Author Topic: Law of the Third -- WORD!  (Read 7038 times)

Reyth

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Law of the Third -- WORD!
« on: June 11, 2015, 07:34:37 PM »
Quote from: Internet User: Bliss
Suppose you want to know how many spins it takes, on average, for a set of numbers (or "events") to arrive completely. ie;

  • How many spins are necessary before all the numbers (0-36) to come in?
  • How many spins necessary before all the streets (1-12) come in?
  • How many spins necessary before all the 3 even money patterns (rrb,  brb, rbb, etc) come in?
and many more.

There is a formula you can use to calculate any of the above, but it can get a little tedious, so a spreadsheet might be useful (or a simple computer program). Rather than just pull the formula out of a hat, I'll show where it comes from, then hopefully it will be more meaningful and easier to apply in new situations.

Also, keep in mind that these numbers are averages, and "raw" averages are not the holy grail. After all, on average, an EC bet occurs once every 2 spins, and if that held on a spin-by-spin basis, you could easily D'Alembert your way to riches. We all know that variance kills bankrolls right? But still, averages can be useful as a starting point.

Ok, an important assumption for the formula to "work" is that the "events" are equally likely; in each of the above examples this is satisfied. Each number has a 1/37 chance of a hit; each street has a 1/12 chance of a hit; each of the patterns (rrb, brb, etc - there are 8 ) has a 1/8 chance of arriving.

Next, a "law" which is needed to derive the formula is the following:

"If the probability of an event occurring is 'P', then the average number of trials needed before "success" is '1/P'."

This is quite intuitive. Using the examples above, it just means that:
if you pick a single number, the probability of a hit is 1/37, and on average you will need to wait 37 spins before a hit; if you pick a street, the probability of a hit is 1/12, and on average you will need to wait 12 spins before it hits; if you pick 'rrb', the probability of a hit is 1/8, and on average you will need to wait 8 patterns before it hits. (Note that this last example uses patterns rather than spins. If you wanted to express the wait in terms of spins you would need to multiply the number of spins making up each pattern, by the number of patterns ie; 3 x 8 = 24 spins)

So now we are in position to begin answering the question (taking the first example) - how many spins does it take to get the "full set" of 0,1,2,3,4,...36?

It doesn't matter what the first number is, say it's 13. So it's one down, 36 to go. This means that the chance that the next number is different from 13 is 36/37 (because 13 has been "removed" from the set). Now, we need to use the above "law" to find out how many additional spins are needed before we get this next (different) number -

"If the probability of an event occurring is 'P', then the average number of trials needed before "success" is '1/P'."

The probability P is 36/37, so the average number of trials needed to get a different number is 37/36 (just turn the fraction upside down).

So far we have that the average number of trials needed to get two different numbers is 1 + 37/36 = 2.0278
Obviously, we can't have a fractional number of trials, so round the number down to 2. This makes sense, because most of the time single numbers don't repeat, so most of the time after 2 spins we will have 2 different numbers hit.

So far, so good. But we want the number of trials necessary to get all the numbers, not just 2, so we carry on. Once you have 2 different numbers, what is the probability that the next spin will produce a number different from the first two? We have now "used up" 2 numbers, so the probability that the next number will be different is (37 - 2)/37 = 35/37. Again, we need to find out how many additional spins are necessary to get this next (different) number. It's always the same procedure, just turn the number upside down using the law, so 37/35 more spins are needed.

So now we have that the average number of trials needed to get 3 different numbers is:

1 + 37/36 +37/35

To get the number of trials needed for 4 different numbers to hit, rinse and repeat:

We have "used up" 3 numbers, so the chance that the next one is different is (37 - 3)/37 = 34/37.
And we have to wait 37/34 spins before we get it. Add this to the total:

1 + 37/36 + 37/35 + 37/34

You should be able to see a pattern developing here.The average number of trials needed to get 5 different numbers is:

1 + 37/36 + 37/35 + 37/34 + 37/33

and 6 different numbers:

1 + 37/36 + 37/35 + 37/34 + 37/33 + 37/32

Like I said, it gets tedious calculating all the sums, so I wrote a little program to finish it off:

