Quote from: Bayes on November 29, 2012, 05:42:08 PM
Hi KR,

Maybe, but that's down to variance. On average, because you need to win once in every 8 games just to break even, it means you don't make enough in the winning runs to offset the losses. But I don't want to stress this too much because it applies to every system.

Winning runs

Rather than posting tedious calculations, I'm just going to present the results in a table. The left-hand side shows the length of the winning run going up in steps of 5 after the first 4 and the right-hand side tells you what the chance is of seeing it. The first 4 results are given in % form and the remainder are in "1 in X" form.

Winning    Chance
  Run

   2          75%
   3          64%
   4          59%
   5          48%
  10         1 in 4.3
  15         1 in 8.9
  20         1 in 18.4
  25         1 in 38.0
  30         1 in 78.6
  35         1 in 162
  40         1 in 337
  45         1 in 697
  50         1 in 1,443
  55         1 in 2,986
  60         1 in 6,181
  65         1 in 12,794
  70         1 in 26,481
  75         1 in 54,809
  80         1 in 113,443
  85         1 in 234,803
  90         1 in 485,991
  95         1 in 1,005,895
  100       1 in 2,081,980

The Pilot had 180 winning run and John had few 100+  Bayes can you calculate the odds of winning 70 times in a row on any step of progression betting on 2 dozens?  John in his run with FIVE when he went 1000/0 (about the same odds like 100/0 for PB) won 70+ consecutive step 4 double dozen bets each time after losing 3 first steps. In my calculations its like 1M multiplied by 1M.