Discussion in 'TurboGenius's Forum' started by Naughty but nice, May 3, 2022.
Only 13 to choose from
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.
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
Separate names with a comma.