How to nullify the negative edge (for free)

[outsider]

To sell what?

here

link:s://:.rouletteforum.cc/index.php?topic=27691.new#new   

[ozon]

If we've already reached one number.
The optimal bet selection will probably be last number for repeat.
If there is someone who is coding in RX, he can check how low the HE will be
If it turned out to be at least a little positive
I would use positive progression cycles of 25 spins
  1-2-4

[outsider]

If we've already reached one number.
The optimal bet selection will probably be last number for repeat.

Unfortunately not. Each spin is new so there is no advantage (in this case and many others) because the probability for the last number will always be 1/37... like for the other ones.

  1-2-4

the Paroli progression? it won't work.