This is a conditional probability problem. We can denote the probability of the amoeba population going extinct as $P(E)$. Let us denote the outcomes: $A_1$ is the amoeba dying, $A_2$ is the amoeba staying the same, $A_3$ is the amoeba splitting into two, and $A_4$ is the amoeba splitting into three. $$ $$ We can simply apply the law of total probability to the outcomes. $$ P(E) = \sum_{k=1}^4 P(E|A_k) P(A_k) $$ Clearly, the probability going extinct if the first amoeba dies is 1 $$ P(E | A_1) = 1. $$ If the amoeba stays the same, the probability of the population going extinct is simply $P(E)$ $$ P(E | A_2) = P(E). $$ If the amoeba doubles, then we have two independent probabilities of the amoeba population going extinct $$ P(E | A_3) = P(E)P(E) = P(E)^2. $$ Likewise, if the amoeba splits into three, $$ P(E | A_4) = P(E)^3. $$ What remains is to solve the equation $$ P(E) = \frac{1}{4} \left( 1 + P(E) + P(E)^2 + P(E)^3 \right). $$ Solving this equation subject to the requirement that $0 < P(E) < 1$ gives $$ P(E) = \sqrt{2} - 1 \approx 0.414. $$