Parrondo's Paradise

[nullified]

Notwithstanding all the slack I am bound to take for starting this thread, I thought it good to get this out on the forum in a distinct and forthright manner.

For those of you who are sick of hearing about "scientific breakthroughs" and pie-in-the-sky ideas, as they apply to the game of roulette, just bear with me.

Yes, I am resurrecting the much maligned concept of using Parrondo's Paradox in a casino game.  I know, I know, everyone's thought of that before... I get it.

I'm also aware that Juan Parrondo himself said that the concept couldn't be used in Casinos.  His partner, David Abbot, however, wasn't so sure:

2. Can I use Parrondo's paradox to win at the Casino? The answer is most probably "no." Parrondo's games rely on exploiting convex linear combinations in a non-linear parameter space. Casino games have a linear parameter space (as far as we know). If anyone can demonstrate any convexity in a casino game, then you are on to something! So the answer is most probably no, because we have not personally exhaustively tested every casino game for convex spaces. Our interest in is scientific study of the Parrondo phenomenon, not in casinos. To find out about convexity and the role it plays in Parrondian phenomena, click on the publications link. The importance of convexity in the Parrondo effect was first pointed out by Moraal in 1999 and then independently by Costa, Fackrell & Taylor in 2000. nolinks://nolinks.eleceng.adelaide.edu.au/Groups/parrondo/faq.html

Interesting.  I also found it curious that, almost as soon as the Paradox was discovered, they were very quick to comment on its' unsuitability to Casino games. 

Really? 

So let's get this straight.  You make a major discovery in physics, and, to prove your theory, you use a simple game of chance to demonstrate your discovery - even going as far as using coin flipping, bias (read: "House Edge"), wagers, and payouts.  And then, after you've proven your theory to death, add a P.S at the end of everything saying, "Oh, and by the way... this doesn't work in a real life Casino - so don't even bother." 

Anyway, I believe the researchers were not very interested in casino games to begin with; so they did not have the time or the intuition to use their discovery to try to bring Vegas to its knees.  Kinda reminds me of Einstein's comments about roulette.  Oh well, to each his own...

In any case, I want to put forth some of my own research into Parrondo's Paradox.  I don't want to hold back anything, including some of my own blind alleys and blunders.  So it seems to me that the best way to do this is to break up this rather daunting topic into several smaller posts.  Which is what I think I'll do.

My goal, right from the very beginning, was to duplicate the actual Parrondo Games. My reasoning was that, if I could accurately duplicate the games, the results would speak for themselves.

I know what you're thinking - "Oh, this is gonna be rich - NOT!"  But, like I said, bear with me.

Sometimes answers come, not from ingenuity, but from plain stubbornness.

[nullified]

Okay, to start off, if you don't know anything about Parrondo's paradox or how it works, you'll probably want to read all about it on Juan Parrondo's official site.  I've found that trying to read up about it on other websites like Wikipedia just confused the crap out of me when I was first starting out.  But the official site itself is pretty self explanatory - nolinks://seneca.fis.ucm.es/parr/GAMES/index.htm

There are also a couple of other posts on this forum as well.  The main post that I followed was only partially thought out, and not at all profitable.  It seemed that the person who posted it was in a real hurry to test out the theory by playing roulette with it, rather than testing it with a simple calculator first.

So, what is this discovery?  And why is it called a paradox?  The discovery was made by a team of physicists, headed by the now notorious Juan Parrondo.  The team, in researching brownian ratchets, applied some new ideas to an old ratchet thought experiment, and then applied their ideas to other areas - most notably, gambling games. 

Their idea was this - to play two losing games in such a way that, when combined, would end up winning.  They put it like this: "Given two games, each with a higher probability of losing than winning, it is possible to construct a winning strategy by playing the games alternately."

Thus, you can see why this is called a paradox.  Two games, both designed in such a way that, when played separately they'll certainly lose, will, when played in a certain manner, end up winning.

The implications of this are profound for roulette players - even if the idea seems a bit vague and out of reach.

In demonstrating the parrondo effect, they set up two coin-flipping games.  Game A is an "unfair game" with a small bias towards losing.  If we determine beforehand to only bet on Heads, then the bias skews the results ever so slightly to land on Tails a half a percent of the time more than Heads.  Thus, Game A is a losing game for us.

Game B is also an unfair game.  It is divided into two games.  Game B₁ and Game B₂.  Our current capital then determines which of the two games we play.  If our current capital is modulo 3 (divisible by 3 exactly) then we play Game B₁.  In Game B₁ we flip a coin that gives us only a 10 percent chance of winning, and a 90 percent chance of losing.  If our current capital is not divisible by 3 exactly, then we play Game B₂, where we flip a coin that gives us a 75 percent chance of winning, and a 25 percent chance of losing.

