I would, but I stay anonymous here.
But to make sure we’re communicating, this is the problem I’m solving:
We start with N people. One person is chosen at random. That person kills someone else chosen at random, and that person dies with 100% certainty before he can do anything.
At each point after that there are k “killers” alive who have killed one other person and i “innocents” alive who have killed no one else. One of the i innocents is chosen at random and he kills one of the other k + i -1 people with 100% certainty.
No corpses do any shooting or are shot a second time.
The procedure continues until there are no innocents left alive only killers. What is the ex ante expected number of people remaining alive.
My answer, which I was not first to propose, is approximately N/e. Your answer N/2.
If that’s the problem we’re agreeing on, then N/2 cannot be correct. Clearly there can never be more than N/2 people left. If there are more than N/2 people left at any time than fewer than N/2 people have shot someone else so there must be fewer than N/2 killers alive or dead. That means there must be more than N/2 innocents alive or dead so there must be some innocents still alive, so the process continues.
As the process can’t stop with more than N/2 people alive, the only way N/2 could be the average number to stay alive is if exactly N/2 people always live; otherwise the average would have to be less than N/2. But the only way exactly N/2 people survive is if each person killed is an innocent so that we’re left with N/2 killers.
That means when there are N/2 + 1 people left alive (N/2 - 1 killers and 2 innocents) one of the two innocents must be picked at random and he must pick the other innocent as his target. The probability of that is 2/(N/2 + 1) * 1/(N/2). That’s a quite low probability, but you must have it happen with certainty.