Author Topic: About virtual losses on Even chances  (Read 11721 times)

palestis

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About virtual losses on Even chances
« on: August 08, 2014, 02:28:08 AM »
Quote
Show me the probability of hitting black in any of 5 consecutive spins
Palestis

There's about a 97% chance of hitting.   (1) - (18/38 x 18/38 x 18/38 x 18/38 x 18/38 ) = 97.275
REAL
Lets deal with that for now.
You admit that commencing a 5 bet series on BLACK there is a 97% probability that you will hit it, at any time during those 5 bets.
And 3% that you will lose. With a Martingale progression the bets will look like this (provided you go all the way to the 5th spin).
$5  $10  $20  $40 $80. Total risk $155 to win $5.
JUST KEEP IN MIND THAT THE PROBABILITY TO WIN WAS DETERMINED TO BE 97%
Question #1: Is there any reason that, if you lost the first 2 spins, the ESTABLISHED probability will change in any way for the 3 spins remaining?  NO. A series bet PROBABILITY does not guarantee that you will hit the target in the first 2 or 3 spins. You have to go thru all 5 spins.
QUESTION #2: What is the probability that you will hit a black in the remaining 3 spins? It still is 97%. Anything else will be contradictory to the principle of probability of series
Now let's say that your betting series won't be like 5 10 20 40 80. Instead it will look like $0.01- $0.01 - $5- $10- $20. Total risk $35.02.
WITH A STIPULATION: That the first 2 bets have to be losing bets (enter VIRTUAL LOSS).
Again will the probability change for the 3 remaining just because we lost the first 2 bets?  NO NO NO Still is 97%.
The only thing that changed is the $ RISK which was reduced from $155 to $35. And some added waiting period waiting for 2 REDS.

ACTUALLY YOUR PROBABILITY OF 97% IS WRONG.
If it was correct, if you played 100 times you will win $485.  (97x$5).
and you will lose $465 (3x$155). So you will always be ahead $20
To makes things simple it's not 97%. Accurately figured, if you play in this fashion (5  10  20  40  80) in the long run you'll will be losing an amount equal to the HE (house edge).
However I just proved to you that if you play with the first 2 spins in the virtual loss mode you will always be winning in the long term.

Even if I downgrade the probability of the series to 90% win rate, if you you play the system 100 times you will be winning $450 (90x$5) and you will be losing $350 (10x$35). A constant profit of $100. Simply by switching the first 2 spins in the virtual loss mode. And if you changed it to 3 virtual losses then the profit will be even more impressive.
The downside is the waiting period. But if I know that I will always be winning, waiting doesn't bother me at all.
We wait from 9 to 5 to make a living. Y not wait when we work with roulettes.

Ps: Any argument that after 2 reds the probability for a B is 50%. is invalid.
Lets not forget that I made a decision in advance to bet Black 5 times. Therefore I REMAIN LOCKED IN THE PROBABILITY OF SERIES. NOT THE PROBABILITY OF EQUAL CHANCES TAKEN 1 SPIN AT A TIME.
which YOU ADMITED it was around 97%.  Meaning that I have a probability of 97% to hit it IN ANY OF THE NEXT 5 SPINS. LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series.

« Last Edit: August 08, 2014, 07:32:46 AM by palestis »


 

BlueAngel

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Re: About virtual losses on Even chances
« Reply #1 on: June 20, 2015, 09:04:13 AM »
Quote
LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series. Palestis[size=78%]  [/size]

