Robert do you have any good maple code for his "another version" of the parrondo game that I could try ?!
Sounds pretty cool. Preferably the code for the equity curves as in the original post ie Game A (red line),
game B (blue line) and game C (green upward sloping line) ?!

Robert do you have any good maple code for his "another version" of the parrondo game that I could try ?!
Sounds pretty cool. Preferably the code for the equity curves as in the original post ie Game A (red line),
game B (blue line) and game C (green upward sloping line) ?!

One thing that would be nice though, when you only have one window open such a
window should still be displayed as one single tab. For example if you have two tabs you can switch
beteween them and you also have that little star indicating that you need to save one of your worksheet.
When you only have one window you can not see that star unless you look in the top left corner. I would
have prefered to look at the same place for an one/two tab window  :-)

One thing that would be nice though, when you only have one window open such a
window should still be displayed as one single tab. For example if you have two tabs you can switch
beteween them and you also have that little star indicating that you need to save one of your worksheet.
When you only have one window you can not see that star unless you look in the top left corner. I would
have prefered to look at the same place for an one/two tab window  :-)

Fair enough :-)

You might be right. I am talking from my UK experience. Here you can get a nice laptop for
400 GBP (approx 650 USD). I have seen laptops that cost 400*2 = 800 GBP, 400*3 = 1200 GBP,
and 400*4=1600 GBP and even more. I promise you that even if you buy the "best" laptop available
in todays market you will want to buy an new computer in 3-4 years time anyway (mostly because
you again will be seduced by cleverly designed marketing campaign) so why even
bother chasing performance figures ?.

"My policy has, more or less, been to buy the best that money can buy and then make it last as long as possible."

I dont not mean to tell you what do nor do I mean any disrespect but I would not personally pursue such a strategy
because there are usually incremental differences between the products at any point in time and such differences
are certainly not worth a double, tripple, or quadruple price tag which we usually see for "high-end" laptops etc.
This is exactly what I mean when it comes to marketing. These companies are so screwed that you have to
micro-manage them. I think everyone wants the "best" unfortunatly the best is more a function of time than
product variety. Major technology improvments are usually stochastic. The micro processor was one of those
major leaps. After that I think the focus has been to make money rather than pushing boundaries.
The speed of development is painfully slow.... 
 

I just have a distinct feeling that these computer companies have been selling the same stuff under different names
for the last 20 years. They are extremely skilled when it comes to marketing / borderline dishonest. They communicate very
aggressively that the users always need to have the latest stuff ie last year we sold XX 50 but now we are selling
sell XX 5000 which is much much much better than previous model. When the reality is that incremental improvements
even though they are branded as major "leaps" have little value to the average user. The value for the computer company is
clear. To make as much money as possible.

Take the brand name intel Core i7 for example. I find the name to be misleading because it is easy to infer that the computer
have 7 cores which is not the case. I am just wondering when we will come to a point when we will have the speed of photonic
computer (speed will not be a function of new models realizes and money will not be made from aggressive and often misleading marketing) what will the computer companies derive their revenue from ?!

This will be the last flag for me today. Feel free to add more flags to this blog post if you feel like it :-)

Here is the French flag

restart:
with(plots):
with(plottools):

display({rectangle([0, 3], [2, 0], color = "DarkBlue"), rectangle([2, 3], [4, 0], color = white), rectangle([4, 3], [6, 0], color = red)}, axes = none);

Here is the Italian flag

restart:
with(plots):
with(plottools):

display({rectangle([0, 3], [2, 0], color = "DarkGreen"), rectangle([2, 3], [4, 0], color = white), rectangle([4, 3], [6, 0], color = red)}, axes = none);

Here is the German flag

restart:
with(plots):
with(plottools):

display({rectangle([0, 1], [6, 0], color = yellow), rectangle([0, 2], [6, 1], color = red), rectangle([0, 2], [6, 3], color = black)}, axes = none);
 

Here is the Swedish flag or at least something that resembles it .

