The Binomial Parachute

[Let Me Win]

I will write a bit about the binomial  parachute and see if there's any interest....

Think of playing a session of roullete as jumping out of a plane at great hight.

If your parachute is good you will survive but if it isn't you will crash to the ground and die.

In roulette if your parachute is good you will win something if it's not then you will lose your bankroll (death)

Here is the full Binomial Parachute and staking plan in all its glory.


EC 1 +1

Dozen 1 +1
Dozen 1 +0

Line 1 +3
Line 1 +2
Line 1 +1
Line 1 +0

Corner 1 +1
Corner 1 -1
Corner 2 +7
Corner 2 +5
Corner 2 +3
Corner 2 +1

Street  2 +3
Street  2 +1
Street  3 +9
Street  3 +6
Street  3 +3
Street  3 +0   
Street  4 +7
Street  4 +3

Split 4 +19
Split 4 +15
Split 4 +11
Split 4 +9
Split 4 +5
Split 4 +1   
Split 5 +11
Split 5 +6
Split 5 +1   
Split 6 +11
Split 6 +5
Split  6 -1
 
Single 3 +8
Single 3 +5
Single 3 +2
Single  3 +1   
Single  4 +31
Single 4 +27
Single 4 +23
Single 4 +19
Single  4 +15
Single  4 +11
Single 4 +7 
Single 4 +3
Single 4 -1   
Single 5 +31
Single 5 +26
Single 5 +21
Single 5 +16
Single 5 +11
Single 5 +6
Single 5 +1   
Single 6 +21 
Single  6 +15
Single 6 +9
Single 6 +3
Single 6  -2

The binomial probability dictate a 50% chance for each location to hit once.

Which is the same playing against seven reds or seven blacks for example.

The secret is how you determine the final selection process....

Let me tell you though it has nothing to do with triggers of any kind!

[Roulettebeater]

I will write a bit about the binomial  parachute and see if there's any interest....

Think of playing a session of roullete as jumping out of a plane at great hight.

If your parachute is good you will survive but if it isn't you will crash to the ground and die.

In roulette if your parachute is good you will win something if it's not then you will lose your bankroll (death)

Here is the full Binomial Parachute and staking plan in all its glory.


EC 1 +1

Dozen 1 +1
Dozen 1 +0

Line 1 +3
Line 1 +2
Line 1 +1
Line 1 +0

Corner 1 +1
Corner 1 -1
Corner 2 +7
Corner 2 +5
Corner 2 +3
Corner 2 +1

Street  2 +3
Street  2 +1
Street  3 +9
Street  3 +6
Street  3 +3
Street  3 +0   
Street  4 +7
Street  4 +3

Split 4 +19
Split 4 +15
Split 4 +11
Split 4 +9
Split 4 +5
Split 4 +1   
Split 5 +11
Split 5 +6
Split 5 +1   
Split 6 +11
Split 6 +5
Split  6 -1
 
Single 3 +8
Single 3 +5
Single 3 +2
Single  3 +1   
Single  4 +31
Single 4 +27
Single 4 +23
Single 4 +19
Single  4 +15
Single  4 +11
Single 4 +7 
Single 4 +3
Single 4 -1   
Single 5 +31
Single 5 +26
Single 5 +21
Single 5 +16
Single 5 +11
Single 5 +6
Single 5 +1   
Single 6 +21 
Single  6 +15
Single 6 +9
Single 6 +3
Single 6  -2

The binomial probability dictate a 50% chance for each location to hit once.

Which is the same playing against seven reds or seven blacks for example.

The secret is how you determine the final selection process....

Let me tell you though it has nothing to do with triggers of any kind!

I like it! I like the idea
But I won’t play it as you suggest.

I will tweak it as follows:

I will create a small RNG and get numbers randomly.

I will use then these random numbers in betting, i won’t use streets, corners, line etc, but only straight numbers

[Mako]

This is a good post for two reasons (more than two of course, but these are the big 2 for me personally).