Number of trials needed to get first number is: 1.00
Number of trials needed to get 2 different numbers is 2.00
Number of trials needed to get 3 different numbers is 3.00
Number of trials needed to get 4 different numbers is 4.00
Number of trials needed to get 5 different numbers is 5.00
Number of trials needed to get 6 different numbers is 6.00
Number of trials needed to get 7 different numbers is 8.00
Number of trials needed to get 8 different numbers is 9.00
Number of trials needed to get 9 different numbers is 10.00
Number of trials needed to get 10 different numbers is 11.00
Number of trials needed to get 11 different numbers is 13.00
Number of trials needed to get 12 different numbers is 14.00
Number of trials needed to get 13 different numbers is 16.00
Number of trials needed to get 14 different numbers is 17.00
Number of trials needed to get 15 different numbers is 19.00
Number of trials needed to get 16 different numbers is 21.00
Number of trials needed to get 17 different numbers is 22.00
Number of trials needed to get 18 different numbers is 24.00
Number of trials needed to get 19 different numbers is 26.00
Number of trials needed to get 20 different numbers is 28.00
Number of trials needed to get 21 different numbers is 30.00
Number of trials needed to get 22 different numbers is 33.00
Number of trials needed to get 23 different numbers is 35.00
Number of trials needed to get 24 different numbers is 38.00 <== This is the "law of the third"
Number of trials needed to get 25 different numbers is 41.00
Number of trials needed to get 26 different numbers is 44.00
Number of trials needed to get 27 different numbers is 47.00
Number of trials needed to get 28 different numbers is 51.00
Number of trials needed to get 29 different numbers is 55.00
Number of trials needed to get 30 different numbers is 60.00
Number of trials needed to get 31 different numbers is 65.00
Number of trials needed to get 32 different numbers is 71.00
Number of trials needed to get 33 different numbers is 78.00
Number of trials needed to get 34 different numbers is 88.00
Number of trials needed to get 35 different numbers is 100.00
Number of trials needed to get 36 different numbers is 118.00
Number of trials needed to get 37 different numbers is 155.00

So you can see that the so-called "law of the third" arises from turning the question around; instead of asking "how many spins are necessary to get all 37 numbers?" you ask "in 37 (or 38) spins, how many different numbers will have not hit" answer - roughly one third.

Notice that although on average, the number of trials needed to get the "full set", is 155, we know that a number can sleep for 300 or more spins, but MOST of the time (68%) they will have all arrived after a similar number of spins to 155.

Ok, back to the formula and the 2nd example of streets. How many spins are needed before you get all the streets? If you guessed the formula would be this:

1 + 12/11 + 12/10 + 12/9 + 12/8 + 12/7 + 12/6 + 12/5 + 12/4 + 12/3 + 12/2 + 12/1

You would be right. Notice that 12/1 = 12! I wrote it like this so you can see that the pattern continues right to the end. Actually I should really have written the first number (1) as 12/12, to be consistent.

Here's the breakdown:

Number of trials needed to get first street is: 1.00
Number of trials needed to get 2 different streets is 2.00
Number of trials needed to get 3 different streets is 3.00
Number of trials needed to get 4 different streets is 5.00
Number of trials needed to get 5 different streets is 6.00
Number of trials needed to get 6 different streets is 8.00
Number of trials needed to get 7 different streets is 10.00
Number of trials needed to get 8 different streets is 12.00 <== this is the "law of the third"
Number of trials needed to get 9 different streets is 15.00
Number of trials needed to get 10 different streets is 19.00
Number of trials needed to get 11 different streets is 25.00
Number of trials needed to get 12 different streets is 37.00

Notice that the law of the third seems to hold for streets (there are 12 streets, and after 12 spins, one third (4) streets will not have hit, on average). Is there any special significance to the law of the third? No. It's not specific to roulette, but is a general rule. All it says is that the fewer the numbers (or events) there are left to hit, the higher the chance of hitting a number (or event) that has already hit. Nor is there any significance to the "third". You could come up with a "law of the half" or a "law of the sixth", and it would apply equally well to any sequence of trials.