This seems straight forward enough.

But there is a catch.  The "catch" is that a win profits us $1.  And a loss causes us to lose $1 also.

And this is where most research into the parrondo effect in roulette stops. 

To win $1 on a 48% probability game is easy enough.  In fact, Game A can very easily be played as an Even Chance bet.  It's bias is rather large, at 2.5%, rather than the half percent used in the original demonstrations.  But, because everyone automatically calculates an even chance bet as 48.64% to begin with, we really don't need to mess around with the bias the way our researchers did.  We just accept the fact that all bets in roulette are biased, and move on.

And even Game B₁, where we play a game that only gives us a 10% chance of winning $1 and a 90% chance of losing $1 is fairly easy to duplicate.  A $1 wager on a corner bet gives us an 89% chance of losing that wager, and an 11% chance of profiting $8.  Heck, that's already better than Parrondo's version of Game B₁.  They only got to win $1 on a win.  In our version of Game B₁, we get to win $8!

The real problem for us, however, is Game B₂.  In Game B₂, we flip a coin that gives us a 75% chance of winning, and a 25% chance of losing.  How do we replicate this in roulette?

Having a 75% chance of winning is not a problem.  75% of the board is 28 numbers out of 37.  If we cover 28 numbers, we have a 76% chance of winning and a 24% chance of losing.

The only problem with this idea is that I can't cover 75% of the table with $1.

I can use $3 maybe.  $1 on dozen 1, $1 on dozen 2 and $1 on line 25-28.

No good.  65% of the time I'll break even, 16% of the time I profit by $3, but 19% of the time I lose $3.

That's a long ways away from a 75% chance of winning $1.

In fact, no matter how clever I try to be, I cannot possibly cover 75% of the table to duplicate Game B₂ of Parrondo's Game.

So, what to do?

In my next post, I'm going to analyze some of my first ideas to address this problem.

[Far-Q]

Very interesting thread 

[nullified]

Continuing on from my last post, we saw that the main crux of the parrondo problem is, in fact Game B₂.  Game B₂ requires us to win $1 with a 75% chance of winning, and lose $1 with a 25% chance of losing.

Admittedly, when I first began tackling this problem, I tried every combination of bet I could come up with to gain this advantage.  And the harder I tried, the worse things got.  It wasn't until I had a couple of insights that I began to make some headway with the problem.

My first real insight was the bet amount itself.  Even though in the official version, to see the parrondo effect in action meant you needed to make $1 per win and lose $1 per loss - I did remember one version where they talked about "units" instead of dollars.

What if I changed the parameters of the game so that, instead of trying to win $1, I tried to win 1 unit?  A unit could be any amount I wanted it to be.

Initially I thought this was brilliant, because I thought that all I had to do was call a "unit" whatever amount I needed to cover 75% of the table.

Here was what I was trying to do: if I called a "unit" 9 chips (i.e $9), then I could use 9 chips to cover 75% of the table, then use 9 chips again to bet on an even chance bet, and another 9 chips to bet on my one corner.  That way, I figured, I was covering all aspects of the games equally with my 1 "unit" of 9 chips.

So I began to work it out.  Let's say I used my 9 chips to cover 27 spots on the table (9 streets), giving me 75% wins and 25% losses.  I win 9 chips and lose 9 chips on Game A; I win (9 x 9) - 9 = 72 chips and lose 9 chips on Game B₁; and - drum roll please - on Game B₂ I lose 9 chips 25% of the time and on the other 75% I win...

$3?

Ummm, $3 is not a full unit.  A full unit is $9, but all I profited back was $3.  So on the game I end up playing the most I either lose my full "unit" or win back 0.3 of a unit.

That doesn't work.

Still, using units instead of dollars seemed promising.

I felt like I was so very close... and yet, unless I could adequately solve Game B₂, I would never get any closer.

So I began my study of parrondo afresh.  Every formula.  Every method.  Every idea.  Every reference, web page, paper, and passing remark I would study... I did.

Brownian ratchets, stachostic processes, markov chains - gahh!

"There has got to be a way!  What am I missing?"

Truth be told, my first "insight" wasn't actually a solution.  It was an inspiration.  It didn't really solve anything at first.  But it kept me driving toward my goal.  If I hadn't had that first insight I know I would not have had the fortitude to carry on as I did.

And that's when I had my second insight.

[revolver]

you´re john gold, true? ...

[nullified]

Alright, by this time in my research I understood that unless I could approach the conditions of the original Parrondo Game, there would be literally zero chance of it being useful in the game of roulette.  So when I began again in earnest, I told my self that if I couldn't duplicate it's methodology, then I was going to try to duplicate it's results.  In other words, maybe I couldn't set up the games exactly the way the Parrondo team did... but maybe I didn't have to.  If I could duplicate the results without having to use their methods, then that might be just as good.