First, because parachutes absolutely provide the exact sort of safety net that Let Me Win is describing. 

Not safety from impact with the ground (BR wipeout) as winning is always about bet selection of course. If we don't have an accurate bet selection, no money management can provide a positive result over time. But a parachute MM will provide some safety from a rapid descent that eviscerates a bankroll almost immediately (ala margingale, etc).

In testing parachutes of various types it's easy to see where the benefit is and why people focus on them.

Second, this is a good post because binomial distribution is also something that plays out as the concept is described if you do large scale testing of Roulette outcomes.

Many have talked about why binomial distribution is an area of focus in the game, Nimo comes to mind as the most recent to me, others on other boards as well.  It's something that happens as its described, meaning in my mind it's "real" so to speak and not hypothesis as say the law of the third is (I personally agree with the concept of LOTT, obvously as anyone does who does thousands of 37-spin cycles and sees how they play out...I'm just using LOTT here as an example of something that isn't proven...while binomial distribution as a concept, is).

Good stuff Law, like it. 

[Normy2000]

I will create a small RNG and get numbers randomly

Here is a RNG generator for roulette and baccarat...

[sturrock]

Binomial Distribution: Formula, What it is and How to use it
Contents:

What is a Binomial Distribution?
The Bernoulli Distribution
The Binomial Distribution Formula
Worked Examples
What is a Binomial Distribution?
A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times. The binomial is a type of distribution that has two possible outcomes (the prefix “bi” means two, or twice). For example, a coin toss has only two possible outcomes: heads or tails and taking a test could have two possible outcomes: pass or fail.
what is a binomial distribution
A Binomial Distribution shows either (S)uccess or (F)ailure.




The first variable in the binomial formula, n, stands for the number of times the experiment runs. The second variable, p, represents the probability of one specific outcome. For example, let’s suppose you wanted to know the probability of getting a 1 on a die roll. if you were to roll a die 20 times, the probability of rolling a one on any throw is 1/6. Roll twenty times and you have a binomial distribution of (n=20, p=1/6). SUCCESS would be “roll a one” and FAILURE would be “roll anything else.” If the outcome in question was the probability of the die landing on an even number, the binomial distribution would then become (n=20, p=1/2). That’s because your probability of throwing an even number is one half.
Criteria
Binomial distributions must also meet the following three criteria:

The number of observations or trials is fixed. In other words, you can only figure out the probability of something happening if you do it a certain number of times. This is common senseâ€"if you toss a coin once, your probability of getting a tails is 50%. If you toss a coin a 20 times, your probability of getting a tails is very, very close to 100%.
Each observation or trial is independent. In other words, none of your trials have an effect on the probability of the next trial.
The probability of success (tails, heads, fail or pass) is exactly the same from one trial to another.

Once you know that your distribution is binomial, you can apply the binomial distribution formula to calculate the probability.


What is a Binomial Distribution? The Bernoulli Distribution.
The binomial distribution is closely related to the Bernoulli distribution. According to Washington State University, “If each Bernoulli trial is independent, then the number of successes in Bernoulli trails has a binomial Distribution. On the other hand, the Bernoulli distribution is the Binomial distribution with n=1.”

A Bernouilli distribution is a set of Bernouilli trials. Each Bernouilli trial has one possible outcome, chosen from S, success, or F, failure. In each trial, the probability of success, P(S)=p, is the same. The probability of failure is just 1 minus the probability of success: P(F) = 1-p. (Remember that “1” is the total probability of an event occurring…probability is always between zero and 1). Finally, all Bernouilli trials are independent from each other and the probability of success doesn’t change from trial to trial, even if you have information about the other trials’ outcomes.

