I noticed the original post was missing and so nobody would really have the foggiest what it was all about.
Here is the 'wayback' link so anybody can read what was originally said in the opening post.
Here is the missing first post of the thread...
This analysis is based on a single-zero wheel flat betting on the even chances.
Suppose you were able to reduce the length of your losing runs - how would your edge vary depending on the longest losing run? To put it another way, what should the length of the longest losing run be to ensure that you would make profit flat-betting?
The analysis assumes that there is no limit to the length of the winning runs. I wrote a program which generated even-chance outcomes and varied the length of the longest losing run from 10 to 1, and for each value I calculated the player's edge.
Here's the code:
procedure get_advantage(maxL: integer);
n = 100000000;
i, Lrun, w, l: longint;
w:= 0; l:= 0;
for i:= 1 to n do begin
if random(36) > 18 then begin
if Lrun <= maxL then
Pw:= w/(w + l);
write('Max losing run = ', maxL);
writeln(', HA = ', (Pw*100 - 50):4:3)
for i:= 1 to 10 do
and here are the results:
Max losing run = 1, PA = 15.451
Max losing run = 2, PA = 5.357
Max losing run = 3, PA = 1.194
Max losing run = 4, PA = -0.764
Max losing run = 5, PA = -1.723
Max losing run = 6, PA = -2.229
Max losing run = 7, PA = -2.493
Max losing run = 8, PA = -2.627
Max losing run = 9, PA = -2.704
Max losing run = 10, PA = -2.736
For losing runs of length 10 (or more), the expected PA (player advantage) of approximately -2.7% applies. As you shorten the longest losing run, the PA increases, but it's not until you get to a max losing run of 3 that it becomes positive!
So, if you can find a way to get your maximum losing run down to 3, you will have an advantage of about 1.2%. Alternatively, you could try to recover all losing runs above 3 by some sort of progression (good luck with that).
It's quite surprising how many losses you need to eliminate in order to get an advantage.