Q: 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?

A: 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:

<< Discrete-Math`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]]

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