What is a Binomial Distribution? Real Life Examples
Many instances of binomial distributions can be found in real life. For example, if a new drug is introduced to cure a disease, it either cures the disease (it’s successful) or it doesn’t cure the disease (it’s a failure). If you purchase a lottery ticket, you’re either going to win money, or you aren’t. Basically, anything you can think of that can only be a success or a failure can be represented by a binomial distribution.

The Binomial Distribution Formula
Binomial Distribution formula
A Binomial Distribution shows either (S)uccess or (F)ailure.



The binomial distribution formula is:

b(x; n, P) = nCx * Px * (1 â€" P)n â€" x

Where:
b = binomial probability
x = total number of “successes” (pass or fail, heads or tails etc.)
P = probability of a success on an individual trial
n = number of trials

Note: The binomial distribution formula can also be written in a slightly different way, because nCx = n!/x!(n-x)! (this binomial distribution formula uses factorials (What is a factorial?). “q” in this formula is just the probability of failure (subtract your probability of success from 1).
binomialprobabilityformula

Using the First Binomial Distribution Formula
The binomial distribution formula can calculate the probability of success for binomial distributions. Often you’ll be told to “plug in” the numbers to the formula and calculate. This is easy to say, but not so easy to doâ€"unless you are very careful with order of operations, you won’t get the right answer. If you have a Ti-83 or Ti-89, the calculator can do much of the work for you. If not, here’s how to break down the problem into simple steps so you get the answer rightâ€"every time.
Example 1
Q. A coin is tossed 10 times. What is the probability of getting exactly 6 heads?

I’m going to use this formula: b(x; n, P) â€" nCx * Px * (1 â€" P)n â€" x
The number of trials (n) is 10
The odds of success (“tossing a heads”) is 0.5 (So 1-p = 0.5)
x = 6



P(x=6) = 10C6 * 0.5^6 * 0.5^4 = 210 * 0.015625 * 0.0625 = 0.205078125

Tip: You can use the combinations calculator to figure out the value for nCx.

How to Work a Binomial Distribution Formula: Example 2
binomialprobabilityformula
80% of people who purchase pet insurance are women.  If 9 pet insurance owners are randomly selected, find the probability that exactly 6 are women.

Step 1: Identify ‘n’ from the problem. Using our sample question, n (the number of randomly selected items) is 9.

Step 2: Identify ‘X’ from the problem. X (the number you are asked to find the probability for) is 6.



Step 3: Work the first part of the formula. The first part of the formula is

n! / (n â€" X)!  X!

Substitute your variables:

9! / ((9 â€" 6)! × 6!)

Which equals 84. Set this number aside for a moment.

Step 4: Find p and q. p is the probability of success and q is the probability of failure. We are given p = 80%, or .8. So the probability of failure is 1 â€" .8 = .2 (20%).

Step 5: Work the second part of the formula.

pX
= .86
= .262144

Set this number aside for a moment.

Step 6: Work the third part of the formula.

q(n â€" X)
= .2(9-6)
= .23
= .008

Step 7: Multiply your answer from step 3, 5, and 6 together.
84  × .262144 × .008 = 0.176.

Example 3
60% of people who purchase sports cars are men.  If 10 sports car owners are randomly selected, find the probability that exactly 7 are men.

Step 1:: Identify ‘n’ and ‘X’ from the problem. Using our sample question, n (the number of randomly selected itemsâ€"in this case, sports car owners are randomly selected) is 10,  and  X (the number you are asked to “find the probability” for) is 7.



Step 2: Figure out the first part of the formula, which is:

n! / (n â€" X)!  X!

Substituting the variables:

10! / ((10 â€" 7)! × 7!)

Which equals 120. Set this number aside for a moment.

Step 3: Find “p” the probability of success and “q” the probability of failure. We are given p = 60%, or .6. therefore, the probability of failure is 1 â€" .6 = .4 (40%).

Step 4: Work the next part of the formula.

pX
= .67
= .0.0279936


Set this number aside while you work the third part of the formula.

