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I have a question ...

Let's say I have a strategy, not for betting but for tracking.
Let's say I want to know the frequency one single number repeats after one hit.

Example nr 19
The number hits and we count how many trails it takes for number 19 to hit again.
Then we get a list with gaps (frequency) 20, 40, 14, 3, 35, 25, 58,

Now can I describe the method and get a module tracking the frequency and get an output file with the gaps of the frequency hits?

This way I can get the RIBOT to become a analyze bot with my own customize modules.

One more question, is the RIBOT click by each spin or can we load a spin file and get the process automated?

Cheers Patrik

Quote from: Patrik on Sep 11, 2022, 01:02 PMNow can I describe the method and get a module tracking the frequency and get an output file with the gaps of the frequency hits?

Hello dear Patrik, this explanation is enough.

A new gaps/frequency tracker module is on the way 👍

Quote from: Patrik on Sep 11, 2022, 01:02 PMOne more question, is the RIBOT click by each spin or can we load a spin file and get the process automated?

Both options are available:

- Manual input.
- File input.

Here is an usage video with automated file input:

Wow looks great, thanks for your effort, now I know you can do anything with your software.
Good to know in the future if I find a way worth coding.

Cheers

Hello dear @Patrik,

There should be an option to discard the first spins in the session, correct?

(To make the first gap value accurate, given the first/initial show may not reflect the correct previous gap. I.e. a long sleeper may come up at 10th spin, but it has a much longer previous gap. This way it counts strict gaps only)

Quote from: VLS on Nov 10, 2022, 02:47 AMThere should be an option to discard the first spins in the session, correct?

Yep. Coming as a configurable option 👍

Even if first spins may not be an accurate measure of the previous gap, it's still a gap that some fellows may want to track/use :D

Here is one example... The binomial probability is 25 attempts for one single number to show with 50% probability... There is as many singles as there is series ... So when I get my number to show it is a trigger to get a serie or a hit for the next 25 attempts... If not I have a single ... So I track hits and see if they become series ... If not you need to track 25 event windows after each other and we can start from spin one

I assumed understanding question correctly

Quote from: Patrik on Nov 23, 2022, 08:03 PMHere is one example... The binomial probability is 25 attempts for one single number to show with 50% probability...

Just to be clear, this means at least one pre-selected number, not exactly one. If it's exactly one the probability is only 35%.

(https://i.postimg.cc/9MJJ4bgG/probability.png)

Quote from: Patrik on Sep 11, 2022, 01:02 PMLet's say I want to know the frequency one single number repeats after one hit.

Example nr 19
The number hits and we count how many trails it takes for number 19 to hit again.

On average it will take 37 trials. The fact that it just hit won't make any difference.

Not sure you look at correct column.
Jules... Bayes calculated this...

Ec has one attempt
Dozen has two attempts
Line has four attempts
Corner has 6
Street has 8
Split has 12
Single has 25

Patrik, yes it's the correct column : P(X >= 1) means greater than or equal to 1. The same applies to the other locations you listed.

Median Joe can you calculate the AROUND 50% likelihood for one single number?

Many Thanks

Not sure what you mean? Do you mean the number of trials needed to get exactly one hit of a number such that the probability is 50%.

If so, you can't get it. The maximum probability is about 37% for exactly one hit and it occurs in a 37 spin cycle. That might seem confusing, but the probability is always higher for a range of values rather than just one.

But the "at least one" probability is fine. I only mentioned it because I thought some might have thought you were referring to exactly one hit.

I talk about the binomial probability that is less than 37 spin cycle to get one number to hit with 50% likelihood

If not then the whole list I posted should be wrong and is not.
Jules or Bayes was a math expert

24 or 25 spins shall give you 50% one hit probability for a single SU number.