You are partly right and wrong Jim.
Let me explain why,perhaps the probability for hitting 1 out of 5 instead of hitting 1 in 1 it's not the same as you rightfully claimed.This fact can be confirmed by everyone and it's been called binomial probability,there are plenty of binomial calculators on the internet free to calculate any possibility within any amount of trials,whether you call it Black/Red or Heads/Tails or Pass/Don't Pass...etc.
BUT you are mistaken non the less, by avoiding to bet the 2 first bets of the series of,let's say, 5 consecutive bets,you are not avoiding only lost bets but also bets that you could have won,this is what you are missing.
This is crucial when you calculate the totality of the possibilities,for example:
If your success rate to win only once within 5 trials is 96%,this means that you are going to 96 times out of 100,thus you would lose 4 x (5+10+20) = 4 x 35 = -140 BUT your profit won't be 96 x 5 because some wins from the total of 96 would be during the first and second trial which you are observing rather than betting.
To be exact,roughly 74 times out of the 96 winning bets would be wasted opportunities because you pass them as virtual bets,therefore 96 - 74 = 22 x 5 = 110 and 140 - 110 = -30
Because of the variance you could do better or worse than the figures above,what I showed is the average expected values.
The aftermath is that you cannot sidestep probability.
Just the facts.
 

palestis

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Re: About virtual losses on Even chances
« Reply #2 on: June 20, 2015, 11:49:20 AM »
Quote
LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series. Palestis[size=78%]  [/size]

You are partly right and wrong Jim.
Let me explain why,perhaps the probability for hitting 1 out of 5 instead of hitting 1 in 1 it's not the same as you rightfully claimed.This fact can be confirmed by everyone and it's been called binomial probability,there are plenty of binomial calculators on the internet free to calculate any possibility within any amount of trials,whether you call it Black/Red or Heads/Tails or Pass/Don't Pass...etc.
BUT you are mistaken non the less, by avoiding to bet the 2 first bets of the series of,let's say, 5 consecutive bets,you are not avoiding only lost bets but also bets that you could have won,this is what you are missing.
This is crucial when you calculate the totality of the possibilities,for example:
If your success rate to win only once within 5 trials is 96%,this means that you are going to 96 times out of 100,thus you would lose 4 x (5+10+20) = 4 x 35 = -140 BUT your profit won't be 96 x 5 because some wins from the total of 96 would be during the first and second trial which you are observing rather than betting.
To be exact,roughly 74 times out of the 96 winning bets would be wasted opportunities because you pass them as virtual bets,therefore 96 - 74 = 22 x 5 = 110 and 140 - 110 = -30
Because of the variance you could do better or worse than the figures above,what I showed is the average expected values.
The aftermath is that you cannot sidestep probability.
Just the facts.
No Angelo.
When we say we will play 100 times, we don't count the times where you win in the first 2 bets. When that happens the system is abandoned. With no loss or gain. It only means lost time.
When we say 100 times we mean the 100 times where the first 2 bets were lost virtually.
That means it might take 500 situations to become 100 playable situations. It takes patience for that to happen.
Betting all 5 times to take advantage of any winning opportunities here is what will happen.
We have to place 5  bets (5+10+20+40+80)= 155 to risk.
At 96% winning rate we expect to win 5x96=480 units.
But the very small 4% losing rate will result in 4x155=620 units lost.
Playing all spins will always result in a loss.  A big loss even with an impressive 96% win rate.
We are only concerned with the situation where the bets will be  0+0+5+10+20. If that's not happening we wait till it happens. Y the pressure to bet?
The risk now is 35 units while all probabilities remain the same.
The probability doesn't change in getting 1 win out of 5 trials, if we see it as  (A) 5,10,20,40,80 or we see it as (B) 0,0,5,10,20. Probability has to do with number of trials. Not money amounts. The difference is that we are only looking to find a situation (B). And we abandon a situation where a win occurs in the first 2 bets. That's how being patient affects the results. We always wait for the roulette to come to OUR TERMS. We shouldn't  follow her terms. That's the secret of success.
Don't forget when you were here, we never lost at Mohegan or Twin River betting the opposite after seeing 6 or more of the same EC. Remember?
Those 6 black on the board and playing red 3 times meant this:  0+0+0+0+0+0+5+10+20.
A 9 bet series where the first 6 were cost free. And I thought you liked this method very  much.
And if you have unlimited bank roll you may never lose. Ever.
How many times do you think you will lose in a row,  betting 3 red after seeing 6 black?
That means in order to lose, every time you see 6 black, they must become 10 black in a row.
I still haven't seen that happening 3 times in a row, never mind indefinitely.
So with a large bank roll you can never lose.
There is nothing wrong to have $15,000 bank roll and aim to win $300.
Most business have to invest a lot more than $15,000 to make less than $300. Y not in roulette?
« Last Edit: June 20, 2015, 11:53:18 AM by palestis »
 