Step 5: Work the third part of the formula.

q(.4 â€" 7)
= .4(10-7)
= .43
= .0.064

Step 6: Multiply the three answers from steps 2, 4 and 5 together.
120  × 0.0279936 × 0.064 = 0.215.

That’s it!

Reference:
WSU. Retrieved Feb 15, 2016 from: :.stat.washington.edu/peter/341/Hypergeometric%20and%20binomial.pdf

[Let Me Win]

An excellent explanation of why devising a roullete system using binomial methods is clearly the sensible and logical way to proceed.

Why wouldn't anyone like it?
It's a wonderful moment of enlightenment when you finally realise that every spin is always exactly the same and all bet selections are always exactly the same.
Never again do you need to talk about nonsense such as virtual spins or triggers of any kind.

You will know that anyone who continues to insist otherwise is truly a person with an intellectual ability of a chimpanzee.

Thats a good thing because it saves hours of wasting time.
One could always go to any casino at anytime and sit down and start playing immediately.

[Let Me Win]

The way I'm playing with great success with my own modifications which anyone could test to independently duplicate my results is as follows.

I play a maximum of 36 spins.
I don't track or any such nonsense I just sit down and play any time any place.

I mainly play online Immersive Roulette if I'm at home because I like the graphics, trust the results and there is enough time for me to place my bets between spins without being unduly rushed.

You could use any method of bet selection you wish as it won't make any difference but what I do is bet so that if number one hits on spin one I win if number two hits on spin two I win if number three hits on spin 3 I win etc etc all the way to spin 36.

A win at any spin will give me an overall profit and I would start again at spin one if I want to continue playing at that time.

I bet the first 4 spins with a standard 1 2 4 8 martingale.

Red
Black
Odd
Even

That's what I do so if number 4 wins on spin 4 I win because it is an even number. Got it?  Make sense?

So I keep winning 1 unit until I eventually lose 15 units or four spins in a row.

Then I play as above with adjusted stakes.
So for example I would stake 8 units on the 1st dozen for spin 5

Etc so I bet 4 times at even money 2 times at dozen 4 times at line 6 times at corner 8 times at street and 12 times at split for a total of 36 spins all together.

I never bet a number straight up I always have a minimum of two numbers on my side.

For me to lose I have to lose the equivalent of nine even money bets in a row which will happen approximately every 655 sessions.

[Let Me Win]

The exact fixed bet I play every single session is as follows.
Remember to stake enough each spin to show an overall profit of at least one unit if you win.

Spin 1 RED
Spin 2 BLACK
Spin 3 ODD
Spin 4 EVEN
Spin 5 First Dozen
Spin 6 Top Row
Spin 7 Natural line two (7-12)
Spin 8 Natural line two
Spin 9 Natural line two
Spin 10 Natural line two
Spin 11 Corner 11/13/10/14
Spin 12 Corner 12/14/11/15
Spin 13 Corner 13/17/14/16
Spin 14 Corner 14/18/15/17
Spin 15 Corner 15/17/14/18
Spin 16 Corner 16/20/17/19
Spin 17 Street 17,25,34
Spin 18 Street 18,22,29
Spin 19 Street 19,15,4
Spin 20 Street 20,1,14
Spin 21 Street 21,4,2
Spin 22 Street 22,9,18
Spin 23 Street 23,8,10
Spin 24 Street 24,5,16
Spin 25 Split 25 and 0
Spin 26 Split 26 and 0
Spin 27 Split 27 and 0
Spin 28 Split 28 and 0
Spin 29 Split 29 and 0
Spin 30 Split 30 and 0
Spin 31 Split 31 and 0
Spin 32 Split 32 and 0
Spin 33 Split 33 and 0
Spin 34 Split 34 and 0
Spin 35 Split 35 and 0
Spin 36 Split 36 and 0

That's how I do it and if you copy the above starting at spin one again each time you win it will be many months before you lose.

