Probability of red-red

You have a cube with two sides painted red. You roll the cube N times. What is the probability that you will have rolled 'red' on at least two consecutive throws?

Bill Ryan

Maybe 1/9?

(2/6 * 2/6 = 4/36 = 1/9)

Luciano

For N=0, it is 0. For N=1, it is 0. For N=2, it is 1/9. For N=3, it is 5/27.

The pattern isn't obvious to me. For N large you have N-1 pairs of consecutive rolls, with a 1/9 chance of each pair being two reds, but these pairs aren't really independent events. Is there a known asymptotoc formula for this sort of thing?

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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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z [ 2/3 ] [ 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z [ 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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) = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 ] . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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 . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . Then X(1) = [ 1/3 ] and X(N+1) = M X(N) where M = . 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)

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, . 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)

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

is the probability that you will have rolled 'red' on at least two consecutive throws?

Georgemw replied:

of consecutive rolls, with a 1/9 chance of each pair being two reds, but these pairs aren't really independent events.

Is there a known asymptotoc formula for this sort of thing? ***** I'm not aware of a 'known asymptotoc formula', but I will give my own ideas on the subject.

As Robert Israeel pointed out, it is easier to examine the probability of NOT getting RR after N rolls. Not surprisingly, it is a decreasing exponential function of N.

For notation purposes, let's paint the non-red sides green (G), Let r = the probability of rolling red and g be the probability of G (=1-r).

The probability of no RR in N rolls is r^N * f(N) where f(N) is a series somewhat akin to Fibonacci, except that each term in the series is a multiple of the sum of the two preceding terms.

The first term is f(1) = 1/r The second term is f(2) = (1/r)^2-1 Each subsequent term[f(N) is equal to: [g/r] * [f(N-2) + f(N-1)]

For a cube with two red sides, r= 1/3 and g/r =2. The first several terms of f(N) are 3, 8, 22, 60, 164, 448, etc. Thus, the probability of no RR after 4 rolls is 60/81.

Using the formula for successive terms, the ratio (k) between successive terms quickly approaches a limit (k) equal to the solution to the equation (for r=1/3): 2(1+k) = k^2; k = sqrt(3) +1.

The value of f(N) can be calculated directly by rounding k^N * 1.07735. I have no idea of what 1.07735 represents except that it works. ***** The above formulas are not too difficult to develop. All that is involved are quadratic equations and the reasonable assumption that, if there is no RR so far, the probability that the last term is G is constant. Among other things,,, you will find that the probability that the last roll is G = sqrt(3)-1 or about .732.

After N rolls with no RR, let : a=probability that last roll is G b= probability that last roll is R a+b = total probability of no RR

Then calculate the results after the next two rolls and you will see the above unfold. ***** For r =1/5, the series of f(N) is 5, 24, 116, 560, etc. The ratio (k) between successive terms approaches the solution to: 4(1+k)=k^2; k=sqrt(8) +2 = 4..828…

***** So, there is a pattern to these situations, but it doesm't leap out at you at first glance.

Bill Ryan