restart:
with(plots):
with(plottools):

display({rectangle([0, 2], [6, 1.5], color = yellow), rectangle([0, 1.5], [2, 0], color = blue), rectangle([0, 3.5], [2, 2], color = blue), rectangle([2, 0], [2.5, 3.5], color = yellow), rectangle([2.5, 2], [6, 3.5], color = blue), rectangle([2.5, 1.5], [6, 0], color = blue)}, axes = none);

I tend to agree with you Robert. My flag (or more specifically Andre Heck's flag) look more like a marijuana leaf
compared to your flag ha ha. I hope the people of canada was not insulted too much  :-)

The first thing:

"The expected value for Game A is not just a weighted sum of the expected values for Game B and Game C, because a turn of one game alters the distribution of the states for the other game"

that is interesting. I have not understod that. Could you please show that in an simple example ?! So what do you suggest conditional probabilities ? or am I forced to learn complicated Markov chain notation ?! I guess it makes sence since mod Capital (mod C) changes depending on your current wealth ( even though it should be the same for + and - wealth) however mod Time (mod T) will remain the the same (not dependent of wealth) according to my understanding.


The second thing I got 0.33 and 0.66 from the below simulation code: 

restart:
with(ListTools):

X := proc (ap) local n, t, x, data1;
n := 5000:

for t to n do
if t mod ap = 0 then x[t] := 1 :
elif t mod ap <> 0 then x[t] := 0 : end if :
end do;

data1 := [seq(x[t], t = 1 .. n)]:

print(["mod T =", ap, "Probability Game B =", evalf(Occurrences(1, data1)/n, 3), "Probability Game C =", evalf(Occurrences(0, data1)/n, 3)])

end proc:

seq(X(ap), ap = 1 .. 10);

["mod T =", 1, "Probability Game B =", 1., "Probability Game C =", 0.]
["mod T =", 2, "Probability Game B =", 0.500, "Probability Game C =", 0.500]
["mod T =", 3, "Probability Game B =", 0.333, "Probability Game C =", 0.667]
["mod T =", 4, "Probability Game B =", 0.250, "Probability Game C =", 0.750]
["mod T =", 5, "Probability Game B =", 0.200, "Probability Game C =", 0.800]
["mod T =", 6, "Probability Game B =", 0.167, "Probability Game C =", 0.833]
["mod T =", 7, "Probability Game B =", 0.143, "Probability Game C =", 0.857]
["mod T =", 8, "Probability Game B =", 0.125, "Probability Game C =", 0.875]
["mod T =", 9, "Probability Game B =", 0.111, "Probability Game C =", 0.889]
["mod T =", 10, "Probability Game B =", 0.100, "Probability Game C =", 0.900]

Another thing that is interesting to note is that the relationship between mod Capital (mod C) and mod Time (mod T) is not linear. For example if you set mod T and mod C =3 the return is + however if you set mod T=4 and mod C =3 then your return will be negative. We can see in the below 3d picture that a higher mod C is better for returns but that the value of mod T is also highly important ( valley vs peak )

restart:
with(plots):
with(ListTools):
with(Statistics):
randomize():

XX := proc (apT, apC) local n, coin1, coin2, coin3, A, Coin1, Coin2, Coin3, t, zA;

n := 5000:
coin1 := Sample(Bernoulli(0.49), n):
coin2 := Sample(Bernoulli(0.05), n):
coin3 := Sample(Bernoulli(0.75), n):

A[0] := 0:

for t to n do

Coin1 := 2*round(coin1[t])-1:
Coin2 := 2*round(coin2[t])-1:
Coin3 := 2*round(coin3[t])-1:

 if `mod`(t, apT) = 0 then A[t] := A[t-1]+Coin1:   zA[t] := 1:
elif `mod`(t, apT) <> 0 then if `mod`(A[t-1], apC) = 0  then  A[t] := A[t-1]+Coin2:   zA[t] := 2:   else A[t] := A[t-1]+Coin3:   zA[t] := 3:  end if end if :
end do:

A[n] ;

end proc:

data1 := [seq([seq([apT, apC, XX(apT, apC)], apT = 1 .. 10, 1)], apC = 1 .. 10, 1)]:

surfdata(data1, axes = boxed, labels = ["mod T", "mod C", "Total Return"], labelfont = [times, roman, 14], labeldirections = [default, default, vertical], title = "mod T vs mod C", titlefont = [times, roman, 16]);





 

The first thing:

"The expected value for Game A is not just a weighted sum of the expected values for Game B and Game C, because a turn of one game alters the distribution of the states for the other game"

that is interesting. I have not understod that. Could you please show that in an simple example ?! So what do you suggest conditional probabilities ? or am I forced to learn complicated Markov chain notation ?! I guess it makes sence since mod Capital (mod C) changes depending on your current wealth ( even though it should be the same for + and - wealth) however mod Time (mod T) will remain the the same (not dependent of wealth) according to my understanding.


The second thing I got 0.33 and 0.66 from the below simulation code: 

restart:
with(ListTools):

X := proc (ap) local n, t, x, data1;
n := 5000:

for t to n do
if t mod ap = 0 then x[t] := 1 :
elif t mod ap <> 0 then x[t] := 0 : end if :
end do;

data1 := [seq(x[t], t = 1 .. n)]:

print(["mod T =", ap, "Probability Game B =", evalf(Occurrences(1, data1)/n, 3), "Probability Game C =", evalf(Occurrences(0, data1)/n, 3)])

end proc:

seq(X(ap), ap = 1 .. 10);

["mod T =", 1, "Probability Game B =", 1., "Probability Game C =", 0.]
["mod T =", 2, "Probability Game B =", 0.500, "Probability Game C =", 0.500]
["mod T =", 3, "Probability Game B =", 0.333, "Probability Game C =", 0.667]
["mod T =", 4, "Probability Game B =", 0.250, "Probability Game C =", 0.750]
["mod T =", 5, "Probability Game B =", 0.200, "Probability Game C =", 0.800]
["mod T =", 6, "Probability Game B =", 0.167, "Probability Game C =", 0.833]
["mod T =", 7, "Probability Game B =", 0.143, "Probability Game C =", 0.857]
["mod T =", 8, "Probability Game B =", 0.125, "Probability Game C =", 0.875]
["mod T =", 9, "Probability Game B =", 0.111, "Probability Game C =", 0.889]
["mod T =", 10, "Probability Game B =", 0.100, "Probability Game C =", 0.900]

Another thing that is interesting to note is that the relationship between mod Capital (mod C) and mod Time (mod T) is not linear. For example if you set mod T and mod C =3 the return is + however if you set mod T=4 and mod C =3 then your return will be negative. We can see in the below 3d picture that a higher mod C is better for returns but that the value of mod T is also highly important ( valley vs peak )

restart:
with(plots):
with(ListTools):
with(Statistics):
randomize():

XX := proc (apT, apC) local n, coin1, coin2, coin3, A, Coin1, Coin2, Coin3, t, zA;

n := 5000:
coin1 := Sample(Bernoulli(0.49), n):
coin2 := Sample(Bernoulli(0.05), n):
coin3 := Sample(Bernoulli(0.75), n):

A[0] := 0:

for t to n do

Coin1 := 2*round(coin1[t])-1:
Coin2 := 2*round(coin2[t])-1:
Coin3 := 2*round(coin3[t])-1:

 if `mod`(t, apT) = 0 then A[t] := A[t-1]+Coin1:   zA[t] := 1:
elif `mod`(t, apT) <> 0 then if `mod`(A[t-1], apC) = 0  then  A[t] := A[t-1]+Coin2:   zA[t] := 2:   else A[t] := A[t-1]+Coin3:   zA[t] := 3:  end if end if :
end do:

A[n] ;

end proc:

data1 := [seq([seq([apT, apC, XX(apT, apC)], apT = 1 .. 10, 1)], apC = 1 .. 10, 1)]:

surfdata(data1, axes = boxed, labels = ["mod T", "mod C", "Total Return"], labelfont = [times, roman, 14], labeldirections = [default, default, vertical], title = "mod T vs mod C", titlefont = [times, roman, 16]);