Finally, the 3rd example of 3-pattern EC bets {rrr,bbb,rrb,bbr,rbr,brb,brr,rbb}

1 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1

Number of trials needed to get 1 pattern is: 1.00
Number of trials needed to get 2 different patterns is 2.00
Number of trials needed to get 3 different patterns is 3.00
Number of trials needed to get 4 different patterns is 5.00
Number of trials needed to get 5 different patterns is 7.00
Number of trials needed to get 6 different patterns is 10.00
Number of trials needed to get 7 different patterns is 14.00
Number of trials needed to get 8 different patterns is 22.00

Remember that in this case, a "trial" is not a spin but 3 spins (to make up each pattern).

The general formula for finding the number of trials needed to get the "complete set" of N equally likely outcomes is:

1 + N[1 + 1/2 + 1/3 + 1/4 + ... + 1/(N-1)]


 

kav

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Re: Law of the Third -- WORD!
« Reply #1 on: June 11, 2015, 07:46:02 PM »
Hello Reyth,

Great post.
I just want to point that these calculations represent  "averages" and not upper limits. This means that the numbers presented are our expectations according to probability and that actual results can differ greatly from these expectations.
 

Reyth

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Re: Law of the Third -- WORD!
« Reply #2 on: June 11, 2015, 07:46:46 PM »
I found this post because I have noticed a new phenomenon associated with equal distribution that I am calling, "Distribution Disparity".  This is where a selected set of numbers have not spun for a disproportionately larger time period which indicates that the Law of the Third has caught those numbers.

The indicator for this theoretically would be the number that has not spun the longest when compared to the number that has spun most recently within the set.

My theory is that if this gap grows too large, you are looking at a "compromised" set of numbers that is made up of mosty sleepers.

I'm not a statistician nor a mathematician but I just know there is a formula that I can use to determine this...

My thoughts were to try and determine average variance for the Law of the Third and use that to compare with the set of numbers.  Maybe its that simple?

I mean, I dunno maybe its just the larger number vs. the average of the set...

Ok, the set of numbers that I am interested in has 10 elements; Number of trials needed to get 10 different numbers is 11.00...

This is simply not helpful in itself. 

Ok, I guess it will simply be:

The average time since all numbers have spun vs. the number with the lowest time in the set.

Pretty sure this will work as a disparity indicator.

Ok, no.  Averages muddle the picture.

Starting to work with the lowest number in the set (which represents the state of the board) & adding 74 to that number (they are splits).

Here's what I have:

184 <=== clogged flow
104 <=== within average flow parameters
229 <=== clogged flow
163 <=== clogged flow (clogs at 169)
95 <=== average state of the board, normal distribution flow

Each of the clogged values are greater than 73 above the average distribution flow of the board.  The idea is to identify this when it happens and end the session because when the Law of the Third captures this set, there will be prolonged miss streaks on these numbers.

Maybe I should work with the total average of all the numbers vs. my selected set...

Ya I think typing this out has helped me alot.  The key is comparing all the numbers on the board, this is where the answer lies!
« Last Edit: June 12, 2015, 01:38:50 AM by Reyth »
 

Jesper

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Re: Law of the Third -- WORD!
« Reply #3 on: June 11, 2015, 08:06:48 PM »
I am not sure average is a number which we can relay on. If the variance is large, the average is
in fact  not so common.

I have found Days when the first hit of  a repeating numbers average 12, in many spins. The math average  and in longer runs it is greater than 7 but less than 8.

We know it is  no chance the first spin, and 1/37 of a repeater second spin and the chance increase as we have more numbers, and become  1/1 when 37 numbers have fallen.

SD 1 is Close to the law of third.

We can use it in our play, but I am not sure, we have use of knowing average.
 

Reyth

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Re: Law of the Third -- WORD!
« Reply #4 on: June 11, 2015, 08:13:25 PM »
I am not sure what you mean, what is "SD 1"?

EDIT: K, thanks.
« Last Edit: June 12, 2015, 01:37:57 AM by Reyth »
 

Jesper

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Re: Law of the Third -- WORD!
« Reply #5 on: June 11, 2015, 08:38:47 PM »
I am not sure what you mean, what is "SD 1"?

