Parrondo's Paradise

[nullified]

P.S.

There are several game sequences you can play, and many of them are posted and can be tested on the web when you search for "parrondo simulator".

I like to use the sequence: BBABBA or maybe even ABBAB

As long as you're playing Game B more than Game A, even a random sequence between A and B will do.

[pins]

as all systems lose at roulette.  I cannot see how you can over come this, but I await your results with interest. good luck

[Mr J]

as all systems lose at roulette.  I cannot see how you can over come this, but I await your results with interest. good luck

I love ya Pins, you keep the board rockin.   

Ken

[Mr J]

@nullified >> Thank you very much for posting your betting approach to roulette, much appreciated sir !!

Ken

[LuckoftheIrish]

I have just tested it on 100 000 results.

I used the LLL trigger for applicable B bets.  I used ABBABB sequence.  All even chance bets were on High.  All the other rules were followed as per your instructions.

Starting Bank...10000
Ending Bank.....9098
End result........-902
A bets...............5795
B even bets......10336
B corner bets....2504

I had tried a method similar to this months ago.

[LuckoftheIrish]

The problem of course is the 75% needed for the applicable B bet.  You can not find a bet anywhere where you risk 1 dollar and have a 75% chance of doubling your money.

[17black]

This analysis is really a work of art - Congratulations. I am however having difficulty coming to grips with the sequence of play.
Do we play Game A until such time as the bank is divisible by 9? Do we then switch to game B1? How long do we continue playing B1before advancing to B2. Is there a trigger required to change? Again how long do we play game B2 before reverting to Game A is there a trigger here again which dictates the change in direction?
I would be grateful for some clarification.

[nullified]

Just some things I want to clear up about this system, since there have been some questions about it, and I think I was maybe not entirely clear.

First, I want to point out to LuckOfTheIrish... yes, you are in fact correct.  My probability calculations had a flaw in them that I had not considered, and that was, after a result comes up, that result is no longer a probability.  It is a certainty.

Somehow I had forgotten all about that, and continued to calculate the probabilities as if the virtual losses were still probabilities when in fact they were not.

I reference a thread started earlier by Kon-Fu-Sed where he addresses mixed chain probabilities nolinks://vlsroulette.com/index.php?topic=2193.15.

My idea for using an Even Chance bet for Game B2 is not going to be as profitable as using a 2 Dozen bet.

I sent this information to another member, but I think I'll go ahead and just post it here, rather than me re-typing it all again.

... Waiting for 3 virtual losses before betting on an Even Chance does in fact give me an 85% chance of winning - but only before there are any results.  So when I start Game B2, but before I've seen any results come up yet, I do in fact have an 85% chance of winning.  But, and here is where I was not thinking clearly... after I've seen my first virtual loss, I can no longer count that result as a probability - I have to count that as a certainty.

So instead of 19/37 (Loss 1) x 19/37 (Loss 2) x 19/37 (Loss 3) = 0.135 or 13.5% expected loss, it should in fact be:

1/1 (Loss 1) x 19/37 (Loss 2) x 19/37 (Loss 3) = 0.264 or 26.4% expected loss.

Then if I wait until 3 virtual losses, I get:

1/1 (Loss 1) x 1/1 (Loss 2) x 1/1 (Loss 3) = 1.0 or 100% expected loss.  So in fact waiting for 3 virtual losses does nothing to increase my odds of winning when I bet, it only delays it.

Interestingly, if I could find a bet where I determined that I was going to bet only 4 times, and in those 4 times I only need to win one time, then things would be different.  But in that case I'm relying on the multiplied averages of missing all 4 wagers in 4 spins - which is not what I'm doing here.

So, all that to say, I've had much better success playing Game B₂ using the "Relaxed Roulette" method, or, even better, to just set a small profit stop of $2 and a profit loss of $4 and I'll still win Game B₂ more than 65% of the time.

As an option, if you can find or figure out any kind of Dozens game to play in Game B₂, you can.  As long as you have a high percentage of winning $2 and only losing between $2 - $4, the rest of the paradox will even out the other losses.

Now I'd like to clarify a few other things.

