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Math Help - calculating P(X>=Y)

  1. #1
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    calculating P(X>=Y)

    hello,

    How do I calculate P(X>=Y) where X,y having the uniform density on {0,1...N}?
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  2. #2
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    Quote Originally Posted by adirh View Post
    hello,

    How do I calculate P(X>=Y) where X,y having the uniform density on {0,1...N}?
    Just make a grid. Y is the vertical axis, X is the horizontal axis.

    Every point has the same probability: \Pr(X = x_i, Y = y_j) = \frac{1}{(N + 1)^2} since there are (N+1)^2 points.

    There are N + 1 points along the diagonal Y = X.

    So the number of points such that X \geq Y is \frac{(N+1)^2 - (N+1)}{2} + (N+1) = \frac{(N+1)N}{2} + N + 1 = \frac{(N+1)(N+2)}{2}.

    Therefore \Pr(X \geq Y) = \frac{\frac{(N+1)(N+2)}{2}}{(N+1)^2} = \frac{N+2}{2(N+1)}.
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    Quote Originally Posted by adirh View Post
    hello,

    How do I calculate P(X>=Y) where X,y having the uniform density on {0,1...N}?
    Mr. F has already given you a first-rate answer, but here's another way to proceed just for the fun of it.

    By the formula for the probability of a union,

    P(X \geq Y) + P(Y \geq X) - P(X = Y) = P(X \geq Y \text{ or }Y \geq X) = 1

    Now P(X = Y) = \frac{1}{N+1}

    and P(X \geq Y) = P(Y \geq X) by symmetry, so

    2 \; P(X \geq Y) + \frac{1}{N+1} = 1.

    Solve this equation for  P(X \geq Y).
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