Roulette Probability Made Easier

[Kon-Fu-Sed]

Roulette Probability Made Easier

By Kon-Fu-Sed

For the VLS forum members

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members


Table of contents:

* Preface
* Uncertainty ... or?
* Probability
* Independent spins
* Three results events - Intro
* All three-results events - Part 1
* All three-results events - Part 2
* Rare events
* Hitting the second or third trial
* Missing the first trial
* Increasing the probability
* Calculate a chain of trials
* More on chains
* Mixed chains
* Deviation from the average
* The standard deviation - Part 1
* The standard deviation - Part 2
* Recorded results
* Fair empirical studies
* Math vs. reality
* Checking my claims
* Favorable situations - Part 1
* Favorable situations - Part 2
* ... where are they?
* It will even out in the long run
* The wake-up
* Misdirected intuition
* The end
* PS - Hit or win
* PPS - Misdirected intuition; a classic
* Appendix - The files
* Appendix - Table #1
* Appendix - Table #2
* Appendix - Table #3

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members


Preface

Before you start reading, I'd like to point a few things out...

* On a personal level:

- English is not my first language. I have spell-checked the text but those programs don't know that I mis-spelled "whole" as "whale", for example. Also they can't tell if I have chosen or inflected words correctly. Point me at any peculiarity and I'll correct it. Thanks.

- I'm not a teacher. I have never been. In this article I have tried to be as clear as I possibly can be, starting with the very, very basics and advancing from there in a slow, slow manner so everyone can follow - the advanced math in these cases are not really advanced at all. And as I REASON you through most of it instead of handling complicated formulas, I really hope I have explained everything in a clear and concise way.


* On a contents level:

The text doesn't express my personal OPINION or something.
Everything I write is the accepted way to calculate general probabilities and roulette probabilities in particular.
Although many have tried to prove this way wrong during the centuries roulette has existed, no one has so far been able to do it. Not one single person!
Not logically. Not mathematically. And not empirically given a sample-size of significance.
Not anyone.
Not ever.

(That I know of.)

The below is not what I THINK is correct; it is not an OPINION.
It is the consensus and - so far - it's accepted as correct among mathematicians and other serious researches in gambling and probability theories. But it's still theories - not laws.
It is however theories that are proven by empirical studies of the real world.

If anyone throughout the centuries had really proven the math, the logic and/or the results of empirical studies to be wrong, the math-world (and in reality our whole world) wouldn't be what we today experience. And if someone in the future proves it, they are in for a Nobel Math Prize - at the very, very least.

So if you think I'm wrong I am willing to discuss it - if you FIRST prove your claims. And I mean prove - not the usual "I have seen..." or "I know..." thing.
A proof is not a few visits to the casino where you've seen some phenomenon. A proof is something anyone is able to replicate anytime on any random sample of at least the same size as yours, and have approximately the same results as you have. Every time. Any time.
If you think I'm wrong; please remember: You are not only suggesting that I'm wrong; you're suggesting that fundamental probability math and the mathematical society as a whole is wrong.
So show your proof before you start a debate, please.
(I show you mine...)


* On a reader's level:

  You will need a few things before you start...
  + A calculator for the four rules of arithmetic and also for Square Roots.
  + Basic math knowledge to the level that you understand and are able to handle and solve fractions like: (4/37)+(12/37)x(18/37)+(7/37)/(12/37), at least on your calculator.
  + Basic knowledge of the square-root is also recommended when we do a bit more advanced calculations. (It's necessary that you at least can handle the Square Root key on your calculator...)
  + A little more than half a brain as A LOT (all of it, in fact) is based on very basic logics... logics that will seldom be spelled out but is there all the time, in the background. All the time...
  + An attention-span that lasts for a bit more than just a few paragraphs. However this post is one of the longest ones...


* And on the wheel-level:

Everything in this text is for single-zero wheels.
There are 37 numbers. For double-zero wheels you have to remember that there are 38 numbers and thus change all "37" to "38" when you calculate and take ZeroZero into account at all times. Otherwise the results will be skewed beyond recognition and the total will not be 100%. Not only does it give you trouble math-wise: Because there's ZeroZero there are lower probabilities for everything - it is "1 / 38" instead of "1 / 37"...
Avoid such wheels if you can.


I hope you will enjoy the reading.

In the title I stress "easier" because it is not easy to understand - but not THAT hard...

And learn at least something new.
If you do; I'm satisfied.

