Puzzle time - shrinking board edition

[Klaus]

Sounds like a prisoner's dilemma-like situation.

I'd say the solution is 2 or 50. I don't think it is anything between.

Should the solution take into account that everyone acts completely rational and knows that everyone else acts completely rational, too? Furthermore, I assume you rule out "deals" among subsets?

[jon-nyc]

You do allow for rational people that assume others are rational. The answer is neither 2 nor 50.

[Klaus]

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Here's the situation for smaller groups, starting with group size 2.

In the following, the number n denotes the n-th oldest board member. "gs" stands for group size. The notation "1,2 vs 3" means "1,2 vote ayes, 3 votes no".

gs = 2 -> termination with 1 and 2 remaining
gs= 3 -> 1,2 vs 3 -> 3 kicked out
gs = 4 -> 1,2 vs 3,4 -> termination with 1-4 remaining
gs = 5 -> 1,2,3,4 vs 5 -> 5 kicked out
gs = 6 -> 1,2,3,4 vs 5,6 -> 6 kicked out
gs = 7 -> 1,2,3,4 vs 5,6,7 -> 7 kicked out
gs = 8 -> 1,2,3,4 vs 5,6,7,8 -> termination with 1-8 remaining

So it looks like the powers of 2 are the termination points.

Based on that reasoning, I'd say that 32 members remain.
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[jon-nyc]

Yep

[Klaus]

So, what do I win?

[jon-nyc]

Bragging rights for beating Horace and Ax to an answer.

[Klaus]

I wondered what happens if a stale mate results in kicking out a member, not termination.

Turns out that in this case the powers of two minus one (1,3,7,15,31 etc.) are the termination points.

[Klaus]

What I meant with my question about "deals" is something like this.

Let's assume group size 4. In the strategy described above it would be a termination point.

But 3 could make a deal with 1 and 2: I'll vote for kicking out 4 if you promise to vote for termination in the next round.

[jon-nyc]

Yeah but they could break their promise and be further ahead on their goals

[jon-nyc]

It’s fun to think about though