SD1   (Standard Deviation 1) is the interval around  average, which is about 68% of the outcomes.
 

dobbelsteen

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Re: Law of the Third -- WORD!
« Reply #6 on: June 12, 2015, 09:07:50 AM »
The experimental outcome of a 37 random sequence is that 12 random numbers on average are not occured. These 12 numbers are together again a true random sequence. How many of these particular sequenses can we expect. A 12 number sequence has a expectation of 1/3.We can expect 3^12 =531441 possible sequences. The 12 not fallen numbers are not so particular as we thought.

Nevertheless these 12 numbers can be the trigger for several systems and strategies. There are a lot of these sequences with special and particular features. Example all numbers are an EC, all numbers of one dozen or colomn, Special Streets or double Streets . The 12 numbers content some neighbours and many others.
These particular features can be a second or third trigger. All the triggers together create hotspots on the table. This is a source for new designs of systems
 
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Reyth

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Re: Law of the Third -- WORD!
« Reply #7 on: June 12, 2015, 10:15:10 PM »
The experimental outcome of a 37 random sequence is that 12 random numbers on average are not occured. These 12 numbers are together again a true random sequence. How many of these particular sequenses can we expect. A 12 number sequence has a expectation of 1/3.We can expect 3^12 =531441 possible sequences. The 12 not fallen numbers are not so particular as we thought.

Nevertheless these 12 numbers can be the trigger for several systems and strategies. There are a lot of these sequences with special and particular features. Example all numbers are an EC, all numbers of one dozen or colomn, Special Streets or double Streets . The 12 numbers content some neighbours and many others.
These particular features can be a second or third trigger. All the triggers together create hotspots on the table. This is a source for new designs of systems

AWESOME POST!  All your Pascal training is amazing!

I feel so comfortable and at home when I am working directly with the The Law of the Third because I know I am actually working with the driving force behind roulette.

The Law of the Third IS the cause of loss streaks.
« Last Edit: June 12, 2015, 11:05:55 PM by Reyth »
 

BlueAngel

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Re: Law of the Third -- WORD!
« Reply #8 on: June 12, 2015, 10:23:09 PM »
Quote
I feel so comfortable and at home when I am working directly with the The Law of the Third because I know I am actually working with the driving force behind roulette.

The Law of the Third IS the cause of loss streaks.

...and of winning streaks too!:-)
 
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dobbelsteen

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Re: Law of the Third -- WORD!
« Reply #9 on: June 13, 2015, 12:20:01 PM »
I have programmed the 1/3- rule in an excel program. The program records the not fallen numbers after 37 spins.
I have for 15 samples these sequences ordened in a tabel. These sequences show partigular events which are usable for betting patterns
You can study these tabel by yourown.



 
 

december

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Re: Law of the Third -- WORD!
« Reply #10 on: June 13, 2015, 12:27:35 PM »
Nice to see Real's comment disappear.
I said to myself: "O, not again...GF and things...!"
 

december

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Re: Law of the Third -- WORD!
« Reply #11 on: June 13, 2015, 12:34:26 PM »
Please Dobbel - this table needs explanations.
Which column says what?!
 

scepticus

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Re: Law of the Third -- WORD!
« Reply #12 on: June 13, 2015, 03:13:05 PM »
Hi Dobbelsteen
Do your 16 samples illustrate the numbers that have NOT occurred in these particular samples of 37 spins  ?
 

TheGenner

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Re: Law of the Third -- WORD!
« Reply #13 on: June 13, 2015, 03:14:35 PM »
Here is something for the 'Law of the Third'

I have done these in 5 cycles of 37 spins. What I wanted to find out is how many numbers come out less and how many come out more than what probability dictates they should. The sum of numbers that come out more than what the average suggests is an average of 12. The sum of numbers that come out less than what the average suggests is an average of 16.

 
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scepticus

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Re: Law of the Third -- WORD!
« Reply #14 on: June 13, 2015, 03:29:12 PM »
Interesting ,The Genner.
The question your analysis  - and  Dobbelsteen's - pose is " Can we make use of these ?"
You give Hot AND Cold numbers .I think an analysis of HOT numbers would be more useful but it would need lots more samples.Food for thought.