I am aware that it's a looong text with lots of (simple) math involved - and you should be aware of that fact too. But I have been trying to write in a light way and in very small doses so I hope I can keep your interest to the end.

The best for you, I think, is to read this in the small portions I post it and digest and recapitulate each part before you go on.

What you actually learn is worth a lot more than the quantity of words you just read...
And what you learn here can be worth money - saved money.

/KFS

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



1. UNCERTAINTY ... or?

... Hmmm ...

...

No...

Probably I shouldn't begin with probability at all, but with CERTAINTY.

Roulette certainty.
Yes.

So I ask you this question:
Can you tell me if there is any event in the game of roulette that is a certainty and that also can be shown to be, without a doubt, a certainty?
Preferably mathematically.

.
.
.

There are TWO (or, rather three):

The FIRST is if you don't make any bet at all...
- Ridiculous!
- No, not at all: It is a certainty that you cannot hit any number at all. Absolutely certain. Isn't it?
- Well, yes. But how, then, is that certainty shown mathematically?

Look at it this way: There are 37 numbers and you bet 0 numbers. You bet, or "cover", 0 numbers of 37.
0 / 37
Enter that into your calculator and what's the result?
0 Zero
Meaning: No, Zero, 0% chance to hit ANY number. That is certainly a certainty.
(Show me the casino that will give you chips when you bet nothing...)

To hit a number when you bet nothing is a 0% chance and a certainty that you don't.
Is there anything that can make you win...? No. It is absolutely a certainty.
And the calculator showed you the probability for you to hit a selected number: 0.

A probability of 0 is the same as certainty that something will NOT happen.


The SECOND event is when you cover all 37 numbers.
Then it is, of course, a certainty that the winning number will be one that you cover.
And you can look at it this way: You cover 37 of the 37 numbers of the wheel.
37 / 37
Enter that into your calculator and what's the result?
1
This "1" says that you have 100% chance to hit the winning number. And you will of course, if you're betting every number there is, 37 of 37 possible.
(We usually multiply the result of a division by 100 and talk about per cent, so the "1" in this case becomes "100%" just as 0.25 becomes 25%)

To hit a number when you cover every one is a 100% chance and ALSO a certainty.
Is there anything that can make you miss...? No. It is absolutely a certainty.

A probability of 1 is the same as certainty that something WILL happen.


Now to the THIRD event and that's a little special because it's the event that has ALREADY happened:

Past results.

Statistics, actually. But often involved when discussing roulette math. (More on that later)
You look at the marquee or your collection of spins and you see a "25" there.
That "25" was the result of that spin at that time at that table. Is there any doubt about that result? Could it be something else?
No, of course not (logics!).
So that is one result, "25", and there can only be one result - it cannot be "4" in some other situation, can it?
The PAST result "25" is 1 result out of 1 possible.
1 / 1
Enter that into your calculator and what's the result?
1
This "1" says that it is a 100% chance that the "25" WAS the hitting number at that spin. And of course it was.

Past results most certainly are certainties.
Is there anything that can make it change...? No. It is absolutely a certainty.

A probability of 1 is the same as certainty that something HAS happened.

Or, as we saw before, WILL happen, whichever happens first...


So now we know this about CERTAINTY:
It can be 0 or it can be 1 in figures, on the calculator. Or 0% or 100% in words.
And nothing in-between.

Usually, when we talk about past spins we talk about STATISTICS.
Statistics are certainties because it reflects events that have already happened and thus cannot be changed.

It is a certainty that it HAS happened.

But...
Will it happen again?? Do we know??

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



2. PROBABILITY

SO ... let's talk about ... <drum-roll here> ... UN-certainty!

Uncertainty. It may happen or it may not happen. Maybe it will happen... It will probably happen... Or not.

Probability.

Now, we do play the roulette, so why don't we place a chip somewhere?
So you place it on a single number - the "13".
What's your chance that the result will be exactly that number? That you hit?

There are 37 numbers and you cover one. No more and no less. 1 number of 37.
1 / 37
If you enter that into your calculator you get the result 0.027027027027
Multiply by 100 and you get 2.7027027 which is the percentage.  2.7%

Likewise, if you place your chip on - for example - the double street 1-6, you cover six numbers. No more and no less.
The chance that you will cover the the result will be 6 / 37 because you cover six numbers. Out of the 37 possible.
And 6 / 37 is 0.162162162 or in other words 16.2%

Last let's look at the RED (or Black, Low, High, Even, Odd) numbers.
If you count them you'll find there are 18 in total and therefore, if you place your chip on Red, you cover 18 of the 37 possible numbers.
18 / 37 is 0.486486486 or 48.6%


The chance that the result will be your selected number(s) is obviously increasing by the increasing number of numbers you cover.
And of course it is - the more numbers you cover the closer you come to certainty that is 37 covered numbers (or 37 / 37 or 100%).
36 numbers are as close as you can get, without betting all, and it is 36 / 37 or 97.3%.

