A viral meme starts in one account, and is re-shared by M other accounts, where M is a non-negative discrete random variable. Assume that the future behaviour following from each of the initial re-shares is independent and distributed identically to the sharing behaviour from the initial creation: after m initial re-shares we have m independent and identically distributed copies of the initial meme process. Let s be the probability that the meme eventually dies out, which is also the probability that any of the m sub-branches of the process dies out. Use the law of total probability to justify why $$ s = \sum_{m = 0}^\infty P(M=m)s^m = E(s^M)$$ and using the fact that now $M \sim Bin(3,1/2)$. Calculate $M_{M}(t)$ and hence find plausible values for s.
Stuck on finding plausible values for s. I thought maybe we can substitute the pmf of M (binom random variable) into the sum. We would then get a sum evaluated from 0 to 3 as P(M=m) would be 0, else where. Then we would get a polynomial equation which we can solve for s. Assuming this approach is correct, I got s = -1.6, s = 0.62, s = 0.25. Now since s is a probability, we exclude -1.6 and now for the other two, I think because we evaluated the sum for 3 accounts, the probability of the meme eventually dying is higher, as I guess intuitively speaking, more accounts having the meme would make it hard for the meme to die out. So i pick s = 0.62. Is this approach correct?
This relates to the classical branching process; see the formula there for $d=h(d)$ where $h(s)$ is the MGF. In your case, $$ h(s)=\sum_{k}\mathbb{P}(M=k)s^k=\sum_{k=0}^3 \binom{3}{k}\left(\frac12\right)^k\left(\frac12\right)^{3-k} s^k=\frac{(1+s)^3}8 $$ so $s=h(s)$ gives the solutions $1,\sqrt{5}-2,-\sqrt{5}-2$. The correct solution, the smallest non-negative one, is $\sqrt{5}-2\approx 0.236$.