So how do I know what the results of the original games are?

Game A is easy enough to figure out.  Its a 50/50 game minus the bias.  In roulette, that's an even chance bet.  I can expect to lose 51 times and win 49 times.  Net loss is -$2.

Game B₁ is easy enough as well.  It's a biased game where 90% of the time I lose $1, and 10% of the time I win $1.  Net loss = $90 - $10 = -$80.

Game B₂ is similar, but it's a biased game where I win $1 75% of the time and lose $1 25% of the time.  Net win = $75 - $25 = +50.

But hold on, this isn't an accurate depiction of the results of the Games.  Remember, in Game B, it's our capital that determines whether we need to play Game B₁ or Game B₂.  That means that I won't be playing both games equally.  In fact, to quote from their own work:

It is clear that by playing Game A, we will almost surely lose in the long run. Harmer and Abbott[1] show via simulation that if M = 3 and \epsilon = 0.005, Game B is an almost surely losing game as well. In fact, Game B is a Markov chain, and an analysis of its state transition matrix (again with M=3) shows that the steady state probability of using coin 2 is 0.3836, and that of using coin 3 is 0.6164.[2] As coin 2 is selected nearly 40% of the time, it has a disproportionate influence on the payoff from Game B, and results in it being a losing game. nolinks://en.wikipedia.org/wiki/Parrondo%27s_paradox

What this means is that Game B₁ and Game B₂ are not played 50/50.  Game B₁, the game wherein we have only a 10% chance of winning, is played approximately 38.36 times out of a possible 100 - or 38% of the time that we play Game B.  Game B₂, wherein we have a 75% chance of winning, is played 61.64 times out of 100, or about 62% of the time.

So when it comes to playing Game B, because our current capital is the deciding factor in which of it's two games we play, we find that in 100 times that we play Game B, 38 times we will play Game B₁, and 62 times we will play Game B₂.

When we play Game B₁ 38 times, we have a 10% chance of winning and 90% chance of losing.  10% of 38 is 3.8, or 4 times that we win.  90% of 38 is 34.2, or 34 times that we lose.  So in our 38 trials of Game B₁ we expect to win 4 times and lose 34 times.  Overall, this amounts to -$30 in our 38 trials.

When we play Game B₂ 62 times, we have a 75% chance of winning and a 25% chance of losing.  75% of 62 is 46.5 or 47 times that we win.  25% of 62 is 15.5 or 16 times we lose.  So in our 62 trials of Game B₂ we expect to win 47 times and lose 16 times.  Overall, this amounts to 47 - 16 = +$31 in our 62 trials.

Now we know that playing Game B for 100 trials we will lose -$30 from Game B₁, and we will win +$31 from Game B₂, netting us +$1 on Game B.

And what happens when we add Game A and B together?  We get -$2 from Game A, and +$1 from Game B, for an overall net loss of -$1.

Unless we play Game B more than we play Game A.  And, in fact, this is exactly what the researchers tell us at the end of their research.  Game B needs to be played more than Game A in order to see the Parrondo effect in action.

It was at this juncture of my research that I was really pulling my hair out. 

I could duplicate Game A.

I could duplicate and improve upon Game B₁.

But the improvements in Game B₁ did not offset the losses I encountered in Game B₂.

Alright. At this point I couldn't duplicate the Game or its' results.  But how close could I get?  If I got close enough to its' methods and/or its' results, would I have a winning bet?

And then it hit me.  Why not make your unit $2?  That way, for Game A I still bet $2 on one of the even chances.  For Game B₁ I would put $2 on a corner bet.  And for Game B₂ I would put $1 on Dozen 1 and $1 on Dozen 2.  Now I've decreased my percentage of winning Game B₂, but my payout for a win is no longer 0.3 of a unit, but 0.5 of a unit (because I bet a full unit - $2, but only win $3 netting me $1 - which is half of my original unit).

Now when I win, at least I'm getting back half of my unit.  But was that going to be enough?  Was my $7 profit from Game B₁ going to be enough to offset the fact that I only win a half a unit back in Game B₂?

Things were beginning to get interesting...

[nullified]

you´re john gold, true? ...

No, I'm not.  I'm not sure who John Gold is exactly.  I know he likes roulette because he sells systems.  I know he's been on this forum before.

But I'm not him... sorry

[nullified]

From my previous posts, if you've been following along (and if you haven't, you should go back to the beginning of this thread to get caught up), you'll know where I left off in my search to make a working roulette system based on Parrondo's paradox. 

I had made two very important discoveries.

First, I could bet with "units" as opposed to single $'s.  This afforded me the opportunity to stretch my bets further along the table, so to speak, in order to approach that magical 75% win rate required in the Games.