@Median Joe do you know of any advanced (binomial) probability calculator, where I can stack multiple blocks of various payouts ..

eg.
a) 2Q 5x spins → DS 4x spins → Q 4 spins
b) or even better .. having a hit on an area, so 3 hits within 9x spins
   2Q first hit secured, thereof → DS 4x spins (2nd hit, if secured) → Q 4x spins (3rd hit)

Quote from: Patrik on Nov 26, 2022, 05:58 PMI talk about the binomial probability that is less than 37 spin cycle to get one number to hit with 50% likelihood

If not then the whole list I posted should be wrong and is not.
Jules or Bayes was a math expert

I didn't say the list was wrong, only that the probabilities are for at least one hit, not exactly one hit. I'm sure if you look at Bayes' original post about it he will say the same, as shown in the screenshot of the binomial calculation which I posted above.

Quote from: HardMan on Nov 26, 2022, 07:52 PM@Median Joe do you know of any advanced (binomial) probability calculator, where I can stack multiple blocks of various payouts ..

eg.
a) 2Q 5x spins → DS 4x spins → Q 4 spins
b) or even better .. having a hit on an area, so 3 hits within 9x spins
   2Q first hit secured, thereof → DS 4x spins (2nd hit, if secured) → Q 4x spins (3rd hit)

I don't think you need a binomial calculator for this. Are you talking about the probability of a sequence of bets like in a parachute?

@Median Joe yes, I've described the scenario(s) in previous post;

where the next & different block is activated or comes in place on the condition of a hit secured, in short a press done with multiple progression dimensions (vertical, horizontal, payouts) ..

.. but it should result in a probability of ≥2 or ≥3.


Also, ideally, since the stacking of blocks comes upon the first hit secured (widespread, multiple positions), as those come when they come .. I wouldn't like the formula to calculate the probability of the first hit, but assume it was secured on the first spin (of such sequence of blocks = the first block type ended on the first spin), then thereof adding the 2nd (≥2 hits) & potentially the 3rd block (≥3 hits).

Ok, so if I understand you correctly, you have a sequence of bets on different locations.

First, 5 bets on 2 Quads, followed by 4 spins on 1 DS, and finally spins on 1 Quad.

You say the outcome of the first spin is assumed to be a win. So no calculation is required for the 2 Quad bet?

Quote.. but it should result in a probability of ≥2 or ≥3.

Not sure what you mean here, because all probabilities are between 0 and 1. Or are you talking about the number of hits on the DS and single Quad respectively?

Either way, no special calculator is needed. You need to calculate the probability of the win(s) on the DS and Quad separately using a standard binomial calculator, then multiply the results together.

e.g. suppose you want the probability of at least one hit on the DS and also at least one hit on the Quad.

https://statisticshelper.com/binomial-probability-calculator/ 

Enter this into the four fields :

Number of trials = 4
probability = 6/37 = 0.1622
Number of successes = 1 (change according to need)
Type of probability = At least X successes (change according to need)

Then click Calculate. Rinse and repeat for the Quad calculate, then multiply the two probabilities together. That will give you the overall probability. The process is the same for any number of "stacked" probabilities.

Since we are calculating the probability of ≥3 hits .. within spins & the first being already secured ..

that first hit definitely impacts the final result, doesn't it?


As you suggested we are calculating the P = ≥1 x ≥1;

but let's put ourselves in the position of having had the first hit, &  at the same time wanting to know the probability of the 3-step press completing .. so =3 hits, or ≥3, whichever result correct; & where each further step is a block of 4-spins.
 

@Median Joe, confirm correctness ..


a)
having a hit, then playing 4x DS → then 4x Q
P(DS)=0.50212864 x P(Q)=0.3672026 = 0.18438296437,

or ≈18.5% chance that the press will conclude successfully


b)
playing a widespread of several positions amounting to 12 numbers per spin & having a hit on the very next spin .. then stacking as above with 4x-spin blocks for each DS→Q = intending to have 3 hits within 9 spins in total

P(DZ)=0.32432 x P(DS)xP(Q) = 0.059799083,

or 6% chance that the whole press of 3 ≈consecutive hits in a row completes within 9 spins (as 3 consecutive block types).

Now, if that is correct, adding a block after block (3 hits, 9-spins attempt) at ..

Quoteat eg. P(9-spin)≈ 6% or 0.6

.. the accumulated probability P(N), where n is the total number of such blocks, should raise up with each additional 9-spin block added ..

