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The Sign of the Four

Started by nullified, February 10, 2014, 12:45:16 AM

0 Members and 1 Guest are viewing this topic.

nullified

Sometimes inspiration can come from very unique places such as a conversation with a cabbie, or a radio commercial. Other times, it just comes.

The following is just such an inspiration I had. I can't recall why I had it, but I suspect it was in connection with an analysis I was doing on a system from the book  "Monte Carlo Anecdotes."

Or it may have been nothing.

Anyways, I say that, not merely to make a point about being prepared to flesh out ideas no matter where you are (I believe I was getting ready to drift off to sleep for the night when the idea hit) - but also as a call for us professional punters to go back to the basics.

There is, I believe, a wealth of treasure in many of the old systems. They merely need to be unearthed again, analized, and reworked.


The Game is Afoot

The idea began, as such ideas often do, with a simple question: how can I survive an impossible run on an Even Chance bet?

Obviously there are many ways to do this; but generally most require nearly limitless reserves of cash, nerves of steel - and a lot of luck.

Since I don't have any of those in abundance I had to figure something else out.

It was as I was mulling this question around in my head: How can I survive an impossible run of an Even Chance bet - that a thought struck me.

What if I didn't have to survive the whole run? What if all I needed was to survive the end?

Immediately I thought that, while that might sound ridiculous, if I divided up this "impossible run" into five or six "mini runs" - I might be able to somehow survive each one.

But how to do this? What would that look like?


It's Elementary, My Dear

In my mind I had thought that if I broke up a run of 24 straight Even Chances into 4 'sessions' of 6-spin games, then all I needed to do was figure out a way to win one game - even if that meant I only won one spin out of six.

Then, having done that, I just repeated this process [whatever it was] 3 more times until the "impossible run" ran it's course and I somehow came out ahead.

Of course at this time these were all just thoughts running around in my mind.

I had no idea how to actually do any of this.

I was playing around with numbers, odds and percentages one day when another thought occurred to me.

What if I created a bet whose win amounts were higher in average than the loss amounts?  Is there some way that I can incorporate this idea into a 6-spin session?

Excited, I began to play around some more with this concept. Using sessions of 6 spins each, working out a bet that had a higher average in profits than losses accrued.

Somehow, I managed to work out the following bet.


The Adventure of the Noble Blacks

I had been playing around on the Blacks with my notepad and calculator when something jumped out at me. If I wagered $1, $3, $7, and then $17 on Blacks my profits would be +1, +2, +3 and +6 respectively. Immediately I saw the significance of this.

This meant that, no matter where in this session of 4 spins that I won, I would always profit a little more as I went on.

My goal then was simple. To win just once in 4 spins - and to come away with profits that outpaced my losses in the long term.

Sometimes I would win on the first spin (+1), sometimes on the second spin (+2), sometimes on the third (+3), and sometimes on the fourth (+6).

Averaged out then, $1 + $2 + $3 + $6 = $12 ÷ 4 = $3.

So no matter where in the session I won, I averaged $3.

But what about when I didn't win at all?

Well, my wagers were $1, $3, $7 and $17. So if I add them all up I get a total loss of -$28.

But were my average wins enough to eclipse my losses? Or would I be seeing them succumb to the usual -2% standard expected loss?

To find out, I needed to know what the odds were of getting 4 losses in a row.  Whatever this was, I would be able to calculate my expected losses in the long term.

4 losses in a row against my Blacks are 0.5135 x 0.5135 x 0.5135 x 0.5135 = 0.0695, or about 7% of the time. In 100 sessions I can expect losses of about 7% x -$28 = -$196.

And what about my wins? How much can I expect from them?

Simple. If 7% represented my losses, then 100% - 7% must be my wins. So my win rate is 93% x +$3 = +$279.

And +$279 - $196 = +$83.

So essentially I'm wagering $28 per session to win an average of about $3 per session.

Doesn't make a lot of sense, but the numbers certainly add up.


A Game of Shadows

So does this trick work all the time? Yes and no

At least, no - in the way I was trying to play it.

I figured that if Blacks were going to profit me $83 for every 100 sessions, then 5 more Even Chances could only be 5 times the profit!

Right?

Wrong.

In fact, playing anymore than one Even Chance never seemed to work because the different Even Chances would always be cancelling each other out. When I tried playing only 3 and then 2 different kinds of Even Chances, it still didn't work. They would always seem to be working against each other.

