The Pendulum

[BlueAngel]

Imagine a parachute which begins from EC betting and reduces betting numbers as it wins, while increasing total of betting numbers as it loses, this is a range of 11 different betting levels with the initial EC of Low or High standing in the middle.

35 numbers betting <--- this is the ultimate level from the negative side, you should arrive here after 5 consecutive losses
17 splits betting <--- this is the 4th level from the negative side, you should arrive here after 4 consecutive losses
11 streets betting <--- this is the 3rd level from the negative side, you should arrive here after 3 consecutive losses
5 six-lines betting <--- this is the 2nd level from the negative side, you should arrive here after 2 consecutive losses
2 dozens betting <--- this is the 1st level from the negative side, you should arrive here after 1 loss

1 EC betting <--- this is the neutral zone, the middle ground so to speak, this is where you start your journey and pass through thereafter for countless times

1 dozen betting <--- this is the 1st level from the positive side, you should arrive here after 1 win
1 six-line betting <--- this is the 2nd level from the positive side, you should arrive here after 2 successive wins
1 street betting <--- this is the 3rd level from the positive side, you should arrive here after 3 successive wins
1 split betting <--- this is the 4th level from the positive side, you should arrive here after 4 successive wins
1 number betting <--- this is the ultimate level from the positive side, you should arrive here after 5 successive wins

By completing either side you finish, when it's very indecisive set your own time and/or money limit(s).
How many units for total BR? 35+17+11+5+2+1 = 71 units win or lose.

[winkel]

Some questions:

1. How do you select the target number?

2. If your target ist "17" which split do you bet?

3. Do you have the probababilities for
-- winning 6 times in a row / or wininng the target in a 6th spin?
-- losing 6 times in a row?

4. Did you notice, that "parachute" is a losing game?

[BlueAngel]

Some questions:

1. How do you select the target number?

By following the last spun number on every betting type.

2. If your target ist "17" which split do you bet?

I prefer the horizontal splits, thus with 14 or 20.



3. Do you have the probababilities for
-- winning 6 times in a row / or wininng the target in a 6th spin?
-- losing 6 times in a row?



If you play something like 6 x 37 = 222 bets 6 days a week you would reach the limits 1 or 2 times, in other words, once per full moon.

4. Did you notice, that "parachute" is a losing game?

Parachute has many variations, so we cannot say that every Parachute variation is the same and a losing proposition. 

I have tried a lot of different Parachute versions and it's relatively conservative betting, I consider the utilization of various betting types under the "Parachute" umbrella makes variance smoother.

[Smith]

I figured out the probabilities.



So you're over 5 times more likely to get to the 6th losing step than the 6th winning step. Of course it's expected that you're more likely to lose than win, but I didn't think the difference would be as high as it is.

[BlueAngel]

I figured out the probabilities.



So you're over 5 times more likely to get to the 6th losing step than the 6th winning step. Of course it's expected that you're more likely to lose than win, but I didn't think the difference would be as high as it is.

That's because of the 0, but your calculations disregard the selection, by quantifying only, this is to say, considering just how many numbers we bet or not then every betting method/system is already doomed before it even starts!
The selection of what to bet could only make the difference between a win and a loss, the "how many" units, numbers...etc. is of secondary importance, quantification matters but not as much as the selection!
Besides probability theory as a whole never meant to be a certainty, but merely an indication, something to get a rough idea of what someone could expect, there are other significant factors at play which probability theory doesn't account for...

[BlueAngel]

Some questions:

1. How do you select the target number?

By following the last spun number on every betting type.

2. If your target ist "17" which split do you bet?

I prefer the horizontal splits, thus with 14 or 20.



3. Do you have the probababilities for
-- winning 6 times in a row / or wininng the target in a 6th spin?
-- losing 6 times in a row?



If you play something like 6 x 37 = 222 bets 6 days a week you would reach the limits 1 or 2 times within a month, in other words, once or twice per full moon.

4. Did you notice, that "parachute" is a losing game?

Parachute has many variations, so we cannot say that every Parachute variation is the same and a losing proposition.

I have tried a lot of different Parachute versions and it's relatively conservative betting, I consider the utilization of various betting types under the "Parachute" umbrella makes variance smoother.

[winkel]

Another question:

The losing betsequence:

do you bet 35 straight numbers at once
or
do you bet 35 times 1 number?

6 steps or 71 steps?

[BlueAngel]

Another question:

The losing betsequence:

do you bet 35 straight numbers at once
or
do you bet 35 times 1 number?

6 steps or 71 steps?

The way it is meant is 35 numbers for 1 time, since we would pick the 35 last spun numbers the other 2 are long sleepers and most likely they are not going to show up in the specific spin we are going to bet against them, for example let's say that those 2 were missing for 100 spins and on the 101st we bet against them, do you think that suddenly, after long missing they are going to hit just because we bet?!
It's 6 steps per side/direction, all together 11 steps and 71 units.

[Smith]

That's because of the 0, but your calculations disregard the selection, by quantifying only, this is to say, considering just how many numbers we bet or not then every betting method/system is already doomed before it even starts!
The selection of what to bet could only make the difference between a win and a loss, the "how many" units, numbers...etc. is of secondary importance, quantification matters but not as much as the selection!

Yes and I'm not saying that the probabilities show the system is poor. Knowing the long run theoretical probabilities can be useful if only as a comparison. For if you don't know them you have nothing to compare empirical probabilities with and therefore won't know if they indicate an anomaly. Besides, I only posted them because Winkel asked. Testing is the final arbiter of the merit of a system.

[BlueAngel]

If we want to make each side exactly equal then all we have to do is to ignore the ultimate step of the positive side, the single number bet, in order to have 5 bets on the positive side:

1. EC
2. dozen
3. line
4. street
5. split

...and 5 bets/steps from the negative side:

1. two dozens
2. five lines
3. eleven streets
4. seventeen splits
5. thirty-five numbers

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Here is an alternative way to play "The Pendulum"

1 EC BETTING, AFTER:

1 LOSS ---> 2 DOZENS WITH 2 UNITS EACH ONCE, RETURN TO EC

2 LOSSES ---> 5 LINES WITH 5 UNITS EACH ONCE, RETURN TO EC

3 LOSSES ---> 11 STREETS WITH 11 UNITS EACH TWICE + 5 LINES WITH 5 UNITS EACH ONCE + 2 DOZENS WITH 2 UNITS EACH ONCE, RETURN TO EC

4 LOSSES ---> 17 SPLITS WITH 17 UNITS EACH FOR 7 TIMES + 11 STREETS WITH 11 UNITS EACH TWICE + 5 LINES WITH 5 UNITS EACH TWICE, RETURN TO EC

5 LOSSES ---> 35 NUMBERS WITH 35 UNITS EACH FOR 7 TIMES + 17 SPLITS WITH 17 UNITS EACH TWICE + 11 STREETS WITH 11 UNITS EACH ONCE, RETURN TO EC

1665 UNITS COST FOR 6 LOSSES IN A ROW (without ANY win)

OPTIONALLY USE THE FOLLOWING }

CONTINUES WITH THE 2 LOST NUMBERS + 1 MORE FOR 3 SUCCESSIVE WINS FROM ANY COMBINATION OF THOSE 3 NUMBERS, APPLYS PARTIAL PARLAYS AS DESCRIBED IN "NO LIMITS!"
5184 UNITS IN TOTAL WITH THE ADDITIONAL PHASE