- Hey! There is, then, a risk of missing the number!
- So...We are UN-certain now, are we? We are not 100% sure we will hit with our 36 numbers.

And that UN-certainty is 1 / 37 or 2.7%.
The chance for 36 numbers to hit is 36 / 37 (97.3%) and to miss is 1 / 37 (2.7%) because there is only one number we haven't covered.

We cover 36 numbers with a 36/37 chance to hit and a 1/37 chance to miss.
We cover 36 numbers with a 97.3% chance to hit and a 2.7% chance to miss.
Add them together and you get 37/37 or 100% - certainty.
And that's correct: You know for certain that you will either hit or miss your numbers (that's logics!). Or? Can there be anything else? No. Of course not.

Once again we have seen that 37/37 or 1/1 or 100% means certainty.


And this is what we do when we deal with chances and probabilities:
We do "Reality Checks" to see that all chances to hit and all chances to miss added together equals 100%, is a certainty.

If it's not 100% there is something wrong in our calculation, because we cannot hit AND miss in neither less nor more than the bets we actually do. Of course.
If we hit in 60% and miss in 30% - what will happen in the other 10%?
Or if we hit 60% and miss 50% - where did the extra 10% come from?

Reality Check = 100%.

Always. That's for certain.


Now you know how to calculate the chance (probability) of a hit on the next spin.
And I have already given you a clue on how to calculate the chance of a miss...

As the chance to a hit, covering 36 numbers, is 36 / 37 the chance of a miss must be 1 / 37 as the sum can only be 37 / 37 = 100%

So the chance of a 6-number bet to miss has to be (37 - 6) / 37...
The parenthesis show you the trick: Subtract the known figure (the numbers you cover, in this case) from the total (37) and you get the chance of the other (the uncovered numbers, in this case): 31 / 37 or 83.8%

And the same for Red: (37 - 18) / 37 = 19 / 37 = 51.4%

To prove that it is mathematically correct we do our "Reality Check" and add the hit-% to the miss-% and we should end at 100%.
Do we, in these examples?

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



3. INDEPENDENT SPINS

And so you know how to calculate everything regarding ONE bet, but only one...
This is valid no matter what, really.
As long as the wheel and ball are fair and the dealer doesn't aim for certain parts of the wheel (with more success than random) it is valid.
Because the wheel cannot remember any past results and cannot influence any future ones and - most important - all 37 numbers are there each and every time so the chance is equal for all numbers in all spins. Always.

We talk about "Independent trials" - each trial has the same probability every time and is, in reality, not dependent on any past results.

Some people claim that because their selected bet hasn't turned up for so-or-so long time, it is "due". But think about it (logics now): How can the ball or wheel have any influence whatsoever on the result?
Do they discuss the matter:
"Red hasn't come for 20 spins so now we have to make it hit!"
"No, the number 24 hasn't hit since we opened - it is really due to come now."
?

If this was the case, empirical studies over significant sample-sizes would have shown that, many, many moons ago.
A lot of people, myself included, have tried endlessly to find proof by studying recorded roulette-spins.
None have so far, to my knowledge, found any evidence. Unfortunately.
(But a lot more on empirical studies later)


So the theory of "Independent trials" is here to stay - at least for a while longer. (Quite a while, I'd think but it IS a theory...)
And this is important, as the rest of this article is based on that phenomenon - that the wheel and ball can NOT influence future results by themselves.

If you think they can, you shall know that I can see your point if you say that mathematics is BS in roulette (you can't think both are correct):
If the wheel has decided to select a certain number to hit, no math, as we know it, can be used. We have to have a way to quantify or "parameterize" the mood of the wheel. And we haven't...
I mean... What if the wheel all of a sudden doesn't care? Or is absent-minded and forgets which number was due? Will the ball be in charge???

As long as there are no formulas including such, we have to cope with the math as it is.
And as I said: No empirical studies, over significant sample-sizes, has ever shown the probability theories or the math wrong. Ever.