Second, I decided to cover only 24 numbers of the table instead of 28.  Although I was taking a slight hit in chances of winning (only 65% compared to 75%), I could now win back more money from Game B₂.

But would this really work?

Its interesting how things play out sometimes.  Because when I had this idea to just use 2 units, and use the 2 Dozen bets as my Game B₂, I was sick in our downstairs bedroom.  It was late at night, and I remember pacing the floor pondering these problems, bent over slightly with a heating pad pressed against my stomach.  I was trying to be real quiet so I wouldn't wake up the rest of the house.  And when the idea hit, instead of racing to the computer to test out my theory, I picked up my calculator and notepad instead and began calculating the new parameters.

I mention that because, had I done what I really wanted to do, which was to go test out my system, I may not have continued.

But because it was late at night, and I was sick but pacing around in the guest bedroom, I decided to do what I considered the next best thing.

I plugged in my new ideas and compared them with the original Parrondo's Games.

Here's what I found:

Game A I kept as previous - a 48.6% chance of winning with $2.  I would win $98 and lose -$102.  Expected loss = -$4.

Game B₁ I changed a bit.  Instead of betting on just 1 corner, why not bet on 2 corners?  I have $2 as my unit to use.  Why put both of them on a single corner, when I can split them up and put $1 on Corner A and $1 on Corner B?  Now I've just doubled my chances of winning on Game B₁, right? So now I've covered 8 numbers on the board, giving me 21.62% chance of winning $7, or about 22%.  Now let's plug in our numbers again and see what this looks like.  I play Game B₁ about 38 times in 100 trials (see previous post for verification).  In those 38 trials, I win $7 22% of the time, and lose $2 78% of the time.  Altogether for Game B₁ I win 22% of time, which is 8 wins, and 78% of the time I lose, which is 30 times.  8 wins x $7 = +$56.  30 losses x $2 = -$60.  Expected loss for Game B₁ = -$4.

This is already much better than Parrondo's original Game B₁.  In his Game B₁, the expected loss was -$30; ours is now only -$4.  That's an improvement of $26 over the original!

But everything hinges on Game B₂.  What happens when we try our new method here?

Game B₂ I bet $1 on Dozen 1 and $1 on Dozen 2.  I play Game B₂ approximately 62% of the time (again, see previous post).  In those 62 trials, I win 65% of the time and lose 35% of the time.  So in this Game 65% of 62 = 40.3 or 40 wins, and 35% of 62 = 21.7 or 22 losses.  With my 40 wins I gain +$1, so I'm +$40.  My losses are 22 x 2 = -$44.  On Game B₂ then, my expected outcome is +$40 - $44 = -$4.

Uh oh.

This isn't going to work like this.

My Game A has an expected outcome of -$4.  My Game B₁ has an expected outcome of -$4.  And my Game B₂ has an expected outcome of -$4.

Although this is very close to what I'm aiming at, there's one small problem.

None of these games is a winning game.  As specified in Parrondo's original research, in order to see the parrondo effect, all of the games cannot be losing games.  At least one of them has to be a winning game.

Now you can see why it was a good thing I picked up my calculator before I rushed off to test out my new theory.  Before even testing out anything, I knew I had a losing system on my hands.

And as I lie there in bed that night, thinking about how close I had come to replicating this infernal formula, I knew that I would have to press on.

I was determined to either completely close the book on Parrondo's Paradox, or bust this whole thing wide open.

Thankfully I went to bed instead.

But inspiration was about to hit again...

[nullified]

By now you've seen my earlier attempts to resolve the Parrondo problem using a 2 Dozen Bet.  If you've been following along, maybe you've seen where this is headed and have attempted to come up with your own solution.  I wasn't ready to give up at this point though.  I had come to within a couple of dollars of having a winning bet.  But being short a few dollars can compound into hundreds or thousands of dollars very easily over the course of even a few thousand spins.

Following are a few more of my attempts to bring Game B into a profitable solution.

Instead of betting on the corner bet (Game B₁) when my capital is wholly divisible by 3, why not do the opposite?  In other words, when my capital is divisible by 3, play Game B₂; otherwise play Game B₁.  Here's how that decision shakes out: I play Game B₂ 38 times, with a success rate of 65%.  So I win 25 times and I lose only 13 times.  The result? I win +$25 - (13 x -$2 ) = -$1.  Okay, that looks like an improvement for sure.  And what happens when I play Game B₁ 62 times, with a success rate of 22%?  Then I win 13.64 times, or 14 times, and lose 48 times.  So I win 14 x $7 = +$98; and I lose 48 x $2 = -$96.  And $98 - $96 = +$2!

Wait a second... did I just do what I think I did?

