Because of the symmetric roles of (u,x) and (v,y) we can without loss of generality assume that
u < x and v < y. A point (A,B) interior to the square will lie within the rectangle defined by the
random numbers u,v,x,y if and only if
u < A < x and v < B < y
Now the probability that u < A is given by A, and the probability of A < x is given by (1-A).
Hence, the probability that (A,B) lying in the given rectangle is Θ = A(1-A)B(1-B). Let
P_{N}(A,B) be the probability that a point (A,B) is painted red after N recolourations.
Then (A,B) must lie within an odd number of rectangles, so
P_{N}(A,B) = N Θ (1 - Θ)^{N-1} + {N (N-1) (N-2)/6} Θ^{3} (1 - Θ)^{N-3} + ..
I.e. P_{N}(A,B) is the sum of the odd numbered terms in the binomial expansion of
[(1 - Θ) + Θ]^{N}. But
[(1 - Θ) + Θ]^{N} = (1 - Θ)^{N} + N Θ (1 - Θ)^{N-1} + {N (N-1)/2} Θ^{2} (1 - Θ)^{N-2} + {N (N-1) (N-2)/6} Θ^{3} (1 - Θ)^{N-3} + ..
[(1 - Θ) - Θ]^{N} = (1 - Θ)^{N} - N Θ (1 - Θ)^{N-1} + {N (N-1)/2} Θ^{2} (1 - Θ)^{N-2} - {N (N-1) (N-2)/6} Θ^{3} (1 - Θ)^{N-3} + ..
So P_{N}(A,B) = [1 - (1 - 2 Θ)^{N}]/2
The measure of the red area is then computed by integrating P_{N}(A,B) over the square. |