But please; read on even if you think future results are influenced in any way by the past - maybe you will learn something you have use for, in the future...?

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



4. THREE-RESULTS EVENTS - Intro

Now, I will use a lot of fraction-math here.
The reasons are that fractions are EXACT and clearly show relations. In only a few cases a decimal-number is not needed to be rounded when divided by 37.

To show the hit-probability for a 6-number bet we can use two ways:
p = 6 / 37 (this is exact)
or
p = 0.162162162 (this is rounded) or 16.2% (this is even more rounded)

The first example entered into a calculator gives the second as the result.
The first example also tells us the EXACT RELATION: It is 6 (somethings) of 37 (somethings).

The second example lacks that exact information. It could as well be 16,216,216,211,199,999 of 100,000,000,000,000,000 because it's rounded.
And rounding is bad especially if we start getting really small figures - the rounding errors will cause us trouble in the long term.

I will use the fraction notation a lot, not so much the decimals.


Now let's focus on THREE RESULTS...

How many combinations are possible?

First you have to take every number 0 - 36 and combine them each with every number 0 - 36 and so you have (37 x 37) 1,369 combinations. This is for two results.
Now combine each and every one of those results with all numbers 0 - 36 and you end up with (1369 x 37 or 37 x 37 x 37) 50,653 three-numbers sequences.
(Did you really do it?)

For three results we have 50,653 combinations and there are no more possible to find. And there can't be less, either, if we want them all. 50,653... This is 100% of all there are.

But we cannot use 50,653 series to show math and probability in a clear way. That's impossible. So first of all we will call those combinations the "Low-Level" combinations. They really are on the lowest level - single numbers.
LL combinations. 50,653 of them...

I could, of course, use only Red and Black for example - excluding the Zero - but that wouldn't be fair. The calculated probabilities wouldn't be true, because it's a simplification.
And it wouldn't be roulette - more like coin-tossing.

And this is about roulette probability.

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



5. ALL THREE-RESULTS EVENTS - Part 1

So we have to use a middle-way - please take a deep breath and hold on to your hat. Here we go:

We will use "High-Level" (HL) combinations: the "even" bets Red and Black, that each have a 18 / 37 probability to hit PLUS the Zero that has a probability of 1/37 to hit.
We will use combinations like "R-B-B" or "R-0-0" or "0-B-R" or... you understand, I guess.

Red + Black + Zero = (18/37) + (18/37) + (1/37) = 37/37 = 100% = Nothing missing.
This will bring down the amount of combinations considerably; from 50,653 "LL" to 27 "HL".

That is because every HL combination in reality contains more than only one of the LL-combinations.
Let's find out how many.

And don't even think about skipping this section - it is VITAL that you KNOW this later!


We can check this one for example: "R-R-R".
In this group you will find...

1  1  1  1  1  1 ... 16 16 16 18 18 18 ... 36 36 36 36 36 36 <--- All the 18 red numbers combined with...
1  1  1  1  1  1 ... 36 36 36  1  1  1 ... 36 36 36 36 36 36 <--- ... all the 18 red numbers combined with...
1  3  5  7  9 12 ... 32 34 36  1  3  5 ... 25 27 30 32 34 36 <--- ... all the 18 red numbers

Each "R" is in reality 18 numbers so what you have to do, to find the number of possible combinations, is to multiply 18 by itself three times (18 x 18 x 18) = 5,832.
This is the same calculation you did to find how many possible combinations there were using ALL numbers - you multiplied 37 by itself three times.
That was three groups of 37 numbers, now you have three groups of 18 numbers. 18 x 18 x 18. OK?

This particular HL combination contains 5,832 LL combinations of the possible 50,653.

Another group is "B-R-B" in which you will find...

2  2  2  2  2  2 ... 17 17 17 20 20 20 ... 35 35 35 35 35 35 <--- All the 18 black numbers combined with...
1  1  1  1  1  1 ... 36 36 36  1  1  1 ... 36 36 36 36 36 36 <--- ... all the 18 red numbers combined with...
2  4  6  8 10 11 ... 31 33 35  2  4  6 ... 26 28 29 31 33 35 <--- ... all the 18 black numbers

Is there a difference in numbers if the three are "B-R-B", compared to "R-R-R"?

