Dr. Patrick T

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15 years, 226 days

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This is a fascinating post, thanks to all the contributors. Robert, this is a very clear explanation of this Parrondo paradox. I wasn't familiar with it (though I'd heard of it), but I'm immediately drawn to it! Like Alex, I also value the verbal descriptions of what's going on.

First question: point of detail. Refer to Robert's post with subject "0.495" above. There it says "Now if you play a turn of game A, even though that is unfavourable in itself, it decreases the probability that your fortune will be 0 mod 3 in the next turns." How does playing game A make it less likely that fortune will be a multiple of 3? Sorry if it's obvious.

And second question: general point. Is it a correct description of the Parrondo game structure to make the following generalization?

Let x,y,z denote state variables (i.e. they evolve over time according to the outcome of the stochastic games). Let x denote wealth and (for simplicity) let y,z denote binary, indicator variables equal to either 1 or 0. Let E(x) denote the expected value of wealth x in a particular subgame.

I write the architecture of the Parrondo game as follows:

if z=0, then                                [game A]

if z=1, then                                [game B]

                  if y=1, then              [subgame B1]

                  if y=0, then              [subgame B2]

Above, the indicator variable z is "time" (turn number), which evolves deterministically and independently, while the indicator variable y is "the property that x=0 mod 3", which evolves stochastically and also depends on z.

Both game A and game B are losing games when played independently of each other. However, game B contains a winning subgame (subgame B1). The Parondo paradox arises whenever the outcome of game A can bias game B in favor of the winning subgame B1. What turns a combination of losing games into a winning one is that the outcome of game A influences the value of the indicator/state variable y in favor of the winning subgame B1.

Is this a fair description of the general mechanism at work in the Parrondo paradox? And if so, doesn't the paradox arise from the "framing" of game B as one single losing game, rather than framing it as two separate (albeit interdependent) games B1 and B2?

By "framing" I just mean the way the Parrondo game is presented, where bundling subgames B1 and B2 into one single game B obscures the existence of a winning subgame that it is actually possible to play more often than it would first seem possible. No?

Do you mean this?

tvalues:= fsolve(subs(curveq, surfeq), t=-2..2);

which will return the same answers as before since they were in the interval specified. However, say:

tvalues:= fsolve(subs(curveq, surfeq), t=0..2);

                        tvalues := 1.291980321


plot understands the discontinuity without the option discont=true:

plot(tan(x), x = -5..5, y = -5..5, style=point, grid=[50,50]);

plot(tan(x), x = -5..5, y = -5..5, grid=[50,50]);

Like Doug I too have had difficulty uploading an animated gif on mapleprimes. I'd like to give it another try. As the gif is too large to be uploaded with mapleprimes, I uploaded it with imageshack:

I saved the gif using the context-menu option "Export As" and select "Graphics Interchange Format (GIF)".

Let's see if it will display here and show the animation:


The animation was created with the code above, using

  x=-1.5..1.5, y=-1.5..1.5,

you may want to start with using "shorthand" notation, like defining:

a := ((R4*l^2+sigma)/R4)^(1/2);
b := 7/6*varpi*(R1-1)*l;

etc., and using subs to substitute the as and bs into the messy expression.

Another thing you may want to do is replace the cosh and sinh by their exponential forms and see if that simplifies to something more compact.

your feedback is urgently expected.

not sure I understand your question ... do the following help?

plottools[reflect](p, [0,0],[1,1]);
plottools[rotate](p, Pi/2, [0,0]);

?pointplot will point you to the help page, does it do what you're looking for?

you will need to define Vectors (not vectors) to use Robert's code, thus:

X:= Vector([-1,0,1]);
Y:= Vector([1+I*2,3+I,4+3*I], datatype=complex);

I don't think you can control the position of the labels automatically. But you can remove them altogether and then redefine text labels that you could place where you wanted to. To remove the labels, you can do something like this:

plots:-implicitplot3d(....., labels=[``,``],.....)


plots:-display(theplot, labels=[``,``],.....)

From the context menu, you may be able to do something (untested):


oh, but then it evaluates the Jacobian at (x,y,z) rather than (x(t),y(t),z(t))...

It all depend on how good you want the figures to look. Currently 2D plots look a lot better than 3D plots. There are also differences between standard and classic gui. I personally save a basic plot in postscript format and add fancy decorations with pstricks and pstricks-add.

Of possible interest:

to get you started,

SteL:=-1: SteV:=-1: h:=1: R1:=1: #select parameters that make sense
is(SteL<0 and SteV<0 and h>0 and R1>0); # this should return true if your assumptions are satisfied
eq := sqrt(Pi)*beta=-1/h*SteL/(exp(h^2*beta^2)*erf(h*beta))
+R1*SteV/(exp((beta/R1)^2)*erfc(beta/R1)); # this is your equation
fsolve(eq,beta); # this is for numerical solutions, different  from solve(eq,beta)
is(%>0); # this should return true if the solution has beta>0

eq := theta0 = 10 * (2.718281828 - exp(S) ) / exp(S);
p := plots:-implicitplot(eq, theta0=-1..1, S=-1..1):

these {} define a set (no order), while these [] define a list (with order), so what method have you tried?

the square brackets define a "list", replace them by round brackets.

[] -> ().

if problems arise, copy-paste your lines of code together with the error message (if any).

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