Jesper

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Re: About virtual losses on Even chances
« Reply #3 on: June 20, 2015, 12:03:42 PM »
It may work well if we find a thoughtful Wheel! It may work anyhow, but better on a thoughtful Wheel.
 

BlueAngel

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Re: About virtual losses on Even chances
« Reply #4 on: June 20, 2015, 12:33:33 PM »
Quote
LOSING THE FIRST 2 OR 3 OR 4 SPINS SHOULD NOT HAVE ANY EFFECT IN THE ALREADY ESTABLISHED PROBABILITY OF SERIES.
It still remains in effect for the entire duration of the series. Palestis[size=78%]  [/size]

You are partly right and wrong Jim.
Let me explain why,perhaps the probability for hitting 1 out of 5 instead of hitting 1 in 1 it's not the same as you rightfully claimed.This fact can be confirmed by everyone and it's been called binomial probability,there are plenty of binomial calculators on the internet free to calculate any possibility within any amount of trials,whether you call it Black/Red or Heads/Tails or Pass/Don't Pass...etc.
BUT you are mistaken non the less, by avoiding to bet the 2 first bets of the series of,let's say, 5 consecutive bets,you are not avoiding only lost bets but also bets that you could have won,this is what you are missing.
This is crucial when you calculate the totality of the possibilities,for example:
If your success rate to win only once within 5 trials is 96%,this means that you are going to 96 times out of 100,thus you would lose 4 x (5+10+20) = 4 x 35 = -140 BUT your profit won't be 96 x 5 because some wins from the total of 96 would be during the first and second trial which you are observing rather than betting.
To be exact,roughly 74 times out of the 96 winning bets would be wasted opportunities because you pass them as virtual bets,therefore 96 - 74 = 22 x 5 = 110 and 140 - 110 = -30
Because of the variance you could do better or worse than the figures above,what I showed is the average expected values.
The aftermath is that you cannot sidestep probability.
Just the facts.
No Angelo.
When we say we will play 100 times, we don't count the times where you win in the first 2 bets. When that happens the system is abandoned. With no loss or gain. It only means lost time.
When we say 100 times we mean the 100 times where the first 2 bets were lost virtually.
That means it might take 500 situations to become 100 playable situations. It takes patience for that to happen.
Betting all 5 times to take advantage of any winning opportunities here is what will happen.
We have to place 5  bets (5+10+20+40+80)= 155 to risk.
At 96% winning rate we expect to win 5x96=480 units.
But the very small 4% losing rate will result in 4x155=620 units lost.
Playing all spins will always result in a loss.  A big loss even with an impressive 96% win rate.
We are only concerned with the situation where the bets will be  0+0+5+10+20. If that's not happening we wait till it happens. Y the pressure to bet?
The risk now is 35 units while all probabilities remain the same.
The probability doesn't change in getting 1 win out of 5 trials, if we see it as  (A) 5,10,20,40,80 or we see it as (B) 0,0,5,10,20. Probability has to do with number of trials. Not money amounts. The difference is that we are only looking to find a situation (B). And we abandon a situation where a win occurs in the first 2 bets. That's how being patient affects the results. We always wait for the roulette to come to OUR TERMS. We shouldn't  follow her terms. That's the secret of success.
Don't forget when you were here, we never lost at Mohegan or Twin River betting the opposite after seeing 6 or more of the same EC. Remember?
Those 6 black on the board and playing red 3 times meant this:  0+0+0+0+0+0+5+10+20.
A 9 bet series where the first 6 were cost free. And I thought you liked this method very  much.
And if you have unlimited bank roll you may never lose. Ever.
How many times do you think you will lose in a row,  betting 3 red after seeing 6 black?
That means in order to lose, every time you see 6 black, they must become 10 black in a row.
I still haven't seen that happening 3 times in a row, never mind indefinitely.
So with a large bank roll you can never lose.
There is nothing wrong to have $15,000 bank roll and aim to win $300.
Most business have to invest a lot more than $15,000 to make less than $300. Y not in roulette?

