You have a cube with two sides painted red.

Let u(N) be the probability in N throws of having no two consecutive 
reds, and ending on a non-red; v(N) the probability in N throws of having
no two consecutive reds and ending on a red.  Let X(N) be the vector

[u(N)]               [ 2/3 ]                               [ 2/3  2/3 ]
[v(N)].  Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = [ 1/3   0  ]. 

Using the diagonalization M = V D V^(-1) where

    [ 3 r1   3 r2 ]         [ r1  0 ]
V = [   1      1  ] and D = [  0 r2 ],

r1 = (1+sqrt(3))/3, 
r2 = (1-sqrt(3))/3.

We get 

       [ (sqrt(3)+3) r1^n/6 + (3-sqrt(3)) r2^n/6 ]
X(N) = [     sqrt(3) r1^n/6 - sqrt(3) r2^n/6 ]

and your probability is 

1 - u(N) - v(N) = 1 - r1^n (1/2+sqrt(3)/3) - r2^n (1/2-sqrt(3)/3)

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2

Here are the first twenty values of this probability:

0: 0/1 = 0
1: 0/3 = 0
2: 1/9 = 0.111111
3: 5/27 = 0.185185
4: 21/81 = 0.259259
5: 79/243 = 0.325103
6: 281/729 = 0.38546
7: 963/2187 = 0.440329
8: 3217/6561 = 0.490322
9: 10547/19683 = 0.535843
10: 34089/59049 = 0.5773
11: 108955/177147 = 0.615054
12: 345137/531441 = 0.649436
13: 1085331/1594323 = 0.680747
14: 3392377/4782969 = 0.709262
15: 10549739/14348907 = 0.735229
16: 32667201/43046721 = 0.758878
17: 100782787/129140163 = 0.780414
18: 309946697/387420489 = 0.800027
19: 950599131/1162261467 = 0.817888

Here is how to solve this problem using Mathematica:

<< DiscreteMath`RSolve`

RSolve[{u[n] == 2/3 u[n - 1] + 2/3 v[n - 1],
  v[n] == 1/3 u[n - 1],
  u[1] == 2/3,
  v[1] == 1/3},
  {u, v}, n]