After reversing the frequencies of Games B₁ and B₂, I'd turned Game B as a whole into a positive win of +$2! Hurray! I did it!

Let the profits roll!

I remember that night pounding my calculator for hours going over the numbers again and again and again.

But the numbers didn't lie.

I had turned the entire Game B into a winning one! Things were going to be different from now on!  My life was changing!  I just knew it!  Finally!

And yet, somehow, in the back of my mind a thought was forming.

A nagging, sagging thought.  Actually it was two thoughts.

The first was this - what happens when I extend the series to 1000 games instead of 100?  I didn't want to think about that. Because I knew in my subconscious what the result was going to be.

Yep.  That 13.64 that I had so judiciously "rounded up" to 14 would not be allowed to be rounded up. And that was a serious problem.

Reluctantly, I got out my calculator again and punched in the numbers. By playing Game B for 1000 games, 380 of those will be Game B₂ and 620 will be Game B₁.  So for 380 times with a success rate of 65%, I expect to win 247 times (not 250); and I lose 133 times (not 130).  The result now is 247 wins minus (133 losses of $2 each, which equals 266) = -$19.  That's a lot different than taking a typical result for 38 spins and multiplying it by 10.  In fact, by extending the session to a realistic 380 spins, I had almost doubled my loss rate!

Things get worse when apply the same concept for Game B₁.  Playing Game B₁ 620 times, with a success rate of 22% means I win 136 times (not 140), and lose 483.6 times, or 484 if I round up.  Now 136 wins times $7 net profit = +$952.  But 484 losses times -$2 each time = -$968.  Now, by simply extending Game B₁ to 620, my losses outweigh my profits by -$16. 

Clearly any profits I obtained in my sessions of 100 were due to rounding of numbers.  And clearly, the more I play, the greater the likelihood is that I'll end up losing more than I win.

The second thought I had was equally as frustrating.  Who's to say that those frequency calculations were correct, anyways?

After reading and studying and searching, I could not find where the Parrondo team came up with their calculation that one Game would be played 38% of the time and the other 62%.  I was using their numbers because I figured they were the experts.  But if those figures were off in anyway, my outcome for Game B could be anywhere from +$100 to -$100.  Were those hard and fast probabilities?  I had know way of checking this out.

All of the calculations that I did seemed to indicate that their numbers were wrong.  Shouldn't it be more like 33% of the time I play one game, and 67% of the time I play another?  For all numbers between 1 and 100, isn't it only 33 of them that are in fact divisible by 3?  Then where did these researchers get their probabilities from?

Now I had gone from feeling ecstatic to feeling depressed - all in the space of one day.

And I would have left it there, resolved to believe that this method was only going to work in the short term.  Another hit and run system.  Just like all the rest.

I decided that it was time to close the book on Parrondo's paradox.  It seemed that I just could not get this thing to work.  I had tried just about everything, and it all continued to come up short.

So I put my calculator away.  I hid my note pad.  I shut off my computer.

Then I picked up my iPad and started reading a book.

And everything began to fall into place...

[progambler]

... Game B2, wherein we have a 75% chance of winning, is played 61.64 times out of 100, or about 62% of the time.

Maybe I'm missing something,but if game B2 is 75% winning,why do you bother with playing other games?
Stick to this one and you will win 75% full time !

[nullified]

Maybe I'm missing something,but if game B2 is 75% winning,why do you bother with playing other games?
Stick to this one and you will win 75% full time !

Progambler,

Thanks for the input.  Unfortunately, the quote you have posted above is from my work at trying to figure out how the original paradox payed out.  In the original, they play a game that wins 75% of the time.

Which is exactly what I was trying to duplicate.  Sure, I can win 75% of the time, but when I lose the other 25% of the time, my loss will amount to a lot more than just a $1 loss.

Besides, if I had a bet that allowed me to win $1 75% of the time and only lose $1 the other 25% of the time, I wouldn't need to bother with Parrondo's paradox at all.  I would just play this little system and laugh all the way to the bank!

Roulette is much crueler than that though.

Fortunately for us, it also has it's own little quirks - at least one of which we will use to apply to our Parrondo system...

[progambler]

If our current capital is not divisible by 3 exactly, then we play Game B₂, where we flip a coin that gives us a 75 percent chance of winning, and a 25 percent chance of losing.

This seems straight forward enough.

But there is a catch.  The "catch" is that a win profits us $1.  And a loss causes us to lose $1 also.

This is what you were saying in your original post
So: 75% of winning 1$ and 25% of losing 1$ is exactly what you told

[nullified]

Personally I don't see what is so interesting about Parrondo's paradox but you are not the first to ask me about it so I'll give you my thoughts on it. The thrust of it is that if you alternate between two particular losing games the player can gain an advantage.
As an example, consider Game 1 in which the probability of winning $1 is 49% and losing $1 is 51%. In Game 2 if the player's bankroll is evenly divisible by 3 he has a 9% chance of winning $1 and 91% of losing $1. In Game 2 if the player's bankroll is not divisible by 3 he has a 74% chance of winning $1 and 26% of losing $1.