You are discarding the winning series when you don't bet when a win occur on the first or second virtual bet.
That's why those lost opportunities must be deductedfrom the total of 96.
Perhaps it's not a problem with time,but it's a practical problem when your expected wins becoming less because of the missed opportunities.
Yes I remember that we have won,it's not bad in lack of a better method.
 

palestis

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Re: About virtual losses on Even chances
« Reply #5 on: June 21, 2015, 02:11:13 AM »
Quote
Quote from: BlueAngel on June 20, 2015, 09:04:13 AM

You are discarding the winning series when you don't bet when a win occur on the first or second virtual bet.
That's why those lost opportunities must be  deductedfrom the total of 96.
Perhaps it's not a problem with time,but it's a practical problem when your expected wins becoming less because of the missed opportunities.
Yes I remember that we have won,it's not bad in lack of a better method.
Betting the entire series with Martingale will always result in a loss. Not necessarily immediately but eventually.
In the example we are discussing If you plan to bet BBBBB (black 5 times), the only time you will continue all the way to the 5th spin is only if the first 2 spins ( that you have  bet) are RR. That means you have already bet $0.01 twice and you lost. Then you bet 3 more times 5-10-20.
But you have to start the bets from the time you go to the roulette table, ignoring  what has come up before. Because it doesn't count. What counts is what happens after you start your guessing.
If you see a roulette with RR already spun,  it doesn't count. And betting 3 more bets  it becomes a 3 series bet. NOT 5. Ready made situations where you had no part in the betting process don't count. (except if you see long run with the same EC repeated).
So for statistical purposes, the 100 spins will only be the ones where RR were the first 2 results, provided you placed those 2 bets with a penny. Any time you get B or BB you don't count it in the 100 spins. You simply wait until you see RR (after betting BB of course in your head, or virtually or 1 penny bet, they are all the same thing). NEVER A READY MADE SITUATION OF RR, showing on a roulette board down the hall.
Because the power of your prediction did not play a roll in that result. It just happened to be there. And therefore it is a past result that doesn't count
So any time you win with a virtual bet doesn't hurt you. You don't win but you don't lose either.
The only cost is your time.
Without virtual losses (where the player had a part in the betting process to produce that loss), 
no system can win. It will eventually fail. The only exception maybe is if you follow a trend, but it requires experience to recognize trends.
Trends have a mind of their own and they come and go when they feel like it. They are their own boss and don't account to nobody. At least while they last. And they don't obey any probability. 
Unless players are willing to endure extreme patience and incorporate virtual losses, they may never see themselves on the winning side.
That's the best advice I can give to anyone.
Thankfully I forced myself to embed those 2 principles, and since then I can't remember the last time I lost.

« Last Edit: June 21, 2015, 02:19:28 AM by palestis »
 

kav

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Re: About virtual losses on Even chances
« Reply #6 on: June 21, 2015, 02:43:46 AM »
Hi Palestis,

Let's take a sample of 1000 spin sequences. If we bet Black, according to probability, we expect:
500 1st spin wins
250  2nd spin wins
125 3rd spin wins
62 4th spin wins
31 5th spin wins
and 31 5th spin losses

Now let's take 250 spin sequences. If we bet Black, according to probability, we expect:
125 1st spin wins
62 2th spin wins
31 3th spin wins
and 31 3rd spin losses

So theoretically there is no difference, between betting on the first two spins or not.

