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.

Hi RH,

Not sure what you mean. I never really understood how FIVE worked. So basically, is this the odds of not losing a 4 step progression on the double dozens?

BTW, a winning run of 100 in PB is equivalent to an EC streak of 21, so it IS possible, but of course not very common.

As for a winning run of 1000, forget it. 

Quote from: Bayes on November 30, 2012, 08:27:04 AM
Hi RH,

Not sure what you mean. I never really understood how FIVE worked. So basically, is this the odds of not losing a 4 step progression on the double dozens?

Its like having no single loss in 70 tries betting on 2 dozens. It happened on 4th step of 1,3,9,27 progression in FIVE. Btw winning streak of 1000+ in FIVE carries about the same odds like 100+ in PB.  For FIVE try to go LLLW  70 times in a row 
L represents loss in each step of 1,,3,9,27 progression. And the other guy reported winning 180 in a row with PB. Few thousands
millions in 1.

Quote from: Robeenhuut on November 30, 2012, 08:56:17 AM
try to go LLLW  70 times in a row 

Ok, so a loss is 13/37 and a win is 24/37. LLLW = [(13/37)³ × 24/37]⁷⁰ = 1 in 3.58 × 10¹⁰⁸



Edit: sorry, made a mistake - now corrected. 

Just in case anyone's wondering, this is a very big number. It would be like winning the UK lotto (odds 14 million to 1 against) about 14 times in a row.

Quote from: Bayes on November 30, 2012, 09:07:12 AM
Ok, so a loss is 13/37 and a win is 24/37. LLLW = [(13/37)³ × 24/37]⁷⁰ = 1 in 3.58 × 10¹⁰⁸

RESULTS UPDATE FOR *****FIVE***** FOR  THE 14/10/2012

TOTAL GAMES PLAYED 1,120

TOTAL GAMES WON 1,119

TOTAL GAMES LOST 1

STRIKERATE 1,119/1

BALANCE 1,040 POINTS PLUS

STEP 1 WINS=442
STEP 2 WINS=412
STEP 3 WINS=191
STEP 4 WINS=74----LOSSES=1

These are the stats for 1,3,9,27 double dozen progression for FIVE.  There were 70 wins in row on a last step....
Nobody sees anything unusual? 

Wow, I would love to get numbers like that. I never tested a four step Marti for doubled dozens.  I wonder how often a 1 loss in a thousand occurs?

Actually RH, how do you know it's 70 in a row? I don't see where JL is claiming that, not in these stats anyway. Of course, 1 loss in 1,120 games looks pretty amazing, but you'd have to know where the loss came in order to know how long the winning run was.

Quote from: Bayes on November 30, 2012, 09:49:24 AM
Actually RH, how do you know it's 70 in a row? I don't see where JL is claiming that, not in these stats anyway. Of course, 1 loss in 1,120 games looks pretty amazing, but you'd have to know where the loss came in order to know how long the winning run was.

He posted it before that he went 70/0 on a last step.  But even 74/1 does not happen. Odds of 1000+ winning run with FIVE overall are peanuts compared with what happened on a last step. Try to go 4th step in 1,3,9,27 and win 70 in a row. 

Peanuts hugh? I have a very new interest. I did extensive experimentation with a three step.

This is moving away from PB into FIVE...new thread perhaps gents?

Moderator's note: This topic has been split. Please confine comments on FIVE to this thread.

It does seem odd that almost every win on the 4th bet was uninterrupted by any wins on the 1st, 2nd, and 3rd bets.

However, the calculations all assume, of course, that the system has no advantage, and since I never got my head around FIVE, I can't do any tests on it, not to mention that apparently it's impossible to simulate hit & run. 

But it's right to be sceptical about these stats until JL shows us that they're plausible. One thing about them is that you WOULD expect to get more wins on the first bet and proportionately fewer on the 2nd, 3rd and 4th bets. The theoretical breakdown over 1120 bets is this:

1st bet -  727
2nd bet - 255
3rd bet -  90
4th bet -  31

JL's stats are pretty out of whack compared to these, but at least they follow the same pattern in that there are increasingly fewer wins on successive bets.

Quote from: Robeenhuut on November 30, 2012, 09:55:25 AM
He posted it before that he went 70/0 on a last step.  But even 74/1 does not happen. Odds of 1000+ winning run with FIVE overall are peanuts compared with what happened on a last step. Try to go 4th step in 1,3,9,27 and win 70 in a row. 

I don't think that's what he means by 70/0 on the last step either. It's not 'in a row'.
In amongst those 70 wins on Step 4, there will be wins on other steps.


I think JL is saying that when he has been forced to go to Step 4, he's won 70 in a row of them.

Quote from: monaco on November 30, 2012, 11:19:28 AM

I think JL is saying that when he has been forced to go to Step 4, he's won 70 in a row of them.

I think that's what he meant. 

Quote from: Bayes on November 30, 2012, 10:46:15 AM
It does seem odd that almost every win on the 4th bet was uninterrupted by any wins on the 1st, 2nd, and 3rd bets.

However, the calculations all assume, of course, that the system has no advantage, and since I never got my head around FIVE, I can't do any tests on it, not to mention that apparently it's impossible to simulate hit & run. 

But it's right to be sceptical about these stats until JL shows us that they're plausible. One thing about them is that you WOULD expect to get more wins on the first bet and proportionately fewer on the 2nd, 3rd and 4th bets. The theoretical breakdown over 1120 bets is this:

1st bet -  727
2nd bet - 255
3rd bet -  90
4th bet -  31

JL's stats are pretty out of whack compared to these, but at least they follow the same pattern in that there are increasingly fewer wins on successive bets.

Bayes will Matt ever let this go? Yes I survived 70 bets on the 4th step, many on your RNG. But they werent consecutive. They were interrupted by wins on the other three steps. Sometimes days apart.