Here's another version.  This time the turn number doesn't matter: which game you play is decided randomly. 

You have three coins, Coin 1 which wins with probability 4/5, Coin 2 which always loses, and Coin 3 (used only in Game C to decide whether to play a turn of A or B) which is fair: probability 1/2 for heads and 1/2 for tails.

Game A: if fortune divisible by 3, flip Coin 2, otherwise flip Coin 1.

Game B: if fortune congruent to 1 mod 3, flip Coin 2, otherwise flip Coin 1.

Game C: Flip Coin 3.  If heads play a turn of Game A, if tails play a turn of Game B.  .

In Game A, the three states (fortune == 0 mod 3, 1 mod 3 and 2 mod 3) have long-run probabilities  21/51, 5/51 and 25/51 respectively. 
Your expected gain per turn is (-1)*21/51 + 3/5*5/51 + 3/5*25/51 = -1/17.  Similarly in Game B.

In Game C, you always flip Coin 1 if your fortune == 2 mod 3, otherwise effectively you're flipping a coin with winning probability 2/5.  The long-run
probabilities of the three states are then 13/55, 19/55 and 23/55, and the expected net gain per turn is (-1/10)*13/55 + (-1/10)*19/55 + 3/5*23/55 = 53/275.

> S0 := [[0,0,0,0],[0,1,1,0],[0,1,1,0],[0,0,0,0]]:
  n:= 4:
  X:= Array(0..n-1,0..n-1,S0);

X := Array(0 .. 3,0 .. 3,{(1, 1) = 1, (1, 2) = 1, (2, 1) = 1, (2, 2) = 1},datatype = anything,storage = rectangular,order = Fortran_order)

> S0 := [[0,0,0,0],[0,1,1,0],[0,1,1,0],[0,0,0,0]]:
  n:= 4:
  X:= Array(0..n-1,0..n-1,S0);

X := Array(0 .. 3,0 .. 3,{(1, 1) = 1, (1, 2) = 1, (2, 1) = 1, (2, 2) = 1},datatype = anything,storage = rectangular,order = Fortran_order)

As in your original post, use phaseportrait if you want solution curves too.

So for example:

> with(plots): with(DEtools):
  hyperb:= implicitplot( y^2 - x^2 - 1, x = -4 .. 4, y = -3 .. 3, colour=gold):
  phasep:= phaseportrait([(D(x))(t) = -.1+x(t)^2-x(t)*y(t), (D(y))(t) = y(t)^2-x(t)^2-1], [x(t), y(t)], t = -10 .. 10, 
   [ [x(-2) = -2, y(-2) = -1], [x(0) = 1, y(0) = 0], [x(0) = -0.04, y(0) = 1],[x(0)=-.14,y(0)=1.012]],
   x=-4..4,y=-3..3, stepsize=0.01, linecolour=[red,green,blue,blue]):
  display([hyperb, phasep], 

What other hyperbola were you expecting?

As in your original post, use phaseportrait if you want solution curves too.

So for example:

> with(plots): with(DEtools):
  hyperb:= implicitplot( y^2 - x^2 - 1, x = -4 .. 4, y = -3 .. 3, colour=gold):
  phasep:= phaseportrait([(D(x))(t) = -.1+x(t)^2-x(t)*y(t), (D(y))(t) = y(t)^2-x(t)^2-1], [x(t), y(t)], t = -10 .. 10, 
   [ [x(-2) = -2, y(-2) = -1], [x(0) = 1, y(0) = 0], [x(0) = -0.04, y(0) = 1],[x(0)=-.14,y(0)=1.012]],
   x=-4..4,y=-3..3, stepsize=0.01, linecolour=[red,green,blue,blue]):
  display([hyperb, phasep], 

What other hyperbola were you expecting?

That's a rather funky Maple leaf.  Here's a more standard flag.  It requires Standard GUI (because Classic doesn't fill polygons properly).

> with(plots):with(plottools):
  G1:= (p1,p2) -> arccos(.3/sqrt((p1[1]-p2[1])^2 + (p1[2]-p2[2])^2)) + arctan(p2[2]-p1[2],p2[1]-p1[1]):
  G2:= (p1,p2) -> -arccos(.3/sqrt((p1[1]-p2[1])^2 + (p1[2]-p2[2])^2)) + arctan(p2[2]-p1[2],p2[1]-p1[1]):
  C:=[31.5,2],[32.5,2], 
    op(op(arc([32.5,8], .3,  G2([32.5,8],[32.5,2])+2*Pi.. G1([32.5,8],[39,7.5])))),
    [39,7.5],
    op(op(arc([38.3,9.5],.3, G2([38.3,9.5],[39,7.5])+2*Pi .. G1([38.3,9.5],[44.6,15])))),
    [44.6,15],
    op(op(arc([43.3,16.1],.3, G2([43.3,16.1],[44.6,15])+2*Pi .. G1([43.3,16.1],[44.1,20.1])))),
    [44.1,20.1],
    op(op(arc([40.5,20],.3, G2([40.5,20],[44.1,20.1])+2*Pi .. G1([40.5,20],[39.5,21.5])))),
    [39.5,21.5],
    op(op(arc([35.8,18.3],.3,G2([35.8,18.3],[39.5,21.5])+2*Pi .. G1([35.8,18.3],[37,25.5])))),
    [37,25.5],
    op(op(arc([34.3,24.9],.3,G2([34.3,24.9],[37,25.5])+2*Pi .. G1([34.3,24.9],[32,29])))),
    [32,29]:
  C:= C,op(ListTools[Reverse](map(t -> [64-t[1],t[2]], [C]))):
  display([POLYGONS([C],COLOUR(RGB,1,0,0)),
    rectangle([0,0],[16,32],colour=red),
    rectangle([48,0],[64,32],colour=red),
    curve([[0,0],[64,0],[64,32],[0,32],[0,0]])],scaling=constrained,axes=none);

Where are you getting this .33 and .66?  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.

