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Gamow
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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 12 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.

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 1 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.

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.
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matega
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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 1 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.