Another problem I encountered was what to do after I won.

After all, shouldn't I just start a new 4 spin session all over again? The wheel didn't have a memory so it shouldn't matter, right?

Wrong again.

I found that, in order to keep the odds accurate, I had to finish playing out the session even after I won. But I did so without wagering on anything.

When I thought about it, it finally made sense. In reality I was betting that, for each session of 4 spins, I would get more wins than losses.

But if I used a 2 spin session, then a 4 spin session, then a 1 spin session, then back to a 4 spin session - and so on and so forth - then my odds of 1 win in 4 spins went completely out the window.

I experimented with other ideas too.

Betting on the Dozens, the progression would be 1, 2, 3, and 7.  The average win was +$4. My win rate was 79% and the loss rate was 21%. My net profit was +$43.

Betting on the Lines, the progression would be 1, 1, 1, 3. The average win was $6. The win rate was 51% and the loss rate was 49%. Net profit for a line was +$12.

Betting on the Streets, the progression would be 1, 1, 1, 2. The average win was $12.75. The win rate was 29% and the loss rate was 71%. Net profit for a street was +$1.

As compelling as all these other games were, I could find no way of combining any of them in order to increase my profits.

It's probably just as well.

I went back to playing just the one Even Chance at a time. Tedious.

Boring.

Disheartening when I would lose a full $28.

But somehow, my bankroll slowly climbed. Mostly up.  Then down drastically a couple times. But always back up.


To play:

1. Choose an Even Chance and follow it.

2. After it hits once, use the progression: 1, 3, 7, 17.

2a. If you win finish off the session to the 4th spin, without wagering. Start a new session using any Even Chance you like.

3. If you lose, your bankroll decreases by $28. Sorry about that. Start a new session. Pray against a monster losing session.

4. Cash out after you've made 20% profit.  Or not.

leroy

Nice concept Nully,

I know 20 spins is hardly a conclusive test, but I finished up 9 units.

For some reason I always have better luck with Odd as my E/C.

E -1
E -4
E -11
O +1

E 0
O +3
E +3
E +3

E +2
E -1
O +6
O +6

0 +7
E +7
O +7
O +7

E +6
O +9

nottophammer

For some reason I always have better luck with Odd as my E/C.

Could it be its distribution round the wheel,

beretta28

nullified
your calculation is completely wrong.
If it is right,all Casinos will be bankruptcy in a few days.

It's wrong to calculate 1+3+7+17,  i.e +1,+2,+3,+6=12:4=3

The good calculation is:
you have 50% probability to win the first spin(+1),25% probability to win the 2nd spin(+2),12,5% probability to win the 3rd spin(+3) and 6,25% to win the 4th spin (+6).
The average of all that is not 3,unluckily for you...and me.
It's ok that you lose all bkr (28 units) 1/15 attacks,i.e 7% roughly.

Calculate again with what I wrote and you'll stop playing like this.
Bye


nullified

Quote from: beretta28 on February 12, 2014, 05:06:46 AM
nullified
your calculation is completely wrong.
If it is right,all Casinos will be bankruptcy in a few days.

It's wrong to calculate 1+3+7+17,  I.e +1,+2,+3,+6=12:4=3

The good calculation is:
you have 50% probability to win the first spin(+1),25% probability to win the 2nd spin(+2),12,5% probability to win the 3rd spin(+3) and 6,25% to win the 4th spin (+6).
The average of all that is not 3,unluckily for you...and me.
It's ok that you lose all bkr (28 units) 1/15 attacks,I.e 7% roughly.

Calculate again with what I wrote and you'll stop playing like this.
Bye

I have no idea what it is you are trying to prove here.  My post was very straight forward and thorough. Please re-read it and make sure you understand the concepts. Otherwise others will be confused.

beretta28

Members of this Forum are confused with your post.
The average win is 1,86(zero effect excluded,otherwise is worse) units and not 3 units as you said.
So the good calculation is 100% -7%=93% x 1,8= + 167,4
7% x 28= - 196
Very simple,even a child could calculate it.
May I ask you what is your education or mathematical background?

nullified

Quote from: beretta28 on February 14, 2014, 08:06:18 AM
Members of this Forum are confused with your post.
The average win is 1,86(zero effect excluded,otherwise is worse) units and not 3 units as you said.
So the good calculation is 100% -7%=93% x 1,8= + 167,4
7% x 28= - 196
Very simple,even a child could calculate it.
May I ask you what is your education or mathematical background?