Where are you getting this .33 and .66?  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.

Oops, silly of me... if f and g are not coprime, the resultant is 0, and then it's easy (you could e.g. take A = g and B=-f). 

Yes, you can use gcdex.  In fact you may as well use the 6-parameter form of gcdex:

> gcdex(f, g, resultant(f,g,x), x, 'A', 'B');

Oops, silly of me... if f and g are not coprime, the resultant is 0, and then it's easy (you could e.g. take A = g and B=-f). 

Yes, you can use gcdex.  In fact you may as well use the 6-parameter form of gcdex:

> gcdex(f, g, resultant(f,g,x), x, 'A', 'B');

Sorry if I didn't explain this enough.  Really, I'm trying to make it as simple as possible.

When your fortune is a multiple of 3, in Game B you use Coin 2 that always loses.  Therefore in Game B, you can't ever increase above a multiple of 3.
From 3*n you always go to 3*n-1, and then you throw Coin 3 that most of the time takes you back to 3*n.

Sorry if I didn't explain this enough.  Really, I'm trying to make it as simple as possible.

When your fortune is a multiple of 3, in Game B you use Coin 2 that always loses.  Therefore in Game B, you can't ever increase above a multiple of 3.
From 3*n you always go to 3*n-1, and then you throw Coin 3 that most of the time takes you back to 3*n.

It may be helpful to consider a more extreme case.  Suppose Coin 2 always loses, while Coin 3 wins with probability 1-e, 0 < e << 1, and Coin 1 wins with probability 1/2 - e.  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, but occasionally Coin 3 would lose twice in a row, so you get to 3*n-3, and then you're in the same position as before but with n replaced by n-1.  So in the long run, you certainly lose in Game B.

Now consider Game C, starting in a turn divisible by 3. 
If your current fortune is 3*n, then with probability almost 1/2 you win with Coin 1 and then twice with Coin 3, so at the start of the next turn divisible by 3 you are at 3*(n+1); with probability almost 1/2 you lose with Coin 1, then win with Coin 3, then lose with Coin 2, resulting in 3*n-1.
If your current fortune is 3*n+1, with probability almost 1/2 you win with Coin 1, win with Coin 3 and lose with Coin 2, resulting in 3*n+2, while with probability almost 1/2 you lose with Coin 1, lose with Coin 2 and win with Coin 3, resulting in 3*n.
If your current fortune is 3*n+2, with probability almost 1/2 you win with Coin 1, lose with Coin 2 and win with Coin 3, while with probability almost 1/2 you lose with Coin 1 and then win twice with Coin 3, in both cases resulting in 3*(n+1).
The result is that only rarely will your fortune be of the form 3*n+1 starting in a turn divisible by 3.  Either it will be 3*n+2, resulting with high probability in
3*n+3 three turns later, or it will be 3*n, resulting with high probability three turns later in either 3*n-1 (which, as I just mentioned, with high probability would go back to 3*n in another three turns) or 3*(n+1).  So in the long run, you win in Game C.   

It may be helpful to consider a more extreme case.  Suppose Coin 2 always loses, while Coin 3 wins with probability 1-e, 0 < e << 1, and Coin 1 wins with probability 1/2 - e.  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, but occasionally Coin 3 would lose twice in a row, so you get to 3*n-3, and then you're in the same position as before but with n replaced by n-1.  So in the long run, you certainly lose in Game B.

Now consider Game C, starting in a turn divisible by 3. 
If your current fortune is 3*n, then with probability almost 1/2 you win with Coin 1 and then twice with Coin 3, so at the start of the next turn divisible by 3 you are at 3*(n+1); with probability almost 1/2 you lose with Coin 1, then win with Coin 3, then lose with Coin 2, resulting in 3*n-1.
If your current fortune is 3*n+1, with probability almost 1/2 you win with Coin 1, win with Coin 3 and lose with Coin 2, resulting in 3*n+2, while with probability almost 1/2 you lose with Coin 1, lose with Coin 2 and win with Coin 3, resulting in 3*n.
If your current fortune is 3*n+2, with probability almost 1/2 you win with Coin 1, lose with Coin 2 and win with Coin 3, while with probability almost 1/2 you lose with Coin 1 and then win twice with Coin 3, in both cases resulting in 3*(n+1).
The result is that only rarely will your fortune be of the form 3*n+1 starting in a turn divisible by 3.  Either it will be 3*n+2, resulting with high probability in
3*n+3 three turns later, or it will be 3*n, resulting with high probability three turns later in either 3*n-1 (which, as I just mentioned, with high probability would go back to 3*n in another three turns) or 3*(n+1).  So in the long run, you win in Game C.   

> with(Statistics):
  X:= Sample(Normal(0, 0.5), 4) + I*Sample(Normal(0, 0.5), 4);
  ArrayTools:-Reshape(X,2,2);