Remember jumping out of planes can cost you your life and here your life will cost you around 750 units.

[MartyV]

Interesting approach, Let Me Win. I love parachute systems so this definitely caught my eye.

I'll be testing this on my end as I find the time. Could you please inform me if I have the following correct, based on your calculations :

- One expects to lose approximately every 655 sessions.
- A complete bust is 750 units.
- Hence, if one plays for one unit per session, one expects to lose 750 units for every 654 units gained, for a net loss of 96 units per 655 sessions.

Do I have this right? If so, do you not expect this approach to predictably yield a net loss ? Is there anything here I am misunderstanding?

Thanks for posting this, Let Me Win. And thanks for your time, if you decide to answer me. Cheers!

[ego]

That is my Parachute and I am known as Sputnik.
I have some selection ideas and my goal is to survive one whole year playing +1 unit each day.

This is theory and you use the information at your own risk ...

Mr. J inspired me into this field of the selection process.
Here is a selection that won 100 sessions using a Parachute Method.

1) One single number that hit within 25 attempts are equal an even money bet.

Hot & Cold

Assume you have two reds in a row or two blacks in a row.
Then for the next two bets, you have 3 in 4 to hit a winning bet.

For example:

RR / RR Win
RR / RB Win
RR / BR Win
RR / BB Lose

That is the Hot selection where one color repeat.

Now reverse.
Assume you have RR and get BB then you lose both bets.
Then you have 3 in 4 ways to get one opposite color.
A sleeping color or Cold selection.

BB / RR Win
BB / RB Win
BB / BR Win
BB / BB Lose

Now numbers and the Parachute Method.

Assume one number hit twice within 50 attempts.
Then that is the same thing getting two reds.
So for the next 50 attempts, you play that this number will repeat once.

Assume you lose, then you have two losing bets during 50 attempts, also equal two blacks.
Now you play 50 attempts that you will get at least one repeat of that particular number.

Hot and Cold selection using one single number equal an even money bet.

Now you bet the Parachute Method until you win once and then start over the tracking and charting process.
So even if the single number would fail to show during 100 attempts, so can the Parachute Method win.

For example, assume number six show twice within 50 attempts, then you Parachute that location on the table.

This is 100 session won.
6 times the single number fail to show but the Parachute method won.

1. 21
2. 7
3. 34
4. 30
5. 7
6. 26
7. 36
8. 32
9. 18
10. 6
11. 6
12. 34
13. 49
14. 43
15. 13
16. 8
17. 11
18. 42
19. 6
20. 50/8
21. 33
22. 5
23. 34
24. 7
25. 14
26. 4
27. 18
28. 24
29. 37
30. 33
31. 36
32. 50/21
33. 50/18
34. 50/50 target number fail but parachute win
35. 34
36. 50/50 target number fail but parachute win
37. 50/6
38. 3
39. 50/50 target number fail but parachute win
40. 1
41. 9
42. 9
43. 3
44. 35
45. 50/50 target number fail but parachute win
46. 40
47. 50/48
48. 50/15
49. 37
50. 21
51. 50/36
52. 50/50 target number fail but parachute win
53. 50/17
54. 38
55. 48
56. 6
57. 50/36
58. 50/49
59. 7
60. 14
61. 21
62. 50/44
63. 7
64. 19
65. 21
66. 50/50 target number fail but parachute win
67. 34
68. 25
69. 7
70. 9
71. 29
72. 11
73. 26
74. 9
75. 50/45
76. 2
77. 15
78. 14
79. 23
80. 50/7
81. 50
82. 12
83. 42
84. 15
85. 18
86. 27
87. 50/26
88. 6
89. 50/20
90. 13
91. 50/41
92. 50/19
93. 11
94. 6
95. 9
96. 3
97. 2
98. 39
99. 50/32
100. 50/12

