Solution
There is indeed a unique such X, and it is 0.60577....
The function p(q, r) is the sum
Sum[q, (1 − q)i (1 − r)j, {i, 0, Infinity}]
This is a geometric series that converges to q / (q + r − q r). Now, q / (q + r − q r) ≥ X is equivalent to q ≥ (r X) / (1 − X + r X), so the probability we seek is the integral as r goes from 0 to 1 of (r X) / (1 − X + r X), and this is ((X - 1)/X) ln(1 - X). Therefore, the desired X satisfies X =((X − 1)/X) ln(1 − X). Easy calculus shows that the right side decreases from 1 to 0 for X between 0 and 1. Hence the equation has exactly one root, which is 0.6057701616316195581....
Of course, I use Mathematica to evaluate the integral in the solution, but it is easily done using elementary calculus.
[Back to Problem 1212]
© Copyright 2015 Stan Wagon. Reproduced with permission.
