strategy - Three voting prisoners

Three prisoners have a brief strategy meeting, and then are not allowed to communicate.

Each night one of the three has steak for dinner, while the other two have fish tacos. Also each night, each of the three prisoners votes for one of the following two options:

1. All of us have had steak at least once.


2. Don't know yet.

If a majority go with option 2, nothing happens that night. If a majority go with option 1, then they are set free if they are right, and executed if they are wrong. The distribution of votes is kept secret, so it is unknown what each of the others voted.

What should their strategy be?

Answer

Another intuitive, no-math (and I believe best) strategy could be as follows:

The prisoner that gets steak the first night should always vote 2 (Don't know). The other two prisoners that get fish tacos the first night should vote 2 until they get steak for the first time, then vote 1 (Steaks) every night from then on.

This ensures that

• The majority won't vote 1 (Steaks) when they would be wrong.


• The majority will vote 1 the first night they all had steak.