While raiding a tropical island, 6 pirates have an incredibly unlucky day.
While they scoured the island in search of treasure, one of them was infected with the Fanatic Fever, a disease which hits the pirate right in his greatest strength. (As everybody knows, a pirates greatest strength is his perfect rationality.)
A healthy pirate will always act in a certain way, following lower number rules first then higher numbered rules:
They will always choose action a over action b if action a means they have a higher chance of survival.
They will always choose action a over action b if action a and b have equal chances of survival, and action a leads to higher expected gold. Note: expected gold is defined as such: an x percent chance of getting y gold is x*y expected gold.
All else equal they prefer to see other pirates die. (They are ambivalent about whether more pirates die after they are dead.)
They randomly choose between actions they are ambivalent about in an entirely unpredictable fashion. In general, if there are n actions, all of which are equally preferable, they will choose each with 1/n probability.
However, a pirate infected with the fanatic fever is different. He will always, and only, vote for plans that involve the oldest pirate getting the most gold. If they are the oldest pirate, they can and will only propose such plans.
As you may know, pirates have a particular method of distributing the gold pieces they get from piracy. The oldest pirate proposes a plan which says exactly how many gold pieces each pirate will receive. Gold pieces cannot be cut, so each pirate must receive a whole number of gold pieces in any given plan.
Next, starting with the oldest pirate, and in order such that no younger pirate votes before an older pirate, each pirate publicly votes yes or no to the plan. Every pirates vote counts once except the oldest pirate's vote which counts one and a half times. The votes are tallied up. If there are more yesses, the gold is distributed as proposed. Otherwise, the oldest pirate is executed and they restart. No two pirates are exactly the same age.
Although all pirates know that exactly one of them is infected with fanatic fever, and all other pirates are healthy, no healthy pirate initially knows who is infected (although each pirate does know if they in particular are infected). Every healthy pirate initially believes that every other pirate has an equal chance of being infected.
Now, these pirates are actually so unlucky, that in addition to one of them being infected, they only found a single gold piece.
Question: When it comes time to distribute the gold pieces, what should the oldest pirate propose assuming he is healthy, and how likely is he to survive assuming each other pirate has an equal chance of being infected? For completeness, give every possible scenario with its probability. Remember that none of the other pirates other than the infected one know that the oldest pirate is not infected.
Challenge: Generalize for N pirates.
Extra hard challenge: Generalize for N pirates and X gold.
Impossible (Literally?) challenge: Generalize for N pirated, X gold, and Z of the pirates are infected with fanatic fever.