# Reverse puzzling 6

(Continuing thematically from here and here)

"Well, George, I've got a tricky one for you today. Lots of people struggle to get into it. They think it's about real human behavior but it's set up for logicians."

George looked at me blanking not seeing the difference.

"Anyway, here's the puzzle. There are 5 people..." I described the puzzle for him and folded my arms with a smile.

He squinted his eyes at me in disbelief that I would ask such a simple question. Then shrugged and said: "Well, 1 is 100; 2 is 0; 3 is 99; 4 is 97; and 5 is 97. So that's the answer."

I was flummoxed that he'd solved it so easily. In desperation, I asked him how it would continue and he immediately said, "It will drop by 1 every other person from now on until it gets to 0."

What was the puzzle I gave him?

I think that the puzzle you gave him is very similar to

The 5 pirates, 100 gold coins puzzle with some small changes.

The wording is possibly as follows

There are 5 people (call them $$A,B,C,D,E$$ in order of seniority) who found $$100$$ gold coins.
They must decide how to distribute them.
The rules of distribution say that the most senior among them first proposes a plan of distribution. Everybody, including the proposer, then votes on whether to accept this distribution. If the majority accepts the plan, the coins are dispersed and the game ends. In case of a tie vote or a majority rejection, the proposer is eliminated from the distribution, and the next most senior person makes a new proposal with the remaining members of the group to begin the system again. The process repeats until a plan is accepted or if there is one pirate left.
Everyone bases their decisions on four factors. First of all, each person wants a cut. Second, each person wants to maximize the number of gold coins they receive. Third, each person would prefer to eliminate another from the group, if all other results would otherwise be equal. And finally, they do not trust each other, and will neither make nor honour any promises between people apart from a proposed distribution plan that gives a whole number of gold coins to each person.
What is the maximum number of gold coins the most senior person can guarantee?

Reasoning the solution

We can look at the problem with different values of $$N$$, the number of people.
For $$N=1$$, the proposer, $$E$$, will obviously gives themselves $$100$$ gold coins.
For $$N=2$$, the more senior member $$D$$, knows that the other, $$E$$, will reject any proposal which gives them less than $$100$$ gold coins so they cannot guarantee anything more than $$0$$.
For $$N=3$$, the proposer $$C$$, knows that $$D$$ will accept any proposal that will give $$D$$ at least one coin (since a rejection would mean they possibly get nothing). Hence, $$C$$ can guarantee a haul of $$99$$ coins by giving $$1$$ to $$D$$ and $$0$$ to $$E$$.
For $$N=4$$, the most senior proposer $$B$$ needs to get two others to accept the proposal. Looking at the $$N=3$$ case they can guarantee $$D$$ and $$E$$ to accept the proposal by giving $$2$$ coins to $$D$$ and $$1$$ to $$E$$, keeping $$97$$ for themselves.
For $$N=5$$, the most senior proposer $$A$$ needs to get two others to accept the proposal. Looking at the $$N=4$$ case, they can guarantee the votes of $$C$$ and $$E$$ by giving $$1$$ coin to $$C$$ and $$2$$ coins to $$E$$.
The general result proceeds by induction and results in the number as stated by George.

• Great job and very quick! – Dr Xorile May 12 at 4:56