Game 1 clearly has an expected value of 49%*1 + 51%*-1 = -2%.


In Game 2 you can not simply take a weighed average of the two possibilities. This is because the game quickly gets off of a bankroll remainder of 1 with a win, and often alternates between remainders of 0 and 2. In other words the bankroll will disproportionately play the game with a 9% chance of winning. Overall playing Game 2 only the expected value is -1.74%.


However by alternating two games of Game 1 and two games of Game 2 we break the alternating pattern of Game 2. This results in playing the 75% chance game more and the 9% less. There are an endless number of ways to mix the two games. A 2 and 2 strategy of playing two rounds of Game 1 and two of Game 2, then repeating, results in an expected value of 0.48%.


I should emphasize this has zero practical value in the casino. No casino game changes the rules based on the modulo of the player's bankroll. However I predict it is only a matter of time before some quack comes out with Parrondo betting system, alternating between roulette and craps, which of course will be just as worthless as every other betting system.


nolinks://wizardofodds.com/ask-the-wizard/149/

The magical craps system featured in GOD DOESN'T SHOOT CRAPS is based on a recent breakthrough in advanced mathematics and physics called "Parrondo's Paradox."


Developed by Professor Juan M. R. Parrondo of the Complutense University of Madrid, the theory states that two "losing" games can be combined into a single winning outcome.


How?


Simply stated, it is accomplished through the use of ratchets – the familiar saw-toothed mechanical device used in elevators and self-winding wristwatches. A ratchet is used to allow movement in one direction but to prevent it in another. Every child, for example, knows that by shaking a can of mixed nuts you can get the heavier Brazil nuts to move to the top because the smaller nuts help "ratchet" their movement upward.


Dr. Parrondo discovered that the same principle can be applied to gambling games. To prove it, he created two coin-flipping games that were designed to lose by virtue of using weighted coins.


When you play these games separately, your bankroll will gradually diminish to zero. But if you alternate or "flash" between the two games, the occasional winnings of one game will be trapped by the ratchets of the second game. In other words, you can lose at both games and still emerge an overall winner!


Can this theory be applied to casino games? Dr. Parrondo says no. But GOD DOESN'T SHOOT CRAPS says yes! And the book-within-a-book included in the novel ("WIN BY LOSING") shows exactly how it's done.


But "Parrondo's Paradox" has implications that go far beyond casino gambling. The theory can be used to explain why some investors make money in a falling stock market. Why random events in the "primordial soup" led to the origins of life. How the chaotic movements of quantum particles at the subatomic level combine to create the logical order of the universe. Even how Bill Clinton's poll numbers seemed to go higher the more he misbehaved!


To learn more about "Parrondo's Paradox," click on the links below to see articles in The New York Times, Nature, and Science News.


nolinks://nolinks.goddoesntshootcraps.com/paradox.html

It was between these two quotes that I found myself.  Part of me feeling rather stupid for having chased the whole paradox thing as far as I did.  But the other part of me still tantilizingly frustrated that I was so close to some kind of a solution.


There were, in fact, several more attempts on my part to resolve the problem with Game B₂.  One of them was to change the modulo to 4 or 5.  That didn't solve it.


A second time I tried changing both Games for B to 2 separate Dozen bets.  Nope.  Then I tried to change them both to single corner bets.


Nothing got me any closer to solving the problem.


And when I had all but given up, I opened up the book that I quoted from above, "God doesn't Shoot Craps".  It's a fictional story, with non fiction inserted in the middle where the author talks about his system.


I didn't know anything about craps.  I didn't know anything about the author.  But I did know from reading his Parrondo method for craps, that the author hadn't solved the problem of Game B₂ anymore than I had.


But he mentioned one thing that gave me my last "Eureka!" moment.


In my next post, I'm going to share with you what it was that came to me.  I'll not keep you in suspense any longer. 


I appreciate all the players who have been reading along with me as I went.  My intention with this thread was not to string anyone along (like others may have done in the past) or to gather some big following.


I wanted to share my creation but also the highs and lows that went along with it.  In fact, that's been half the fun.


Thanks for being so patient...


On my last post... 2 new insights that solved the paradox.

[LuckoftheIrish]

Well I find this thread very interesting and entertaining.

I myself have tried Parrondo's Paradox in numerous ways.  I even contacted a Slovak man who claims he had the solution. His responses were in Slovakian and I enlisted the help of a translator to properly translate his messages.  I understood his system clearly, but it still failed the medium-long term test.

