Answer to old chestnut: rotating table

Here's the winning method:

  1. Turn two adjacent glasses right-side-up. (If they're already that way, leave them. The purpose of this step is simply to ensure there are at least two adjacent glasses that are turned the same way, and 'right-side-up' is chosen arbitrarily.)
  2. Turn two non-adjacent glasses right-side-up. (Now we know at least three glasses are right-side-up. If we don't win here, we know that the remaining glass is upside-down. However, there is no guarantee we can choose that glass blindfolded in any finite number of tries, so...)
  3. Take two non-adjacent glasses. If one is upside-down, turn it right-side-up (winning the game); otherwise, turn one upside-down (this makes two upside-down glasses adjacent and two right-side-up glasses adjacent).
  4. Take two adjacent glasses. Turn both of them over. (If these were initially the same way, you win; otherwise it gives you two non-adjacent glasses right-side-up and the other two upside-down.)
  5. Take two non-adjacent glasses and turn both over. (Now, it doesn't matter which pair you choose; both possible pairs of non-adjacent glasses are turned the same way, and the others are the other way.)

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