Product Over One Half

Draw $X$ and $Y$ independently, each uniform on $[0,1]$ . What is the probability that their product exceeds $\tfrac{1}{2}$ ?

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#A region in the unit square

The pair $(X,Y)$ is uniform on the unit square, so the probability is just an area. The product clears $\tfrac{1}{2}$ only when both draws run large. Since $Y \le 1$ , we need $X > \tfrac{1}{2}$ , and then $Y > \tfrac{1}{2X}$ . The favourable points form the thin wedge above the hyperbola $y = \tfrac{1}{2x}$ .

#Integrate the wedge

\PP\!\left(XY > \tfrac{1}{2}\right) = \int_{1/2}^{1}\left(1 - \frac{1}{2x}\right)dx = \left[x - \tfrac{1}{2}\ln x\right]_{1/2}^{1} = \frac{1 - \ln 2}{2}. \tag{1}

The product beats one half only in the small wedge above the hyperbola, where both draws run large. Its area, and so the probability, is (1 - ln 2)/2, about 0.153.

#Area above the hyperbola

\PP\!\left(XY > \tfrac{1}{2}\right) = \frac{1 - \ln 2}{2} \approx 0.153. \tag{2}

A little under one chance in six. The favourable region sits above the hyperbola $xy = \tfrac12$ , and integrating its area is what produces the logarithm.