After the first turn there are four possibilities for the number of amoba left, 0, 1, 2, 3 with probabilites of eventually dying out of 0, p, p², and p³. With each outcome being equally likely the probability of all amoba eventually dying out is 1/4*[ 1 + p + p² + p³ ]. So p = 1/4*[ 1 + p + p² + p³ ] or 1 -3p + p² + p³ = 0. This reduces to (p - 1)(p² + 2p - 1) = 0.

The solutions for p are 1, (sqr(2)-1), and (-sqr(2)-1). The only one which satisfies the constraints of the problem is sqr(2)-1.