TurboGenius I take my hat off to Turbo

Only 13 to choose from

 

Hi Notto,
you like to show the numbers and/or the game on RS ?

 

Herb; are you trying to find how Turbo selects his numbers?
Remember he throws in a riddle; reply 9 of adv 3. This will widen the selection.

To all Riddler follows; over at RF-forum, JEK, is it Eddy, got a reply from Saint Steve; who's losing the plot over at RF.
The reply in a thread was, betting 30 numbers is okay, but can be messy. Then he says 15 numbers is okay.
For someone who don't like KTF thread that starts with 27 or 28 numbers and reduces as the stream flows.
You wonder why he said 30 numbers is ok.

Now Chief Riddler says he had only 11 numbers to choose from, some go and some come back.
Top3 leave sometimes and comeback. But what of reply 9? They pay better than 35-1.

Leave it there for you all to do some thinking.

 

Thank you, working on it.

 

That s the problem there we do not know when the 3 leaders will strike again.
I have not ran many simulations to see how they change but it may be worth it you never know.

 

John Have you looked at You'd call this random; Teorulet over at RF forum.
You can see how some leave and come back.
If you track ADV: 3 it's like GUT all laps want to go left to right and get to all 37 hit.
Winkel's pic shows this.
So, if lap 1x shows 12-34-25; why wouldn't they show as lap R1. Reply 9 you'd stretch your top group to maybe top7 or 8. Why?

Teorulet or luck of the irish shows 7 or 8 numbers for the repeat.
Data also shows spins 1-10 usually has 1 repeat.
You can see this over at RF

 

Everything is trying to go to the right.

 

Winkel is the one he was the great universal theory website and selling the system?
I do not remember clearly because that was years ago but i think that is it.

 

9 unique numbers requires the following number of spins:
37×(1÷37+1÷36+1÷35+1÷34+1÷33+1÷32+1÷31+1÷30+1÷29) = 10.15

So we see that in 10 spins on average we will have 1 repeater as we only expect 9 unique numbers.

I'd we aim for 12 unique numbers we see we need 37×(1÷37+1÷36+1÷35+1÷34+1÷33+1÷32+1÷31+1÷30+1÷29+1÷28+1÷27+1÷26) = 14.26 spins

This implies by spin 14 we should expect 2 repeaters.

And 18 unique numbers requires
37×(1÷37+1÷36+1÷35+1÷34+1÷33+1÷32+1÷31+1÷30+1÷29+1÷28+1÷27+1÷26+1÷25+1÷24+1÷23+1÷22+1÷21+1÷20) = 24.19 spins which implies 6-7 repeats.

18 unique numbers is the 2/3 point in terms of spins, with half the wheel covered yet in the remaining 12-13 spins we only expect 5 new numbers and 7-8 repeats "on average" according to the math.

Covering all 18 numbers is a fools errand, but let's indulge the idea..

You miss the next spin and now you need to cover 19 numbers and a win cannot net profit without a progression and it gets expensive as you're covering 18+19+20+21+22+23+24+25.. numbers over the next 12-13 spins.

A progression that "works" is 18×1u, 19×2u, 20×4u, 21×10u, 22×26u, 23×72u, 24×216u which is 7,758 units and hope it doesn't go to 25 unique numbers.

A more limited progression and then give up would be the next four spins, once 18 numbers have appeared: 18×1u, 19×2u, 20×5u, 21×12u

This needs 408u and will be bet near spins 25-29. This has a 95% chance of success (5% failure rate), and nets +18, +16, +24, +24 so we expect an average win of +20.5u. it's still negative EV when you factor the expected loses.

Not recommending this Morely exploring as a reference point to the practical limits when betting many numbers. Clearly the solution lies with less numbers.

 

TwoUp
In pale green is my repeat average to hit. Comes from thousands of spins even 10'330 from The unicorn hunter the Dr Sir anyone.
So, obviously 1+3=4 repeats by 20th spin.
This game on the unreliable MPR. 3 repeats in spins 1-10, making repeats fast,+2.
I stopped betting in spins 11-20 as repeats fast; and missed 2 repeats.
Commenced after 20 spins, you can see in red box there are 10 repeats for 26 spins. Even now they are +1.
Profit made, why carry on?

I could post the excel, but the cat would be out the bag. Riddle!!! all want to get to 37. Left to right

 

Twoup

If I had 268 spins and sometimes could be betting between 8-11numbers.

Have 58 wins. I think it shows 21.64%

You can see it won the Turbo quote, take 5’000 to win 5’000.

Now like the time machine, you know a number is going to show for 11 laps. The question is all about progression.

To fast a progression or even to slow, is the problem; length of gaps between hits. That’s the some go but can comeback.



You see the #5, 5th place on row 3. But better still #21, gaps?

So, if a number goes out, let it go. When comes back, this is the dilemma, what do you do? Pick up on the progression where left, or start again.

That 2nd #3, wants you to drop #13. Okay drop it, but 2 spins later it hits. But the time machine told you to let it go. Good job that time machine as it doesn’t show till 102nd spin and we don’t see it again. Some go, some comeback or definitely don’t comeback.

Reply #9; that’ll make you think.

 

Reply #9 adv 3

 

How would you quatifify the balanced or progression proper?

 

That's then applicable to anything .. first principles.

 

TRD

I know or hope you are not falkor, anyway.

Reply 7. Winkles fancy graph. Everything starts at the left, 0 and wants to get to 37.

What starts lap 1 has the potential to start lap 2; here you need to think about reply 9.

And that’s all there is to it.

Tip.

If you can’t solve all the riddle; don’t try and make thousands like the chief, you’ll be better of taking 1st profits.

 

Hmmmm ... what I was getting at .. is the progression in itself, a balanced one,
& what makes it balance -- first principles onward ..so that it can be applied to
anything (including Turbo-play alike).

Quantify what makes progression balanced .. I am only attempting to see if
there's anything else that I do not know & exercise already.

 

The suggested 1-2-3-4-5 even 6-7-8-

Disaster, 20, 1x’s. happens once in a while. Be -4 for repeats. If next 10 had no repeat, that would be 30, 1x’s. Never seen it happen. But all you’d be betting is 3 units for the 1st 3 of lap1.

So, that would be -9 for repeats.

If next 10 spins are repeats, 40 spins will be -6 for repeats; now I’ve seen that on R-sim, -6.

Would you allocate units as 1x-1unit, R1-2unit, R2-3unit, R3-4 unit.

Or have you a favourite choice of progression?

But you can see X might have hit R1-3rd. And X could start R2 or Y. But each new lap has the potential to be started by Top3 from previous lap. Or think about reply 9, better than 35-1. But all are going left to the right.
--------------------------------------------------------------------------------------

If this exaggerated example was true; then the next 20 spins usually has 30 repeats for the 60 spins.

All 20 spins could be repeats. Then you’d have the 30/30.

Non-hit average for 7 numbers is 5 spins.


If we take the average to hit of 5-6-8 that’s 19 spins; potentially 33 non-hits for 60 spins. But look at their maximum to hit. The 31st non-hit could take the known 29 spins.

Beyond some members understanding.
But all they need is watch those going left to the right.

 

Jono

Why the long red?
was it the -3,-3,-3,-3 for repeats?
The use of the slow prog did the job. 10% made.

 

Think about it.

Red losing over 37 spins.
What was reason for betting #19?

 

Think about it.
Game 9 was a losing game placing a unit every spin.

A clue #6
Chief would have it.