Okay, let me explain again.

The idea is that for every 4 spins, I want to win once, and only once. The "session" is 4 spins, win or lose. If I bet $1 on the first spin and win, my profit is $1 - then I continue spinning 3 more times without betting and finish off the session.

If I lose that first bet (so I'm down -$1) I then bet $3.  If I win on this second bet, my profit is $3 - $1 = $2 - and I continue to spin 2 more times without betting and finish off the session.

If I lose both the first and second bets (so I'm down -$1 + -$3 = -$4) I then bet $7.  If I win on this third bet, my profit is $7 - $4 = $3 - and I spin one more time without betting to finish off the session.

If, however, I've lost the first 3 bets (so I'm down -$1 + -$3 + -$7 = -$11) I then bet $17.  If I won on this last bet my profit is $17 - $11 = $6.

And if I lose my four bets in a row, I'm down -$28; I then start a brand new session of 4 spins and I use the same progression of 1,3,7,17.

========================================
Now then, since I'll never know at which spin I'm going to win, I need to know what to expect as an average win amount. To do this, I add the four actual profit amounts together (not the win amounts or the bet amounts) and divide by 4 (because there are 4 spins in a session). So, $1 + $2 + $3 + $6 = $12.  And $12 profit divided by the 4 spins = $3.  So my average profit is $3 per session.

But how often can I expect to win my average of $3 compared to how often I'll lose my $28?

Remember, just 1 win in any session nets me profit of either $1, $2, $3 or $6.

How do I figure out the expected number of times I going to lose?

I take the number of losing outcomes - in this case, 19 (because there will be 18 losing even chances + the zero) and divide by 37 (numbers on the wheel) and I get 0.5135 (just over 51%). That is the odds of me losing one spin.  Now what about the odds of losing all 4 even chances? I take that same number - 0.5135 - and times it by itself 4 times (for each of the 4 spins). Thus, 0.5135 x 0.5135 x 0.5135 x 0.5135 = 0.0695.  This is 6.95% or, rounded up, about 7%.

This obviously means that for any given session of four spins, I can expect to lose my $28 about 7% of the time.

What this also means is that, if I'm only losing 7% of the time, then I'm winning 93% of the time. And when I win, I average about $3.

Now then, how much do the wins and the losses come to?

I take my full session loss amount - $28 - and multiply it by the expected loss rate - 7% - and I get -$196.

Then I take my average win amount (since my profits are going to vary between $1, $2, $3 and $6) of $3 and multiply it by the expected win rate - 93% - and I get +$279.

So my expected wins - my expected losses = $279 - $196 = +$83.

========================================

I hope this clears things up.



beretta28

I had well understood.
You insist in your basic mistake.
The probability to win at the first spin is not the same(it's higher) to win  at the second,the probability to win at the second is not the same(it's higher) to win at the third and so on...
The odds are the same for the four spins,you are right,but not the probability of winning.
So to sum 1+2+3+6 and divide by four ,it's a mathematical ,better it's a big "a statistical mistake".
The good calculation is (+1 x 50%) + (+2 x 25%) + (+3 x 12,5%) + (+ 6 X 6.25%) :4= much less than your 3
All your other calculations are correct,but the final result is that ,at the end,you'll lose money with your approach.
But it doesn't matter,trust in your conclusions,if you like.

ehtelgaeb

And on top of all this, how many of us can really win one in four on a regular basis?

nullified

edu/~thiriot/Math1040/notes/probability_notes.html

Scroll to the bottom and read the section called "The At Least Once Rule."

This is what I am doing in my method... Trying to win just once.

beretta28

I have read the section you mention.
It's says exactly what I'm saying and,I'm afraid,you have not understood.
To win at least once in four spins is 15/16,that is 93,75%( roughly less Zero in a roulette spin).
The article confirms that probabilities are different of winning at the 1st,2nd,3rd and fourth spin

nullified

QuoteThe article confirms that probabilities are different of winning at the 1st,2nd,3rd and fourth spin

Hmmm... I can only assume that the part of the article to which you are referring is this:

The Multiplication Rule

Now we'll learn how to determine probabilities for two events happening by multiplying probabilities.



Multiplication Rule (finding Joint Probabilities)

The chance that two things will both happen equals the chance that the first will happen, multiplied by the chance that the second will happen (given that the first has happened).



Example:  What is the probability of getting two heads in a row when you flip a fair coin?