There could be a difference though, if we believe that somehow, if we lost the first spins, then Black has a slightly higher chance to return to equilibrium than Red to continue the deviation. (what Real would call Gambler's fallacy, but anyway). It his case however, how can we explain the difference between a "ready made situation" and virtual bet situation?  If we believe in equilibrium both cases are equal, no?
« Last Edit: June 21, 2015, 02:55:19 AM by kav »
 

BlueAngel

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Re: About virtual losses on Even chances
« Reply #7 on: June 21, 2015, 03:30:32 AM »
Yes Kav,I agree with you.This is just the average expectation regarding the theory.
Someone could do better or worse because of the variance and deviations.
This example inspired me to think another matter,let's say that 100 players are playing exactly the same system simultaneously in different locations around the world OR the same amount of gamblers are applying the same system with 1 hour difference on the same casino.
Would the results be the same for everyone because they are using the same method/system?
Would it be correct to assume that everyone or nobody would lose because they are using the same method/system??
My conclusion regarding this matter is more or less,the most of the matters are subjective because there are not absolute definitions which could include every possibility.
« Last Edit: June 21, 2015, 03:32:06 AM by BlueAngel »
 

kav

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Re: About virtual losses on Even chances
« Reply #8 on: June 21, 2015, 03:44:05 AM »
BA
What you describe, is a well known way to explore different possibilities that can result from the same system. There is one category of statistical programs that is called Monte Carlo simulators that do just that. Instead of testing billions of "spins", or one long sequence of outcomes, they test the same initial conditions in many sequences. They are mostly used in finance. And yes the results can vary greatly even if we use the same "system".
 

BlueAngel

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Re: About virtual losses on Even chances
« Reply #9 on: June 21, 2015, 04:11:22 AM »
BA
What you describe, is a well known way to explore different possibilities that can result from the same system. There is one category of statistical programs that is called Monte Carlo simulators that do just that. Instead of testing billions of "spins", or one long sequence of outcomes, they test the same initial conditions in many sequences. They are mostly used in finance. And yes the results can vary greatly even if we use the same "system".

Therefore a method/system/strategy could not transform everyone in to winner.
Despite Mr Blackey had success by using his system,which I don't doubt,cannot transform everyone to long term winner.
Besides the huge variety of possibilities,there are more specific elements one has to consider,for example if I want to back up 16 sleepers,why this 16 numbers have to be in 4 corners and not in 8 splits or 5 streets and one single or 2 lines and a corner or a dozen and a corner or 16 straight up numbers or...etc
I'm not criticising the fact that his method preferred inactive numbers rather than active (hot),but the fact that he formed his principle in a very certain way which could be an unnecessary obstacle towards success.
The path to financial well being is full of hurdles,we don't have to make it harder.
 

palestis

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Re: About virtual losses on Even chances
« Reply #10 on: June 21, 2015, 11:49:10 AM »
Hi Palestis,

Let's take a sample of 1000 spin sequences. If we bet Black, according to probability, we expect:
500 1st spin wins
250  2nd spin wins
125 3rd spin wins
62 4th spin wins
31 5th spin wins
and 31 5th spin losses

Now let's take 250 spin sequences. If we bet Black, according to probability, we expect:
125 1st spin wins
62 2th spin wins
31 3th spin wins
and 31 3rd spin losses

So theoretically there is no difference, between betting on the first two spins or not.

