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 Seeing red (Posted on 2008-06-07)

The interior of the square [0,1] x [0,1] is initially coloured white.

Four random numbers: u,v,x,y in the range [0,1] are selected and the points inside the rectangle formed by the corners (u,v), (x,y), (u,y), (x,v) are recoloured: areas painted white are repainted red, and areas painted red are repainted white.

This recolouring procedure is repeated N times. Show that the expectation value of the red area is given by the formula:

1/2 - 1/2 ∫ dA ∫ dB {1 - 2 A (1-A) B (1-B)}N

where both integrals go from 0 to 1.

 Submitted by FrankM Rating: 5.0000 (1 votes) Solution: (Hide) 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.

 Subject Author Date re: error in solution Harry 2009-03-19 20:00:02 re(2): error in solution Eigenray 2008-06-10 19:53:05 re: error in solution FrankM 2008-06-10 12:28:17 error in solution Eigenray 2008-06-09 16:26:14 Power in independence Gamer 2008-06-08 21:43:26

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