Solution: The chance of a head on the first toss equals 1/2.  The first toss doesn't in any way affect the outcome of the second toss.  On the second toss, the probability of heads is also 1/2.  So the probability of two heads in a row is:



P(get head on first toss and head on second toss) =

= P(heads on first toss)*P(heads on second toss)

= (1/2)*(1/2) = 1/4 = 0.25

Example:  Let's pretend that each of you have an envelope containing three different colors of paper squares  (red, blue, and green).  What is the probability of drawing a red square and then a blue square (don't replace the squares after you draw them)?

            Solution:  What is the probability of drawing a red square on the first draw?  About 1/3 of you would draw a red square (the probability is 1/3).

P(draw red on first draw) = 1/3

Don't replace the red back in the envelope.  Now, what is the probability of drawing a blue square from what is left?  (There are only two squares left now.)

P(draw blue on second draw) = 1/2



P(draw red on first draw and draw blue on second draw) =

            = P(red on first draw)*P(blue on second draw given that red was drawn first)

            = (1/3)*(1/2) = 1/6

(The outcome of the first draw affects the outcome of the second draw.)


But I'm not calculating a joint probability here.  The Multiplication Rule is "The chance that two things will both happen equals the chance that the first will happen, multiplied by the chance that the second will happen (given that the first has happened)." I'm not calculating 4 wins in a row. Just one in 4.

Or perhaps you were referring to this:
Either/Or Probability for Mutually Exclusive Events

Consider two mutually exclusive events, A and B.  The probability that either A or B occurs is:

P(A or B) = P(A) + P(B)

This principle can be extended to any number of mutually exclusive events.  For example, the probability that either event A, event B, or event C occurs is:

P(A or B or C) = P(A) + P(B) + P(C)

As long as A, B, and C are all mutually exclusive events.


Either/Or Probability for Not Mutually Exclusive Events

Assume events A and B are not mutually exclusive.  The probability that either A or B occurs is:

P(A or B) = P(A) + P(B) – P(A and B)



My calculation for these probabilities are mutually exclusive.

So when you said So to sum 1+2+3+6 and divide by four ,it's a mathematical ,better it's a big "a statistical mistake".
The good calculation is (+1 x 50%) + (+2 x 25%) + (+3 x 12,5%) + (+ 6 X 6.25%) :4= much less than your 3


You seem to be thinking that these probabilities are not mutually exclusive, but which of course, they are. If I was trying to win 4 in a row, then you would be perfectly correct. But to say that I have a 25% chance of profiting $2 on the second spin after the first spin lost makes no sense. There is an equal probability profiting $1 as there is $6.

I understand (I think) what it is you are trying to prove.  If I wanted to win 4 in a row the probability is (18/37) x (18/37) x (18/37) x (18/37) =0.48654 =0.05601, or 5.6%. That is the probability of winning 4 EC's in a row. In which the multiplication rule comes into effect, since the probability of the subsequent spins must be multiplied, given that the spin before it has happened.

Which is not what I am doing.

Your concern over my "averaging" the profits is, I think, where your real problem lies.

Why do I insist on averaging the profits, and not attempting to calculate the individual probabilities for profiting $1, or $2, or $3, or $6?

Because I'm not trying to win exactly $1, or $2 or exactly anything.

I'm content to win anything.

Even if I wanted to calculate the probabilities for exact profits, I couldn't because a) all my wagers are mutually exclusive(a 4 spin session is either a win or loss - not 2 wins or more, just one), and b) I'm stopping my wager as soon as I win and wagering on nothing until the session is ended. If I continued to wager on each spin, then all your calculations and concerns would be well warranted. But wagering on nothing for 1 to 3 spins in a session has serious consequences for any long term probabilities.

Thus, the only way to calculate wins versus losses is to find the average win per session. And because I stop wagering on 1 win, I can do exactly that, without being overly concerned about how much I've won.

Now, I've gone to a lot of trouble to make my case here. If you're still not convinced, would you be kind enough to give me your basis for this:
QuoteSo to sum 1+2+3+6 and divide by four ,it's a mathematical ,better it's a big "a statistical mistake".
The good calculation is (+1 x 50%) + (+2 x 25%) + (+3 x 12,5%) + (+ 6 X 6.25%) :4= much less than your 3
?

Richard

This has defiantly got something going for it  :)

nottophammer

is this played only live wheel ?

nottophammer

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