No, each group have the same 18/37 chance each to hit so this HL combination also contains 5,832 LL of the 50,653.
In fact: ALL HL combinations that contains only R and/or B is the same. Here they are:

R-R-R (18 x 18 x 18 = 5,832)
R-R-B
R-B-R
R-B-B
B-R-R
B-R-B
B-B-R
B-B-B

There are 8 of them so they together contain 8 x 5832 = 46,656 LL combinations.
Reality Check: 50,653 - 46,656 = 3,997

We are missing 3,997 combinations!
Where are they?

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



6. ALL THREE-RESULTS EVENTS - Part 2

This is roulette. There is a ZERO also!

"R-R-0" - there is a HL with a Zero. In this group you will find:

1  1  1  1  1  1 ... 16 16 16 18 18 18 ... 36 36 36 36 36 36 <--- All the 18 red numbers combined with...         
1  3  5  7  9 12 ... 32 34 36  1  3  5 ... 25 27 30 32 34 36 <--- ... all the 18 red numbers combined with...
0  0  0  0  0  0 ...  0  0  0  0  0  0 ...  0  0  0  0  0  0 <--- ... the Zero

How many LL combinations does it contain?
Red (and Black) is 18 numbers and Zero is... well, only 1 number of the 37.
So the calculation is: 18 x 18 x 1 = 324. Each one of those HL that contains one (and only one) Zero contains 324 LL of the 50,653 possible combinations.

Another group is "B-0-R" and there you will find:

2  2  2  2  2  2 ... 17 17 17 20 20 20 ... 35 35 35 35 35 35 <--- All the 18 black numbers combined with...
0  0  0  0  0  0 ...  0  0  0  0  0  0 ...  0  0  0  0  0  0 <--- ... the Zero that is combined with...
1  3  5  7  9 12 ... 32 34 36  1  3  5 ... 25 27 30 32 34 36 <--- ... all the 18 red numbers

And all those "one-zero" groups are:

R-R-0 (18 x 18 x 1 = 324)
R-B-0
B-R-0
B-B-0
R-0-R
R-0-B
B-0-R
B-0-B
0-R-R
0-R-B
0-B-R
0-B-B

There are 12 of these so they altogether contain 12 x 324 = 3,888 LL of the original 50,653.

Reality Check: 46,656 + 3,888 = 50,544 - still not 50,653
Still missing some...

There are also HL combinations with two Zeros like "0-B-0" in which you will find:

0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 <--- The Zero that is combined with...
2  4  6  8 10 11 13 15 17 20 22 24 26 28 29 31 33 35 <--- ... all the black numbers combined with...
0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 <--- ... the Zero

That's all: 18 combinations in this group. It has only 1 result with 18 numbers but two with only one number (the Zeros).
Therefore the calculation is: 18 x 1 x 1 = 18 for all such groups.
And they are:

R-0-0 (18 x 1 x 1 = 18)
B-0-0
0-R-0
0-B-0
0-0-R
0-0-B

6 of them, each having 18 combinations, give 6 x 18 = 108 LL.

Reality Check: 46,565 + 3,888 + 108 = 50,652.
100% = 50,653.

ONE is missing!


This one: "0-0-0"
As it is only one number in each result the calculation is quite simple: 1 x 1 x 1 = 1


And there you are: 27 HL combinations containing each and every one of the 50,653 possible LL.
And this is the complete table, showing how many each contains and their probability to hit:

B-B-B  -  5,832 p = 5832/50653 = .115136 (11.5%)
B-B-R  -  5,832
B-B-0  -    324 p = 324/50653 = .006396 (0.64%)
B-R-B  -  5,832
B-R-R  -  5,832
B-R-0  -    324
B-0-B  -    324
B-0-R  -    324
B-0-0  -     18 p = 18/50653 = .000355 (0.036%)
R-B-B  -  5,832
R-B-R  -  5,832
R-B-0  -    324
R-R-B  -  5,832
R-R-R  -  5,832
R-R-0  -    324
R-0-B  -    324
R-0-R  -    324
R-0-0  -     18
0-B-B  -    324
0-B-R  -    324
0-B-0  -     18
0-R-B  -    324
0-R-R  -    324
0-R-0  -     18
0-0-B  -     18
0-0-R  -     18
0-0-0  -      1 p = 1/50653 = .00002 (0.002%)
---------------
    Sum: 50,653 = (37 x 37 x 37) = Correct

I will refer to this table a lot later and therefore it is VITAL that you understand what each group contains.

Please recapitulate if you are uncertain of anything... (there is a high probability that you are)

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



7. RARE EVENTS

As each combination in the above table is three results, the total of results are 3 x 50,653 = 151,959!

