How I Saved The World!
My solution to my second FiveThirtyEight Riddler Puzzle
You can find solutions to my puzzle that Riddler column editor Zach Wissner-Gross selected toward the end of this link. I still prefer my solution:
Let N be Earth's population and let M be the number of people who answer “Yes”. Let θ be a tiny positive number. Then if you ask “Is your number greater than 1-θ?”, M will follow — almost precisely — a Poisson distribution with expected value lambda = N θ; for each non-negative integer m:
Pr[M=m|λ] = λ^m Exp[-λ] / m!
M=1: your question has identified the person with the highest number and humanity is saved!
M = 0: you will need to randomly choose one person out of everyone on Earth, which will be pretty bleak. But since this happens with probability Exp[-λ], the larger your choice of theta, the less likely this is to happen.
M > 1: you will need to randomly choose one of these M people and the probability of saving humanity is 1/M.
So over all, your probability of saving humanity is
S[λ] = Exp[-λ] Sum[λ^m / (m * m!), {m, 1, ∞}]
= -Exp[-λ] (EulerGamma + Gamma[0, -λ] + Log[-λ])
where EulerGamma is Euler's constant ɣ = 0.57721566... and Gamma[a,x] is the incomplete gamma function.
S[λ] is maximized by setting λ = 1.5028610173354528..., giving a value of 0.51735143689437297...
My numbers are a bit different from those appear in the Riddler solution because in my problem submission I set N to the world population estimate at Our World In Data for the end of 2023! For the column, Zach rounded N to 8 billion.
P.S. It didn’t occur to me that people would be able to use a fairly direct numerical computation to solve the puzzle. In retrospect I’m glad I didn’t come up with a formulation of the puzzle that used a much larger value of N: Moore’s law is wonderful!