[nullified]

"Nothing is so powerful as an idea whose time has come" - Edmund Burke

I remember back four years ago when I first started fiddling around with roulette.  I found the game fascinating and exhilarating all at the same time.  I was playing online when I first started, so it was a completely new experience for me.  I just could not believe that I could make money by simply pressing buttons... it was crazy!

That wore off after a couple months.  I had turned $200 into about $1700, and was considering how tempting it would be to quit my day job.

And then I lost it.

All.

In one day.

"Sick to my stomach" does not even do justice to how I felt for several weeks after that happened.  And that's when I began to look a lot more seriously at trying to beat the game.

But after reading that novel "God Does Not Play Craps," I felt something different.  It wasn't a foreboding or a knot-in-your-stomach kind of thing.  It was an assurance.

An assurance that what I was about to do was going to level the playing field.

You see, one thing that the author talked about in his version of Parrondo's paradox was that he had to play more than one game to get his system to work.  At first I thought, "Well yeah, you have to play all three games to see the paradox in action...".  But that's not what he meant.  He meant that for the one problem Game, Game B₂, he had to play more than one 'session.'

That's when it finally hit me.  Why do I have to play for just one spin?  There's no "rule" that says that I must play each game for one and only spin or the system breaks down.  If I decide to play Game A for 1 spin, Game B₁ for 1 spin, and then play Game B₂ for 5 spins, why should that matter?  It's not as if playing extra spins on Game B₂ is going to some how screw up the rest of the system.

We all often talk about how roulette is a game of independent spins, and how roulette doesn't remember anything because it's random.  So if I randomly insert 5 extra spins in between Game A and Game B2, so what?  As long as I profit my full unit back in the end, I could play an extra 50 spins to win my unit, and it won't screw up the effects of Parrondo's Paradox one bit!

Here's when I realized that all I really wanted to do anyways was profit one full unit (whatever that may be) on my Game B₂ - regardless of how many spins that may take.

At last!  That revelation freed up so much constraint from my Game B₂.  There are thousands of potential ways that one could set a small profit goal of $2 or $4, and make that within a few spins to a few minutes.

But what was the best way?

Initially, since I was playing a 2 Dozen flat bet as my Game B₂, I thought, "Well, probably the best thing to do is to find a bet that is designed for dozens play, and where winning a small amount like $2 or $4 is relatively quick and easy."  So my first attempt at doing this was to play Game A as a $2 bet on an Even Chance like High, play my double corner bet for Game B₁, and then start a full "game" where I would begin by wagering $1 on Dozen 2 and $1 on Dozen 3 and play using the "Relaxed Roulette" method.

"Relaxed Roulette" is a dozen system where you always play the last 2 different dozens that hit, and use a simple cancellation calculation to make sure that you hit your target for the game.  In my case, all I wanted to do was end each mini-session with a $4 profit.

Actually, this is a very stress free way to play since you've changed the probabilities of Game B₂ from a 75% winning rate up to over 98%!

Relaxed Roulette was very appealing to me because when I began my study this past summer of the "Midas Method" (for those of you who know what I mean), I loved how simple and straight forward it was.  Certainly powerful, you could literally play for hours and slowly build up your bankroll in a consistent, determined way.

There's only one problem with the "Midas Method."  Your bankroll has to be able to withstand a monster losing session.  And "monster" losing sessions happen often enough that you feel it immediately when your bankroll is under funded.

And here was my problem with using "Relaxed Roulette" as my Game B₂ system.

It works almost all the time.  But when that 2% hits when it doesn't work - hang on to your wallet because you could be on the slow but steady decline to Broke-ville.

Can you use "Relaxed Roulette" for Game B₂.  Sure.  You'd have great success with it.  But you know my reservations with it - not to mention that sometimes a "session" can be a grind.

As I was playing this way, and using the "Relaxed Roulette" method for my Game B₂, I had another thought.

I was almost in a kind of daze as I was playing - not in a bad way or anything, more like a detached gawk - but I remember kind of sighing inwardly and thinking to myself, "I wish I didn't have to play Game B₂ so long.  I wish I could just play one spin and be done with it and move on"

So I stopped playing.  I had only been hovering around +$9 or $10 mark anyways, so I figured I needed a break.

I was thinking rather jealously of the original Parrondo Games where they played Game B₂ and "only" got a 75% win rate.

I couldn't think of any kind of mini-game to plug into my system where I could wager just one time and have those kinds of odds at winning.

And as I sat thinking about odds, I realized that I had been thinking about the odds completely backwards!

There was my answer. 

My odds of winning an Even Chance bet are set at 48.6%

And what's the opposite of that?  Well, we all know that, don't we?  But, in probability theory, they would describe the probability of winning like this using set theory notation: p(w) = 18/37 which reads "The probability of winning equals 18 divided by 37."

