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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
PN(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
PN(A,B) = N Θ (1 - Θ)N-1 + {N (N-1) (N-2)/6} Θ3 (1 - Θ)N-3 + ..
I.e. PN(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} Θ2 (1 - Θ)N-2 + {N (N-1) (N-2)/6} Θ3 (1 - Θ)N-3 + ..
[(1 - Θ) - Θ]N = (1 - Θ)N - N Θ (1 - Θ)N-1 + {N (N-1)/2} Θ2 (1 - Θ)N-2 - {N (N-1) (N-2)/6} Θ3 (1 - Θ)N-3 + ..
So PN(A,B) = [1 - (1 - 2 Θ)N]/2
The measure of the red area is then computed by integrating PN(A,B) over the square. |