- Hey! Why bother to have that 0-0-0? It happens only once in 152K spins? That's a rare event if I've ever heard of one! I will never bet 152 thousand times in my life!
- First of all it HAS to be there - without it the table would be incomplete and not be covering 100% of all possible combinations.

Second: That particular combination has the exact same probability to hit as the combination 25-3-31. You wouldn't exclude that one, would you?
Or any other three-numbers combination? They are all 1/50653 chances. 1-2-3 or 7-34-22 or 36-36-36... All of them are equal - the 0-0-0 is no exception in that context. It's just painted green on the wheel.

The ONLY reason 0-0-0 is distinguished in the TABLE, is because 25-3-31 is contained inside the R-R-B combination and the other examples inside their respective groups.
0-0-0 doesn't fit anywhere (it's bullied because it's green!) but it has exactly the same probability as any other combination.
Why should we exclude it?
Why should we bully any...one?


As a side-note: The 0-0-0 combination doesn't come once in 152K spins, it comes in 1/50653 combinations of three spins, and that's quite a difference. Suppose this sequence of spins:

R-B-0-0-0-R-B-B-R

There is a 0-0-0 combination, obviously. But then you split the sequence into 3-results events:

R-B-0
0-0-R
B-B-R

None of these is a 0-0-0 combination... So the probability to find a 0-0-0 combination in a three-results event is a lot lower (1/3, actually) than simply find it anywhere.
In three-results combinations: 1 in 50,653 COMBINATIONS (times 3 = 151,959 results in total)
Anywhere: 1 in 50,653 SINGLE RESULTS when we treat the results as one looong continual chain.

(Isn't math beautiful?)


Also; 0-0-0 has a probability (mathematical average) to happen like 2.16 times a year at a 300 spins/day table.
Maybe, just maybe, you happen to be there at the time that happens - why not, really?
And you know what they say:


If s**t can happen, s**t will happen.
- But when it hits the fan, probability theory is like having an umbrella.

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



8. HITTING THE SECOND OR THIRD TRIAL

OK now. Order in class!

Let's go back to the table.

We saw before that the probability to hit at THE NEXT trial, if we cover 18 numbers, is always 18/37. Now; how about the trial after that? The second trial? Is it the same probability?
We can use the table for that but first we have to decide where to bet. Let's say we are going to bet Red.

We also HAVE to check that the 18/37 for the FIRST Red-bet trial is correct:
To do that, you look for all combinations that contains a "R" in the first position. Note the number of LL combinations and add them together:

R-B-B  -  5,832
R-B-R  -  5,832
R-B-0  -    324
R-R-B  -  5,832
R-R-R  -  5,832
R-R-0  -    324
R-0-B  -    324
R-0-R  -    324
R-0-0  -     18
---------------
    Sum: 24,642 / 50,653 possible. 24642 / 50653 = 18 / 37. We were correct.

- OK that's fine but what about the second?
- Do the same thing: Collect all combinations where "R" is in the second position.

B-R-B  -  5,832
B-R-R  -  5,832
B-R-0  -    324
R-R-B  -  5,832
R-R-R  -  5,832
R-R-0  -    324
0-R-B  -    324
0-R-R  -    324
0-R-0  -     18
---------------
    Sum: 24,642 / 50,653 possible. 24642 / 50653 = 18 / 37. Correct?

And for the third position:

B-B-R  -  5,832
B-R-R  -  5,832
B-0-R  -    324
R-B-R  -  5,832
R-R-R  -  5,832
R-0-R  -    324
0-B-R  -    324
0-R-R  -    324
0-0-R  -     18
---------------
    Sum: 24,642 / 50,653 possible. 24642 / 50653 = 18 / 37. Correct?

Yes, it's correct. What we've seen here is that it doesn't matter if you bet at the first, second or third attempt of three: The probability to hit is always 18 / 37.

But that's on one IMPORTANT condition: That we have THREE TRIALS LEFT.

Remember this!

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



9. MISSING THE FIRST TRIAL

- So if we DON'T have all three left... Like, what if I miss the first bet - it's Black or Zero? Then for sure Red will have at least a little higher probability to hit.
- That's a good question. We will check that.

What combinations can possibly end our three-trials combination AFTER the first trial?
These ones:

B-B
B-R
B-0
R-B
R-R
R-0
0-B
0-R
0-0

That's all there are!

