Product of Two Uniforms

Draw $X$ and $Y$ independently, each uniform on $[0,1]$ . What is the probability that the product $XY$ exceeds $\tfrac12$ ?

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#A region in the unit square

The pair $(X,Y)$ is uniform on the unit square, so any probability is just the area of the matching region. The event $XY > \tfrac12$ is the set of points lying above the hyperbola $xy = \tfrac12$ . Since $xy \le x$ , the event forces $x > \tfrac12$ , and for each such $x$ it forces $y > \tfrac{1}{2x}$ .

Both draws are uniform on the unit square, so the chance is an area. The product beats one half only in the small region above the hyperbola, pinched off near the corner at x equals one half. Its area works out to one minus the natural log of two, all over two.

#Integrating the area

For each $x$ in $(\tfrac12, 1]$ the admissible $y$ run from $\tfrac{1}{2x}$ up to $1$ , a strip of height $1 - \tfrac{1}{2x}$ . Sweeping $x$ across the interval,

\PP\!\left(XY > \tfrac12\right) = \int_{1/2}^{1} \left(1 - \frac{1}{2x}\right) dx. \tag{1}

The antiderivative is $x - \tfrac12\ln x$ , so

\PP\!\left(XY > \tfrac12\right) = \left[\,x - \tfrac12\ln x\,\right]_{1/2}^{1} = (1 - 0) - \left(\tfrac12 - \tfrac12\ln\tfrac12\right) = \tfrac12 + \tfrac12\ln\tfrac12. \tag{2}

#Result

With $\ln\tfrac12 = -\ln 2$ ,

\PP\!\left(XY > \tfrac12\right) = \frac{1 - \ln 2}{2} \approx 0.153. \tag{3}

So a little under one chance in six. The product is small far more often than intuition suggests, since multiplying two numbers below $1$ drags the result down.