Amoebic Generation

I count 144 ways. I'll use the following diagram, so A is the original amoeba, B and C are its offspring, etc.

https://i.imgur.com/SHQJ1br.png

We take cases, based on the last common ancestor of the two surviving amoeba.

The last common ancestor of the two surviving amoeba is third generation .

There are 4 ways of choosing the surviving amoeba. For example, the surviving amoeba could be H and I. In this case, the rest of the amoeba in the fourth generation are all dead, A and B have to be alive. E can be dead or alive, and there are 5 ways of choosing the states of C, F, and G, so the number of ways in this case is $4 \cdot 2 \cdot 5 = 40$ .

The last common ancestor of the two surviving amoeba is second generation .

There are 8 ways of choosing the surviving amoeba. For example, the surviving amoeba could be H and J. In this case, B, D, and E have to be alive, and again there are 5 ways of choosing the states of C, F, and G, so the number of ways in this case is $8 \cdot 5 = 40$ .

The last common ancestor of the two surviving amoeba is first generation .

There are 16 ways of choosing the surviving amoeba. For example, the surviving amoeba could be H and L. In this case, B, C, D, and F have to be alive. E can be alive or dead, and G can be alive or dead, so the number of ways in this case is $16 \cdot 2 \cdot 2 = 64$ .

The total number of ways is then $40 + 40 + 64 = 144$ .

In the possibility of 16 alive is imposible so 1/2 16 = 8, then with symmetry we can do 8 2=16 and 4 1= 4 (16-4)^2= 12^2= 144 or 2 8^2 so that 8^2 is the two probability of maximum and minimum range of symmetry with 2(minimum) and then + 1*4^2 as sames as the first probability of 1/2.