There could be a difference though, if we believe that somehow, if we lost the first spins, then Black has a slightly higher chance to return to equilibrium than Red to continue the deviation. (what Real would call Gambler's fallacy, but anyway). It his case however, how can we explain the difference between a "ready made situation" and virtual bet situation?  If we believe in equilibrium both cases are equal, no?
Kav
This is a very important subject as it may hold the answers to winning consistently. Not only in this particular situation, but in many other situations. Theory and reality can be 2 different things when money enters the picture.
What I'm describing here is the betting of a series of bets, in this case 5 bets. 
What I know is that the probability to win AT LEAST ONE TIME in those 5 bets is 1-(1/2 x1/2x1/2 x 1/2x1/2) or about 96% of the time. And the probability to lose all 5 bets about 4%.
I don't know when I will hit the target. It could be the 2nd bet or it could be the 5th bet. I don't see this process as a 50:50 chance taken individually each time. It's a serial process that has a beginning and an ending. Where the END RESULT counts, and not each result individually.
The only thing I know is that during those successive 5 bets, at least one of them will be a hit by 96% of the time. If that happens we stop. The aim is one win. Not more.  But to make a profit, we have to double after each loss, otherwise it will not produce a profit if there is a hit during those 5 bets. Except if it wins with the very first bet. Yet if we play that way all the time one thing is for sure. That eventually we will lose money. Even at the impressive 96% winning rate. The math behind it combined with the house edge are elementary.
Without the variable of money we don't lose. In fact we will be winning by an overwhelming percentage (96%).
Enter the money and the winning turns into losing. Because it is the $ value that becomes the deciding factor and not the probability. That's y money management is so important.
VIRTUAL BET  to me means one thing and one thing only. That it is an actual bet. You actually have to stand right in front of the roulette and make the bet. Whether you make it with your mind or with a penny (if it was allowed ). The only difference is that this bet has no substantial money value. However it has the same statistical value as if money was wagered.
The power of human guessing cannot be underestimated. But you have to be involved in the guessing to qualify as a virtual bet. Seeing RR in some roulette and use that as 2 virtual losses doesn't count. Therefore placing 3 more bets after seeing that, is a 3 bet series. And it has another statistical value. The number of bets in the series is the actual number of guesses you will act on. Including bets that have no money value.

To make things even simpler, after a long empirical research in live roulette numbers (not the software testing garbage), you will find that a random guessing of an EC will most likely produce a profit between guesses 3-5. That is spin 3-4-5. This is the most frequent winning range when money is involved.  Not 1-3 and not 2-4.. That's what years long observation is telling me. And I can't deny history. And acting on empirical experience I can easily say that winning in roulette is a piece of cake if you are willing to endure the required patience. 
Now there is a great difference between seeing 2 reds in a row and 8 reds in a row on a score board.
2 reds is not a great imbalance that you can take advantage of, as it happens much more frequently than a single red. But 8 is a different story. And after 8 EC's in a row  (that's an already made situation that has value), you will find hat most of the time it turns to the opposite within the next 3 spins.Coming again from empirical research.
The moral of the story is that long time empirical research is what counts.
Those who bypass this lengthy process, by demanding easy money with the least amount of brain labor are destined to lose. From experience these people that claim to have the perfect system (and I know a lot of them), are the ones that are standing with empty pockets. And they all blame it on low bank roll. Gambling has evolved into a big business because it's based on a simple premise. The false hope of easy money, without any labor involved. That is the real "GAMBLER'S FALLACY" and not what some other people are talking about.
 

kav

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Re: About virtual losses on Even chances
« Reply #11 on: June 21, 2015, 12:13:41 PM »
Palaistis,

According to your last post, if I read it correctly, we don't need to lose the first two virtual bets on Black and then seriously bet Black for the next 3 spins.
Any series of two lost guesses (if for example we virtually bet R B and lost both bets) can be followed by any 3 (RRR, RBR, RBB or whatever) serious bets with Martingale.

Is this correct? Do you refer to series of the same color for ease of understanding or does that play a role in your way of thinking?
 

palestis

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Re: About virtual losses on Even chances
« Reply #12 on: June 21, 2015, 02:27:26 PM »
Palaistis,

According to your last post, if I read it correctly, we don't need to lose the first two virtual bets on Black and then seriously bet Black for the next 3 spins.
Any series of two lost guesses (if for example we virtually bet R B and lost both bets) can be followed by any 3 (RRR, RBR, RBB or whatever) serious bets with Martingale.