---

.. however, multiplying the probability of blocks

QuoteP(N)= P(n-(n-1)) x P(n-(n-2)) x .. x P(n)

simplified to

QuoteP(N)= P(1) x P(2) x .. x P(n)

& since Ps are equal even further to

QuoteP(N)= P(1)ⁿ

gets actually smaller & smaller, which would be the cumulative probability of each block completing (=3 hits) again & again in a row.

---

Now, again, what's the formula for one of them blocks completing, as a series of 9-spin blocks pass by?

Your calcs in reply #21 are correct, but

Quote.. the accumulated probability P(N), where n is the total number of such blocks, should raise up with each additional 9-spin block added ..

No, because multiplying a number less than 1 will always result in a smaller number; if you're requiring that you get wins in each section of course the probability will decrease as you add more sections. The way to find the probability of at least one win is to subtract from 1 the probability that all the bets lose.

To do this you can use the same probability calculator and put 0 successes into the 3rd field. But if you are betting on different locations it's easier to find the probability that each bet loses, multiply them all together and subtract the result from 1.

ie, probability that DZ bet loses  = 1 - 12/37 = 25/37
probability that a DS bet loses is 1 - 6/37 = 31/37, so the probability that all 4 bets lose is (31/37)⁴
probability that a Quad bet loses is 1 - 4/37 = 33/37, so the probability that all 4 bets lose is (33/37)⁴

Probability of at least one win = 1 - 25/37 × (31/37)⁴ × (33/37)⁴ = 79%

But for anything more specific, such as the chance that you get multiple wins in a particular section and all other sections lose, the calculations would be different, but always you are multiplying the probabilities together.

Let's get to the nuts & bolts

2Q→DS→Q
with the prerequisite to activate the 2nd block type being a hit

---

thereof, let's say the 1st hit is on the 4th spin, of the 1st block type (could be 1st, 5th, 8th .. whatever)

QuoteP(1st-hit)= 1- (29/37)⁴= 0.62261459927

That gives us 62% chance of the 1st hit, out of 3 .. coincidentally on the 4th spin.

---

Now, with the secured 1st hit & its gains, press comes in place with activating the 2nd block type.

QuoteP(2nd, 1st spin) = 1- 0.377385401 x (31/37)= 68%
P(2nd, 2nd spin) = 1- 0.377385401 x (31/37)²=..
P(2nd, 3rd spin) = 1- 0.377385401 x (31/37)³=..
P(2nd, 4th spin) = 1- 0.377385401 x (31*37)⁴=..

Now, the probability of having more than 1 (≥1) or 2 hits exactly, should abruptly descend & then slowly increase with each new spin till the 4th of the second block.

Obviously, it doesn't .. so let's get the next hurdle off the way; frankly, we could have done all that in one go.

---

I think the best is to specify formulas for
• exactly 1
• ≥1
•  2
• ≥2
•  3
• ≥3
• etc.

& why not use exactly, when to use ≥.

Quote from: Patrik on Sep 11, 2022, 01:02 PMI have a question ...
Let's say I have a strategy, not for betting but for tracking.
Let's say I want to know the frequency one single number repeats after one hit.
Example nr 19
The number hits and we count how many trails it takes for number 19 to hit again.
Then we get a list with gaps (frequency) 20, 40, 14, 3, 35, 25, 58,
Now can I describe the method and get a module tracking the frequency and get an output file with the gaps of the frequency hits?
This way I can get the RIBOT to become a analyze bot with my own customize modules.
One more question, is the RIBOT click by each spin or can we load a spin file and get the process automated?
Cheers Patrik

I have bigrobens sheet for this its 5.5mb too big to upload here...

I have it in media fire ..heres the link

https://www.mediafire.com/file/0j5xm9j1k93u1wy/Gaps%252BDistanceCalculator.xlsx/file

Quote from: 6th-sense on Jan 03, 2023, 09:48 PMits 5.5mb too big to upload here...

Hi! Forum's upload limits have increased (25MB).

https://rouletteideas.com/index.php?msg=646

Many thanks for your helpful uploads! 👍

thanks Vic