1. EC

2. Dozen
3. Dozen

4. Line
5. Line
6. Line
7. Line

8. Corner
9. Corner
10. Corner
11. Corner
12. Corner
13. Corner

14. Street
15. Street
16. Street
17. Street
18. Street
19. Street
20. Street
21. Street

22. Split
23. Split
24. Split
25. Split
26. Split
27. Split
28. Split
29. Split
30. Split
31. Split
32. Split
33. Split

34. Single
35. Single
36. Single
37. Single
38. Single
39. Single
40. Single
41. Single
42. Single
43. Single
44. Single
45. Single
46. Single
47. Single
48. Single
49. Single
50. Single
51. Single
52. Single
53. Single
54. Single
55. Single
56. Single
57. Single

58. Single
59. Single
60. Single
61. Single
62. Single
63. Single
64. Single
65. Single
66. Single
67. Single
68. Single
69. Single
70. Single
71. Single
72. Single
73. Single
74. Single
75. Single
76. Single
77. Single
78. Single
79. Single
80. Single
81. Single
82. Single
83. Single

84. Single
85. Single
86. Single
87. Single
88. Single
89. Single
90. Single
91. Single
92. Single
93. Single
94. Single
95. Single
96. Single
97. Single
98. Single
99. Single
100. Single

[ego]

This means that if you can lose the selection six times and still win and a total of 100 sessions, then is close to one year.

There are 365 days for one year and the interest banks give you equal zero.
365 Euro playing 1 Euro minimum, pretty good interest.
3650 Euro playing 10 Euro minimum, closer to the stock market return for the average investor.

What I would like to do is to calculate a higher base bet and then when winning half the amount regress and using half of the original unit size.
That way you only risk your money for a one-half year and then operate with casino money for the next half year.

Problems with this idea are that the cost is very expensive for the whole parachute
One could reduce the cost and win half the amount the last steps of the parachute.


Cheers

[ego]

I will announce this as Sputnik's HG and will report my success.
Did more testing and come up with tweaks that improve the method even more.

Cheers

[ego]

1. 27
2. 8
3. 7
4. 50/17
5. 50/24
6. 18
7. 8
8. 9
9. 29
10. 21
11. 19
12. 3
13. 8
14. 7
15. 47
16. 50/42
17. 40
18. 47
19. 12
20. 50/40
21. 2
22. 50/39
23. 15
24. 35
25. 20
26. 50/32
27. 28
28. 50/7
29. 12
30. 20
31. 20
32. 44
33. 50/32
34. 42
35. 3
36. 50/32
37. 21
38. 44
39. 4
40. 50/22
41. 25
42. 9
43. 7
44. 2
45. 13
46. 29
47. 14
48. 50/5
49. 39
50. 35
51. 20
52. 3
53. 16
54. 7
55. 44
56. 50/46
57. 7
58. 50/50 target number fail, but parachute win
59. 12
60. 50/38
61. 9
62. 11
63. 7
64. 50/14
65. 16
66. 15
67. 42
68. 28
69. 44
70. 50/2
71. 7
72. 50/15
73. 3
74. 44
75. 50/33
76. 50/50 target number fail, but parachute win
77. 50/7
78. 17
79. 8
80. 10
81. 50/50 target number fail, but parachute win
82. 9
83. 6
84. 1
85. 14
86. 50/4
87. 41
88. 11
89. 14
90. 30
91. 21
92. 11
93. 2
94. 20
95. 8
96. 41
97. 21
98. 4
99. 3
100. 4

[Mako]

Sputnik, nice posts, interesting parachute variant...will definitely test when time permits.

Keep up the good work. 

[ego]

Thanks, Mako

I will test another selection process and see if I can get similar results.
Wait for 50 spins/trials and pick one sleeping number with no hits.
Target the target number for the next 100 spins/trails.

Random parameter.
Pick Low or High and alternate after each new session with a new fresh number.

Maybe just pick any number from the beginning.

Cheers