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, p^{2}, and p^{3}. With each outcome being
equally likely the probability of all amoba eventually dying out is
1/4*[ 1 + p + p^{2} +
p^{3} ]. So p = 1/4*[ 1 + p + p^{2} + p^{3} ] or
1 -3p + p^{2} + p^{3} = 0. This reduces to
(p - 1)(p^{2} + 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.

Michael Shackleford, A.S.A.