There are two things to keep in mind in order to make things work.  First, there is the Games themselves.  Second, there is a betting sequence wherein we play one of the three games.

I've found the best sequence to be : BBABBA (you may find another works, in which case, go for it).  Obviously with this sequence we end up playing Game B twice as much as we play Game A, which is what we want.  I will never deviate from this sequence.

Keep in mind that this sequence has nothing to do with "spins" - these are "Games".  So our sequence above is not 6 spins, it is 6 separate "Games".

Two of our three games however, will in fact only be 1 spin.  The other game will range anywhere from 2 to 20 spins, depending on how long it takes you to win your $2 profit.

So here is what this looks like.  I'm going to start with a bankroll of $1,000 and walk you through 6 separate games so you can see how they ought to play.

My sequence is BBABBA.

Game B - Because my betting sequence starts with "B", I play Game B.  I first check to see if my bankroll is divisible by 9.  If it is, I play Game B₁, otherwise, I play Game B₂.  $1000/9 = 111.111  My bankroll is not divisible by 9 (that's why there is a decimal), so I play Game B₂.  Game B₂ I play as a 2 Dozen bet.  I choose 2 Dozens and stick with them for Game B₂.  I choose Dozens 2 and 3.  I place $1 each on them.  Result:  I win $3-(my $2 wager)=+$1.  Now I've won, but only half the amount I need.  I'll need to play this again.  Result: I lose -$2.  Now so far for Game B₂ I've won $1 and lost $2... so right now I'm -$1.  But I'm not done.  My goal is to get to either +$2 or -$4.  I spin again.  Result: I win $1.  Now I've broken even (good thing Roulette won't remember any of this!)  Now I spin again - remember, this is all still Game B₂ - and I win $1.  I spin one more time and win, making my profit goal of +$2.

Playing Game B₂ as a mini-series of 2 Dozen bets helps me to keep my percentage of wins, and the small stop loss is the higher price I pay for playing this way.

Since I won Game B₂, my bankroll is $1002.

After playing my first "Game", I move on in my Betting Sequence (BBABBA).

My betting sequence again calls me to play Game B again.  I divide my bankroll again by 9 to determine which game I play.  So $1002/9 = 111.33.  It doesn't divide by 9 evenly, so I play Game B₂ again.

Game B2: I place a $1 bet on my 2 Dozens.  I lose, so my bankroll is now $1000.  But since I'm not done playing this Game, I carry on, until I reach +$2 or -$4.  At the moment I am -$2 on this Game.  I wager $1 each on my Dozens again.  I lose again, so I've reached -$4 for this Game.  So I've lost Game B₂ this time, and my bankroll is now at $998.

Now, even though my bankroll is evenly divisible by 9, my betting sequence tells me to bet using Game A.  Remember, I only check to see if my bankroll is divisible by 9 when my betting sequence tells me to play one of the two Game B's. 

Game A - I play an Even Chance bet of $2.  It's important to play the Even Chance that is opposite of the 2 Dozens that you play in Game B₂.  I like to choose Low.  I win this bet, so I've profited +$2 and my bankroll is now $1000.

That was the first 3 games of my betting sequence :BBABBA.  I've got 3 more bets for this sequence, which is, 2 more Game B's and one more Game A.

Game B - Again, I need to know if my bankroll is divisible by 9.  $1000/9 =111.11, and Again, I'll be playing Game B₂.  I place $1 each on my Dozen bets.  I lose this bet, so my bankroll is $998.  I do not change my bet at this point.  I'm still in the middle of this game, where I'm trying to get to +$2 or -$4.  I bet my $1 on my 2 Dozens again and win.  Now I'm at -$1 for this game.  I play my Dozen bets again.  I win again, so I've broken even for this game and I'm at $1000 again.  But my game is not done.  I have not reached either goal for Game B₂.  I bet my 2 Dozens again.  I win again.  Now I'm at +$1 for this game.  I have yet to reach +$2, so I need to keep going.  I bet on my 2 Dozens and I win again.  I've profited my $2, and now I can carry on in my betting sequence. 

