Three Voting Prisoners

Each night one of three prisoners has steak for dinner, while the other two have fish tacos. Also every 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. Also, it is known that every prisoner eventually will get a steak.

The three prisoners can have a brief strategy meeting and after that, they are not allowed to communicate.  What should the prisoners’ strategy be?

Source: Puzzling StackExchange

The prisoner who gets a steak the first night should always vote 2, whereas the other two prisoners should vote 2 until the night they get a steak, and 1 every night after.

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