Converting prediction error to % Adv

[Firefox]

Say that one is predicting winning numbers using visual ballistics or another method during the spin and there is a good dominant diamond.

If one predicts randomly, maths suggest the mean error would be +\- 9.25 pockets (on a European) wheel.

Is there a way to correlate mean prediction error with % advantage on the wheel in order to establish how well the prediction  system has adapted to prevailing conditions.

For example
9.25 pockets mean = -2.7%
9 pockets mean= ??%
8 pockets mean= ??%
7 pockets mean= ??%

Etc etc

Is this a good way of indicating worthwhile conditions?  Or is there a better way?

Sorry if this has been answered elsewhere. I took a look and couldn't  see it.

[Bigbroben]

Say that one is predicting winning numbers using visual ballistics or another method during the spin and there is a good dominant diamond.

If one predicts randomly, maths suggest the mean error would be +\- 9.25 pockets (on a European) wheel.

Is there a way to correlate mean prediction error with % advantage on the wheel in order to establish how well the prediction  system has adapted to prevailing conditions.

For example
9.25 pockets mean = -2.7%
9 pockets mean= ??%
8 pockets mean= ??%
7 pockets mean= ??%

Etc etc

Is this a good way of indicating worthwhile conditions?  Or is there a better way?

Sorry if this has been answered elsewhere. I took a look and couldn't  see it.

Are you looking for a mathematical formula?

9,25 comes from the average of the half, so 37/4, right?  You get to -2.7% by finding the percentage of payouts/odds*100.  Since your payout is 36, and your odds are 37/4 (9.25), it goes: (36/4-9.5)/9.25*100=2.7%

This is what you mean?

[Firefox]

Brilliant. Thanks for the information.

[nottophammer]

See what forum is all about.

[Mako]

See what forum is all about.

Ben and Shaun = Excel black belts. 

[Bigbroben]

Ben and Shaun = Excel black belts. 

Who is Shaun?