### Author Topic: LOTT  (Read 1050 times)

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##### LOTT
« on: August 19, 2018, 07:40:47 PM »

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

This is for streets.
Does anybody have the same for LINES and for SPLITS?

#### kav

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##### Re: LOTT
« Reply #1 on: August 20, 2018, 01:06:04 PM »
Hi,

How did you calculate these stats? They seem arbitrary.
I mean theoretically you can have the same street for 5 spins.

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#### Joe

##### Re: LOTT
« Reply #2 on: August 20, 2018, 03:45:44 PM »
I guess they're supposed to be averages. Intuitively, the law of the third for a double street would be 4 uniques after a cycle and for a split it would be 12 uniques after a cycle.

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##### Re: LOTT
« Reply #3 on: August 20, 2018, 04:54:51 PM »

I did not calculate them myself. They were part of a post on the old VLS forum.

##### Re: LOTT
« Reply #4 on: August 21, 2018, 08:07:53 AM »
Here KAV, I found the rest.
I copied and pasted it to my computer years ago.
Unfortunately I don't know who the original poster was and cannot give credit where it is due.

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)]

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#### kav

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##### Re: LOTT
« Reply #5 on: August 21, 2018, 09:18:19 AM »
Interesting!

#### GIAJJENNO

##### Re: LOTT
« Reply #6 on: August 21, 2018, 11:08:34 AM »
Can we made some very very good system based on these informations?

Thansk!

#### Joe

##### Re: LOTT
« Reply #7 on: August 21, 2018, 01:08:05 PM »
Also, keep in mind that these numbers are averages, and "raw" averages are not the holy grail.
Very true. On average it takes 155 spins for all numbers to hit, but we all know that a number can sleep for 400 or more spins. But still it's a nice post and formula. Maybe there's one for the variance too, then we could find the probabilities of seeing some number of standard deviations from the average.

#### dobbelsteen

##### Re: LOTT
« Reply #8 on: August 21, 2018, 02:43:02 PM »
This is the Bayes coupon theory. In my old blog you can find this computation also for a 10 step Martingale.There are many strategies and bet selections for the 2/3 rule.