Game B - Since my sequence is "BBABBA", I need to play game B again.  I calculate my capital, and $1002/9 = 111.33.  I play Game B₂ again.  I am fortunate to win my $2 goal in just 2 spins, so my bankroll is now sitting at $1004.

Game A - This is the last of my sequence, and I place a $2 wager on my Even Chance bet, which is Low.  I lose, so my bankroll is now at $1002.

The only game I did not play in this sequence was Game B₁.  I only play game B1 when my betting sequence tells me to play Game B, and my bankroll is divisible evenly by 9.  And Game B₁ I place $1 on 2 different Corner Bets.

So, that is how the Betting Sequence interacts with the 3 different Games.

By the way, you may choose some other type of system for Game B₂.  I like to play this one because it has clear goals and parameters, and it let's me "fight" for a profit when I think I deserve one!

Hope that clears some things up!

[Far-Q]

Just some things I want to clear up about this system, since there have been some questions about it, and I think I was maybe not entirely clear.

First, I want to point out to LuckOfTheIrish... yes, you are in fact correct.  My probability calculations had a flaw in them that I had not considered, and that was, after a result comes up, that result is no longer a probability.  It is a certainty.

Somehow I had forgotten all about that, and continued to calculate the probabilities as if the virtual losses were still probabilities when in fact they were not.

I reference a thread started earlier by Kon-Fu-Sed where he addresses mixed chain probabilities nolinks://vlsroulette.com/index.php?topic=2193.15.

My idea for using an Even Chance bet for Game B2 is not going to be as profitable as using a 2 Dozen bet.

I sent this information to another member, but I think I'll go ahead and just post it here, rather than me re-typing it all again.

... Waiting for 3 virtual losses before betting on an Even Chance does in fact give me an 85% chance of winning - but only before there are any results.  So when I start Game B2, but before I've seen any results come up yet, I do in fact have an 85% chance of winning.  But, and here is where I was not thinking clearly... after I've seen my first virtual loss, I can no longer count that result as a probability - I have to count that as a certainty.

So instead of 19/37 (Loss 1) x 19/37 (Loss 2) x 19/37 (Loss 3) = 0.135 or 13.5% expected loss, it should in fact be:

1/1 (Loss 1) x 19/37 (Loss 2) x 19/37 (Loss 3) = 0.264 or 26.4% expected loss.

Then if I wait until 3 virtual losses, I get:

1/1 (Loss 1) x 1/1 (Loss 2) x 1/1 (Loss 3) = 1.0 or 100% expected loss.  So in fact waiting for 3 virtual losses does nothing to increase my odds of winning when I bet, it only delays it.

Interestingly, if I could find a bet where I determined that I was going to bet only 4 times, and in those 4 times I only need to win one time, then things would be different.  But in that case I'm relying on the multiplied averages of missing all 4 wagers in 4 spins - which is not what I'm doing here.

So, all that to say, I've had much better success playing Game B₂ using the "Relaxed Roulette" method, or, even better, to just set a small profit stop of $2 and a profit loss of $4 and I'll still win Game B₂ more than 65% of the time.

As an option, if you can find or figure out any kind of Dozens game to play in Game B₂, you can.  As long as you have a high percentage of winning $2 and only losing between $2 - $4, the rest of the paradox will even out the other losses.

Now I'd like to clarify a few other things.

There are two things to keep in mind in order to make things work.  First, there is the Games themselves.  Second, there is a betting sequence wherein we play one of the three games.

I've found the best sequence to be : BBABBA (you may find another works, in which case, go for it).  Obviously with this sequence we end up playing Game B twice as much as we play Game A, which is what we want.  I will never deviate from this sequence.

Keep in mind that this sequence has nothing to do with "spins" - these are "Games".  So our sequence above is not 6 spins, it is 6 separate "Games".

Two of our three games however, will in fact only be 1 spin.  The other game will range anywhere from 2 to 20 spins, depending on how long it takes you to win your $2 profit.

So here is what this looks like.  I'm going to start with a bankroll of $1,000 and walk you through 6 separate games so you can see how they ought to play.

My sequence is BBABBA.