- But what about the first, lost, trial? It isn't there!
- That's OK, I can put it there. It was a non-Red, right?
- Yes. 24.
- OK, here's the table including the number of LL combinations within each HL:

24-B-B  -  324 (1 x 18 x 18 = 324)
24-B-R  -  324
24-B-0  -   18 (1 x 18 x 1 = 18)
24-R-B  -  324
24-R-R  -  324
24-R-0  -   18
24-0-B  -   18
24-0-R  -   18
24-0-0  -    1 (1 x 1 x 1 = 1)
--------------
   Sum:  1,369 / 1369 possible LL combinations = 100%. Correct.

- How can that be correct? It's not 50,653 as it was above!
- First of all: The 24 is a past spin and is KNOWN. If you remember the beginning of this article; a known result is a CERTAINTY. Because it cannot be anything else.
So it cannot possibly be ANY Black or the Zero as it WAS, you said it yourself, number 24...

It was number 24 and it is one number of one possible. 1 / 1. The probability is 100% or p = 1.
And so, the calculation for the number of LL in 24-B-B is: 1 x 18 x 18 = 324.
The Zero is one number out of 37 so for the combination 24-B-0 the calculation is: 1 x 18 x 1 = 18.

Now, this is important that you understand: A known result is a certainty and therefore it cannot have any other probability but p = 1.
The probability to hit cannot be 1/2 or something else. Right? It really did hit.
I mean; can it be possible that all of a sudden it DIDN'T hit and what will happen then? It was just last spin and it WAS number 24... And now it's ...
No, it can not be anything else but the 24 we saw hit.

A certainty is a certainty and it's 100% sure - no more and no less - so p = 1.

So, Because we have only TWO FUTURE events, the total of POSSIBLE combinations is calculated as: 1 x 37 x 37 = 1369.

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



10. INCREASING THE PROBABILITY

And your question was...?

Aaah - will the chance to hit Red increase (at least a little bit) at the second attempt if the first was missed?
Just look at the table for [two trials (remember there are only two left in our three-trials series - one, number 24, is gone and is not in the future) and count the number of LL in each HL combination where "R" is in the second position.

24-R-B  -  324
24-R-R  -  324
24-R-0  -   18
--------------
      Sum: 666 / 1369 = 18 / 37

It's the same old 18 / 37 chance, I'm afraid. And it doesn't matter if you bet Black instead - check the table.
(Of course it is another thing if you bet Zero because it's one single number instead of 18. Can you do the calculation for a 0-hit instead of this Red-hit?)


Now, I hope you understand why we have to put "24" in the beginning of our three-trials combinations.
The "24" isn't part of "R-R-B", for example. It IS part of "B-R-B" (and some others) and if we use them (still calculating the chance for Red to hit in the second trial) we would use

B-R-B  -  5,832
B-R-R  -  5,832
B-R-0  -    324

But that would be very wrong as "B-R-B" also includes 2, 4, 6, 8, 10, 11... - not only 24 - as a start in all its 5,832 LL combinations. And we know that 24 hit - not 11 or 4.
So we have to eliminate all combinations with another number but 24 in the first position, to have the correct number of possible combinations in the TWO REMAINING trials: 1 x 37 x 37 = 1,369.


For the third trial, it's exactly the same: If the second result was also a non-Red, 33 for example, the only combinations that can finish our three-trials series are:

24-33-B  -  18
24-33-R  -  18
24-33-0  -   1
--------------
       Sum: 37 / 37 possible numbers to finish the series.

So for both Red and Black it is a 18 / 37 chance to hit that last, third, trial after the two first have been seen.
I mean; the end-result cannot possibly be 12-33-1 or 24-12-2 or 11-15-3, can it?
No, of course the end-result must start with 24-33 and nothing else, regardless the last result.


SO, the answer to your question is:
No, the probability does NOT increase if you miss one or more bets.


Have you noticed something? Until now, I've only talked about the NEXT trial.

First, before anything happened, I showed that there is a 18/37 probability for Red to hit at the first trial - the "next" trial in that context.
Then I showed you that the probability to hit the second trial (the now "next") is also 18/37 and that is REGARDLESS if you saw the "24" or not.
And the same goes for the third trial when it is the "next": The probability to hit is 18/37. And it doesn't matter if you missed the two first trials or not.

The fact is, that you can expand this for as long as you like and to any roulette bets you like; the probability to hit at the next trial is always the same:

"The number of covered single numbers divided by 37".
It's never anything else.

