ok yes that makes sense. From my investigation this is the probabilities that I get after
5000 simulations ( I have renamed the games slightly because I dont like starting with a
game called game C but that is just my personal preferences)



EV[Game*C] = .38*(0.5e-1*(+1)+.95*(-1))+.62*(.75*(+1)+.25*(-1));
                            EV[Game C] = -0.0320

 EV[Game*B] = .49*(+1)+.51*(-1);
                             EV[Game B] = -0.02

EV[Game*A] = .33*(-0.02)+.66*(-0.0320);
                           EV[Game A] = -0.027720

The problem is that as you can see that the expected value for Game A is also negative which
is not according to our theory. We can therefore use an alternative way to calculate the
expected value for Game A. We can look at how many times coin1, coin2 and coin3 occured in
game A. The data is as seen below where PC1 is the probability coin 1 etc :             
                     
                                 PC1 = 0.333
                                 PC2 = 0.204
                                 PC3 = 0.462
                                Total = 0.999

So we get:

EV[Game*A] = .33*(0.49*(+1)+0.51*(-1))+0.204*(0.05*(+1)+0.95*(-1))+0.462*(0.75*(+1)+0.25*(-1));

                            EV[Game A] = 0.04080

Which is posistive. Maybe I have done something wrong here ...hummm

ok yes that makes sense. From my investigation this is the probabilities that I get after
5000 simulations ( I have renamed the games slightly because I dont like starting with a
game called game C but that is just my personal preferences)



EV[Game*C] = .38*(0.5e-1*(+1)+.95*(-1))+.62*(.75*(+1)+.25*(-1));
                            EV[Game C] = -0.0320

 EV[Game*B] = .49*(+1)+.51*(-1);
                             EV[Game B] = -0.02

EV[Game*A] = .33*(-0.02)+.66*(-0.0320);
                           EV[Game A] = -0.027720

The problem is that as you can see that the expected value for Game A is also negative which
is not according to our theory. We can therefore use an alternative way to calculate the
expected value for Game A. We can look at how many times coin1, coin2 and coin3 occured in
game A. The data is as seen below where PC1 is the probability coin 1 etc :             
                     
                                 PC1 = 0.333
                                 PC2 = 0.204
                                 PC3 = 0.462
                                Total = 0.999

So we get:

EV[Game*A] = .33*(0.49*(+1)+0.51*(-1))+0.204*(0.05*(+1)+0.95*(-1))+0.462*(0.75*(+1)+0.25*(-1));

                            EV[Game A] = 0.04080

Which is posistive. Maybe I have done something wrong here ...hummm

Robert it is very interesting but I must say I am struggling to follow along. This was very advanced :-)
Is it possible to illustrate your points in a much simpler way for the weak brains. I am including myself :-)

"Now in Game B it is impossible to increase your fortune above a multiple of 3; starting with 3*n,
most of the time you would alternate between 3*n-1 and 3*n,"

why is that ?

Robert it is very interesting but I must say I am struggling to follow along. This was very advanced :-)
Is it possible to illustrate your points in a much simpler way for the weak brains. I am including myself :-)

"Now in Game B it is impossible to increase your fortune above a multiple of 3; starting with 3*n,
most of the time you would alternate between 3*n-1 and 3*n,"

why is that ?

Even though I am considering myself to be well educated this is the kind of things that
make me regret that I did not take any mathematics courses at uni. Extremely cool and fascinating stuff :-)

Even though I am considering myself to be well educated this is the kind of things that
make me regret that I did not take any mathematics courses at uni. Extremely cool and fascinating stuff :-)

all right thanx. Yes I came to that conclussion as well after I had been looking at some other webpages.
Now the only question that remains is why does the binary entropy function have the form   H := -p*log(p)-(1-p)*log(1-p);
I read somewhere that is it -1 * likelihood function but I am not sure it is correct..

all right thanx. Yes I came to that conclussion as well after I had been looking at some other webpages.
Now the only question that remains is why does the binary entropy function have the form   H := -p*log(p)-(1-p)*log(1-p);
I read somewhere that is it -1 * likelihood function but I am not sure it is correct..

Thanx Axel :-)  I appreciate your input but it all seem to be very "add hoc". Does there not exist a much easier way ?

Thanx Axel :-)  I appreciate your input but it all seem to be very "add hoc". Does there not exist a much easier way ?

Exactly my point :-) That would be sweet ! 

Exactly my point :-) That would be sweet ! 

No I agree the picture is not a good representation of what I want. It is just that I could not find a
picture of a cool looking cube. I know I can loop over a 3D array ie A[x,y,z] but it would be nice if I could visualise
the data structure which is a cube where the data is stored  

No I agree the picture is not a good representation of what I want. It is just that I could not find a
picture of a cool looking cube. I know I can loop over a 3D array ie A[x,y,z] but it would be nice if I could visualise
the data structure which is a cube where the data is stored  

First I thought that the lack of the ability to delete a file was bull crap because every user should be able
to delete file that they have uploaded. However, when I think about it a little more it does make sense
because if all users could delete file that they had uploaded then MaplePrimes would become
very fractionalized. You could read one forum post where the user referes to a picture or Maple file
that no longer exists because the user has deleted it .