Game B - Because my betting sequence starts with "B", I play Game B.  I first check to see if my bankroll is divisible by 9.  If it is, I play Game B₁, otherwise, I play Game B₂.  $1000/9 = 111.111  My bankroll is not divisible by 9 (that's why there is a decimal), so I play Game B₂.  Game B₂ I play as a 2 Dozen bet.  I choose 2 Dozens and stick with them for Game B₂.  I choose Dozens 2 and 3.  I place $1 each on them.  Result:  I win $3-(my $2 wager)=+$1.  Now I've won, but only half the amount I need.  I'll need to play this again.  Result: I lose -$2.  Now so far for Game B₂ I've won $1 and lost $2... so right now I'm -$1.  But I'm not done.  My goal is to get to either +$2 or -$4.  I spin again.  Result: I win $1.  Now I've broken even (good thing Roulette won't remember any of this!)  Now I spin again - remember, this is all still Game B₂ - and I win $1.  I spin one more time and win, making my profit goal of +$2.

Playing Game B₂ as a mini-series of 2 Dozen bets helps me to keep my percentage of wins, and the small stop loss is the higher price I pay for playing this way.

Since I won Game B₂, my bankroll is $1002.

After playing my first "Game", I move on in my Betting Sequence (BBABBA).

My betting sequence again calls me to play Game B again.  I divide my bankroll again by 9 to determine which game I play.  So $1002/9 = 111.33.  It doesn't divide by 9 evenly, so I play Game B₂ again.

Game B2: I place a $1 bet on my 2 Dozens.  I lose, so my bankroll is now $1000.  But since I'm not done playing this Game, I carry on, until I reach +$2 or -$4.  At the moment I am -$2 on this Game.  I wager $1 each on my Dozens again.  I lose again, so I've reached -$4 for this Game.  So I've lost Game B₂ this time, and my bankroll is now at $998.

Now, even though my bankroll is evenly divisible by 9, my betting sequence tells me to bet using Game A.  Remember, I only check to see if my bankroll is divisible by 9 when my betting sequence tells me to play one of the two Game B's. 

Game A - I play an Even Chance bet of $2.  It's important to play the Even Chance that is opposite of the 2 Dozens that you play in Game B₂.  I like to choose Low.  I win this bet, so I've profited +$2 and my bankroll is now $1000.

That was the first 3 games of my betting sequence :BBABBA.  I've got 3 more bets for this sequence, which is, 2 more Game B's and one more Game A.

Game B - Again, I need to know if my bankroll is divisible by 9.  $1000/9 =111.11, and Again, I'll be playing Game B₂.  I place $1 each on my Dozen bets.  I lose this bet, so my bankroll is $998.  I do not change my bet at this point.  I'm still in the middle of this game, where I'm trying to get to +$2 or -$4.  I bet my $1 on my 2 Dozens again and win.  Now I'm at -$1 for this game.  I play my Dozen bets again.  I win again, so I've broken even for this game and I'm at $1000 again.  But my game is not done.  I have not reached either goal for Game B₂.  I bet my 2 Dozens again.  I win again.  Now I'm at +$1 for this game.  I have yet to reach +$2, so I need to keep going.  I bet on my 2 Dozens and I win again.  I've profited my $2, and now I can carry on in my betting sequence. 

Game B - Since my sequence is "BBABBA", I need to play game B again.  I calculate my capital, and $1002/9 = 111.33.  I play Game B₂ again.  I am fortunate to win my $2 goal in just 2 spins, so my bankroll is now sitting at $1004.

Game A - This is the last of my sequence, and I place a $2 wager on my Even Chance bet, which is Low.  I lose, so my bankroll is now at $1002.

The only game I did not play in this sequence was Game B₁.  I only play game B1 when my betting sequence tells me to play Game B, and my bankroll is divisible evenly by 9.  And Game B₁ I place $1 on 2 different Corner Bets.

So, that is how the Betting Sequence interacts with the 3 different Games.

By the way, you may choose some other type of system for Game B₂.  I like to play this one because it has clear goals and parameters, and it let's me "fight" for a profit when I think I deserve one!

Hope that clears some things up!

I would like to thank you for the clear explanation AND for taking the time to type it all out...thank you.

[Far-Q]

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