And for losing we might say: p(l) = 1 - p(w) which reads, "The probability of losing equals 1 minus the probability of winning."  And this amounts to p(l) = 1 - 18/37.  In probability theory percentages and fractions are always described as being a number between 0 and 1.  So our notation reading p(l) = 1 - 18/37 would be translated as p(l) = 1 - 0.4865.  So to solve this is p(l) = 1 - 0.4865 = 0.5135.

And we all know what 0.5135 is.  Yep.  That's our probability of losing on an Even Chance bet (19/37).

In probability theory, sometimes the easiest way to figure out a probability for a complex situation is to determine what the opposite situation is first, calculate that, and then invert it quickly to determine your answer - just like we did above.

For example if I'm a basketball player with an 80% free throw percentage, what is the probability of me hitting at least one free throw in three attempts?  This sounds complicated, until we look at what this really means.  If I know what the probability is of me missing all three free throws, then the opposite of that will tell me what the probability is of hitting just one of those three free throws.

So what is the probability of me missing all three free throws?  That's easy.  My free throw percentage is 80%.  Which means I have only a 20% chance of losing any given shot.  So I take 20% and multiply it three times (because I have three attempts).  20% x 20% x 20% = 0.20 x 0.20 x 0.20 = 0.008 = 0.8%.  So the probability of me missing all three free throws = 0.8%.  Now if I do the opposite of that, I know exactly what the probability is of me hitting at least one of my free throws.

p(hitting 1 of 3 free throws) = 1 - p(missing all 3 free throws) = 1 - 0.008 = 0.992 = 99.2%.  So the probability of me hitting at least one of three free throws is 99.2%.

Simple, huh?

The reason I'm pointing this out is because I wanted to figure out what the probability is for getting only one win in four spins, when betting on the Even Chances.

So to figure this out, we approach it the very same we did above.  If I know what the probability is for me losing 4 spins in a row, then I can calculate the inverse of that to determine the probability of me getting at least 1 win in 4 spins.

On Even Chance wagers, there is a 19/37 probability of losing = 0.5135 = 51.35%.  To lose 4 times in a row, the probability is 0.5135 x 0.5135 x 0.5135 x 0.5135 = 0.0695 = 6.95%.  So the probability of losing all 4 spins is 0.695%.  Now when I do the opposite of that, I know exactly what the probability is of winning at least once in 4 spins.

p(winning 1 of 4 spins) = 1 - p(losing all 4 spins) = 1 - 0.0695 = 0.9305 = 93.05%.  So the probability of me getting at least 1 win out of 4 spins when playing an Even Chance wager is 93.05%.

Okay, that is definitely promising.  In fact, even if I wanted to win once in just 3 spins the probability would still be p(l) = (0.5135)3 = 0.1354.  Then p(w) = 1 - p(l) = 1 - 0.1354 = 0.8646 = 86.46%.

But how can I make sure that I can set up such a scenario?  The last thing I want is to be caught in a vicious cycle where I'm betting larger and larger amounts to chase a win.

Well, the easiest way to handle this is simply to wait until one of the following happens: LL or LLL, and then wager 1 (or 2) units on your Even Chance bet.

So your "event" that your betting for is 1 win in 4 spins.  And to set that up, you wait until 3 virtual losses, and then place your wager one time.  That way your still trying to win just one in 4 spins, but your placing yourself here : Spin 1 (L), Spin 2 (L), Spin 3 (L), Spin 4 *Bet Here*.  And the probabilities remain the same.  That is, the probability of winning at least one even chance bet in 4 spins is over 93%, whether you decide to break in on the first spin or the fourth spin.  In our case, we'll take 3 virtual losses and then place our bet.

My last upgrade to the Parrondo paradox is to change the modulo from 3 to 9.  That is, when your Capital is divisible by 9, you play Game B₁, otherwise play Game B₂.  I changed this because every online simulator I ran showed that there were greater profits to be had, and greater potential for wins in general when the modulo was 9 instead of 3.

The full bet in it's entirety looks like this:

Game A - Bet $2 on an even chance bet.

Game B₁ - When your capital is divisible by 9, place a $1 wager on 2 Corner Bets.  I like to use the 10-11-13-14 Corner and the 28-29-31-32 Corner, because they are the only 2 Corners on the table that are relatively close together on the wheel.

Game B₂ - Wait for 3 consecutive virtual losses, then place a $2 wager on another Even Chance bet.  Option 2 - Place a $1 wager on 2 of the dozens and play a session to win $4 with "Relaxed Roulette"

There you have it.  The system that I've created based on Parrondo's Paradox.

I have replicated the conditions of the Parrondo games as close to the original as possible.

Now I just have to finish coding it...!

[nullified]

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