Is this correct? Do you refer to series of the same color for ease of understanding or does that play a role in your way of thinking?
Good question.
Yes you need to lose the first 2 virtual bets. Then bet another 3, That completes a 5 series bet.
It doesn't have to be the same color.
It all depends what series of bets you decided in advance for your next 5 bets.
If you chose in advance to place 5 bets like RBRRB, then your first 2  virtual losses to qualify have to be a B and then R. Then the next 3 bets will be RRB just as planned. If there is win in the first or second bet the case is dropped and then we start all over again. We only continue for the next 3 actual bets, if the first 2 bets are lost. The reason being, to avoid a 5-10-20-40-80 type of bet (risky and sure to result in a loss in the near future when you run into the black swan), and replace it with a $0-$0- $5-$10-$20 type of bet.
The same logic applies in a single dozen or column bet. But the winning range is different, due to lesser numbers. Empirically when you randomly  chose to bet a dozen, most of the winning happens in the range of 4-8 spins. So if you have preplanned to bet dozen #1 for a series of bets, the first 3 bet results have to be other than the 1st dozen but with no actual money lost.  Then you bet that dozen for the next 5 spins, with the required progression that produces a profit.
The bottom line is that I'm using my 50% or 33% (for a single dozen), power of guessing, while making sure I avoid possible variance in the first few bets, and reducing my total exposure by playing only the most frequently observed winning range. A series of random bets cannot constantly be annulled by a series of opposite results. Especially in the 50%, 33% range.
It's a time consuming process, but who said playing roulette is not work? In a regular job you are there for 8 hours to earn a salary. You don't demand a day's pay after 1 hour's work.
The concept can be taken even further, if one is gifted with extreme patience.
You can increase the number of the virtual losses and reduce the number of the real bets.
Like in the case of BBBBB, 3 virtual losses and only 2 real bets.
This way of thinking has worked just fine for me for many years. I limit my risk, I win many  more times than I lose , but I pay it with very long waiting time. But I enjoy the process, knowing that this monster called casino cannot beat me. Because I don't give them the opportunity to beat me.
Lost winning opportunities is a trap. YOU HAVE TO RISK TO GET  WINNING OPPORTUNITIEs, and pushes you to bet more often. The results of such  behavior are well known. Many casinos would be closed if every winning opportunity turned into actual winning.
« Last Edit: June 21, 2015, 02:32:15 PM by palestis »
 

Real

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Re: About virtual losses on Even chances
« Reply #13 on: June 21, 2015, 03:19:23 PM »

Quote

f you chose in advance to place 5 bets like RBRRB, then your first 2  virtual losses to qualify have to be a B and then R. Then the next 3 bets will be RRB just as planned.

Can you show us the probability of winning with and without the virtual bets?
How does the virtual bet change the probability of winning on the next spin or series of spins?
If we wait for two virtual bet losses, then what is the new probability of winning?
verses...
If we don't wait for the virtual bets, what is the probability of winning?

Can you show the math?
« Last Edit: June 21, 2015, 03:22:07 PM by Real »
 

scepticus

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Re: About virtual losses on Even chances
« Reply #14 on: June 21, 2015, 10:07:26 PM »
Clearly, Palestis is right here .
 He has established his parameter of  5 spins one of  which he expects to win  WITHIN his parameter of 5 spins. GIVEN his parameter he is EXPECTED  to win 97 times in a hundred  " trials " ..  Clearly then if he waits for one, two , three, or even four spins  and his chosen   colour has  not appeared then he STILL has a 97% chance of winning.
His argument  centres on his given parameters !
He does not argue that he is CERTAIN to win, only that he EXPECTS to win.
You can argue anything when you first set your parameters.