And the risk to miss is always "(37 minus the number of covered single numbers) divided by 37".
If your calculations are correct, the two results added together equals 37/37 or 1.0 or 100%.

Always.

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



11. CALCULATE A CHAIN OF TRIALS

Before, you have learned to how to calculate the probability for ONE trial - now we will calculate the probabilities for a CHAIN of trials.

- Yes! What's the chance to hit three Reds in three trials?

You just have to look in the three-trials table and see how many LL there are in the "R-R-R" HL. No other HL group gives you three Red numbers in a row, right?
5,832 LL combinations of the possible 50,653. So the chance to hit ALL THREE is 5832 / 50653 = .115136 or 11.5%
We already did the calculation when we calculated the number of LL: 18 x 18 x 18 = 5832. Remember?

It's the same for all combinations of only Rs and Bs, for example "R-B-R" or "B-B-R" or any other, as long as they don't include any Zeros.


- So what's the probability to have one hit in the three trials if we exit right after the hit?

We can do this in two ways:
* The long and tiresome way, or
* The short and quick one

The former includes lots of adding while the latter includes some, but not a lot, of logic and multiplication. And one subtraction.
Your choice...

Joke; I'll do them both.

The logic way: If you don't care when you hit, during the three trials, and you will be satisfied by the first hit...
In that case, there is only one occasion we need to calculate: When you miss all three attempts.
Think about it: If you don't miss ALL, obviously you MUST have hit once... (logic!)
Right?
The probability to miss one trial is 19/37 so for three trials the calculation is: (19/37) x (19/37) x (19/37) = 6859 / 50653
You have 6,859 chances to miss all three events, so you have (50,653 - 6,859 =) 43,794 / 50,653 chances to hit. (Like 86.5%)

The long way: Check the three-results table and add the LL values from all HL combinations that include one or more "R":
(The "X" shows the eXit-point)

B-B-R X  -  5,832 (3 bets to eXit)
B-R X B  -  5,832 (2 bets to eXit)
B-R X R  -  5,832 (2 bets)
B-R X 0  -    324 (2 bets)
B-0-R X  -    324 (3)
R X B-B  -  5,832 (1)
R X B-R  -  5,832 (1)
R X B-0  -    324 (1)
R X R-B  -  5,832 (1)
R X R-R  -  5,832 (1)
R X R-0  -    324 (1)
R X 0-B  -    324 (1)
R X 0-R  -    324 (1)
R X 0-0  -     18 (1)
0-B-R X  -    324 (3)
0-R X B  -    324 (2)
0-R X R  -    324 (2)
0-R X 0  -     18 (2)
0-0-R X  -     18 (3)
-----------------
      Sum: 43,794  / 50,653 (Exactly the same as the above result)

Some simple logic can help a lot, sometimes

[Kon-Fu-Sed]

Roulette Probability Made Easier
By Kon-Fu-Sed, 2008, for the VLS forum members



12. MORE ON CHAINS

Also: If you want to know the probability for you to hit IF you miss the first time (but you are not SURE to miss as the first result has NOT come yet)... how's that calculated?

Remember in this calculation, the first trial of the three has no result YET - we are BEFORE the first spin. (Otherwise it's of course the same conditions and result as above, when 24 hit the first trial)

Let's do it the long way first: We have the following HL combinations WITHOUT a "R" in the first position but WITH a "R" at the second or third (or both).

B-B-R X  -  5,832 (3 bets to eXit)
B-R X B  -  5,832 (2 bets)
B-R X R  -  5,832 (2)
B-R X 0  -    324 (2)
B-0-R X  -    324 (3)
0-B-R X  -    324 (3)
0-R X B  -    324 (2)
0-R X R  -    324 (2)
0-R X 0  -     18 (2)
0-0-R X  -     18 (3)
-----------------
      Sum: 19,152 / 50,653 (Some 37.8%)

That's the probability for you to miss at the first trial AND THEN hit (at least) one of the two remaining.
IF you STILL have ALL THREE trials WAITING for you!

- Can this be solved by some logic, then?

- Well, look at it this way: You ALWAYS have a 18/37 probability to hit the next - in this case the first - trial. That is equal to 24,642 of the 50,653 possible combinations. (This was calculated above)
We also calculated the probability to hit ANYTIME during the three trials and that was 43,794 / 50,653.
So if we exclude all those first-trial hits, because we calculate to miss that one, we get 43,794 - 24,642 = 19,152...
19,152 / 50,653.

The same as above...

[Kon-Fu-Sed]

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