definitely!
A lot.

VIDEO: https://odysee.com/thereddiamanthe:6
@admin support for ↑


When progression level 5 no-hit, I'd drop the progression back to 3
& only go back to 5 once the combo (wide+focused) has been realized.
Once exposition again in a combo-closing range, shift over again & adjust to only as much as necessary.
Pendulum →1235•53←.

As above, wide+focused, here exclusively ≈2Q→DS.
BS -- using furthest back &! recently appeared - scanning ds areas ..

Adjusting the progression level as the game asks to finish
+
positions as a new level is reached, &or as new ds areas come to the fore.
(occasionally -- when only recent repeats, use repeats; or one of each ..
whatever available .. updating them with new ones as soon as).


Stop-loss -- abolished.
Minimum betting.
Nominal profit (+1) or nearby.
Session = gamechain.

100u goal -- avg. 300 spins
0.3u/spin avg profit ratio

 

& another view
VIDEO: https://odysee.com/@empfire:a/mobizen_20220320_031254:0

watch ds ..

 

& yet another ..
VIDEO: https://odysee.com/@empfire:a/thereddiamanthe:6

a bit longer

 

For some reason the link for the first game three posts above;
thereof the updated link.

VIDEO: https://odysee.com/@empfire:a/mobizen_20220320_000905:8

 

To square it off I am gonna post the video to the 4th session.

 

VIDEO: https://odysee.com/@empfire:a/YouCut_20220322_001728341:d

Another a bit longer session.

The divisor came in place, here requiring more than two hits or combo to finish;
effectively the 2nd level of the recovery as described on the top of the page (•53←).

 

So, I finally got to testing this one. I was wondering how the accuracy would increase. First I extended the multiple laps for dozens. Here are the results. At the top are the results of the average drops of dozens in the order of dropout. The bottom is the same, only I tried to make the order (sequence) of the dropout.

 

It wont make a dent trust me, look at the info from left to right (spin count) and from top to bottom (spin count) from number to number. What will you see? The odds are 1:3 on average. I did this test years ago.

 

then I increased the horses, for a dozen on the lines (I hope I increased the horses correctly)
at the top there are also average hits in the order of the lines(ds) drop. It's the same at the bottom, and I also tried to make a dropout order based on the averages.

Remarkably, I tried to combine the order of dozens and lines, and it turned out well. I don't know if it can be used.
For example: let's say the first dozen that falls will be the first dozen (1-12). Then 2 dozen appears in the same way, let's say also 2 dozen (13-24). Then 3 times we have not 3 dozen, as we know (on average, of course), but one of the first 2, i.e. repeat. Let's say 1 dozen are repeated. Further on, according to my calculations, most likely (on average) there will be a repeat of this dozen again, i.e. again 1 dozen dropped out. then on the 5th spin there is a repeat of 2 dozen, and only on the 6th spin there is a 3 dozen appear. Then, as you can see, 7 spins will be 1 dozen again, then 8 spins 2 dozen, and 9 spins again 1 dozen, 10 spins 3 dozen.
Now if we take the lines and see how they appear at the same time, then at the beginning at least it turns out very interesting. There is 1 line. Because in the dozens, the first one who appeared was 1 dozen, then, accordingly, we will take one of the 2 lines in this 1 dozen. Let it also be 1 line (1-6). Then we have 2 dozen, respectively, we take one of the two lines in 2 dozen. Let it be line 3 (13-18). Then in the dozens we have a repeat of one of the two dozen. And the data in the lines we see that on the 3rd spin a new line appear again. Therefore, we can bet not on two dozen, but on two missing lines in these two dozen. Let's say, as it was written in dozens, 1 dozen was repeated, and accordingly, then 2 lines (7-12) appear. Next on the 4th spin, we have a repeat of this dozen in the dozens again, and there is also a repeat in the lines. And we are already choosing not from 3 lines, but from two that are in 1 dozen.
Now the 5 and 6 spins don't match between the dozens and the lines. Therefore, I would still assume that 5 spins will be 3 dozen and let's say 5 lines (25-30) in it. Then 6 spins in dozens will then be a repeat of 2 or 3 dozen. In the lines we also have a repeat of three lines. But since line 2 refers to 1 dozen, and we have a repeat of is 2 or 3 dozen, then we choose from 3 or 5 lines. Let it be line 3 (13-18). Etc.
In general, somewhere it strongly coincides that you can just guess which line exactly will fall out, and somewhere some averages do not match. And moreover, these are average values, and how to use them, I can't even imagine.

 

I also increased the number of horses to 12. That is, the same tests are already on the streets. And this increase in horses and laps, personally, does not give me any increased accuracy. I don't see where the accuracy can be increased here.
Yes, there are still internal numbers, but this is a different method of the game, which is based on completing the pattern first. But even then, this is also all just an analogy. As those "who found" said, you need to look for patterns in repeaters, but I have no idea how you can find any pattern in repeaters at all?
And here I tried to apply it to 4 parts. Which of course can be based on the third part, but part 3 I don't understand at all what can be done with it? At 4 it's a little different. For example, if we take the average appearance of the first 3x, then if we take all the numbers, this is 21 spins. And if you take dozens separately, then this is already 33 spins on average. And this means that there is a place to be these 2/3, as Turbo wrote. Where after 1 and 2 you need to reset. I thought that we could somehow increase the accuracy by testing this. It turned out to be in vain.