radio1
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if all 4 glasses are up or all 4 down, the game is over at the start.
I see from BMAD's earlier respose that this is his intended interpretation, but to me the wording of the puzzle doesn't make that clear. As stated, the puzzle seems to have the bartender blindfolded and engaged in a minimum of one "round" of turning over either 1 or 2 glasses, as only "After each round, the bartender is told if all glasses are oriented the same and the game is over.".
Fun puzzle though either way -- and the logic in plasmid's solution extends to either interpretation.
Thanks to all.
-radio1
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Very neat problem. The answer is certainly unintuitive.
There are three possible orientations that the glasses could be in that will not end the game.
DIAGONALS) Two glasses that are diagonal from each other are in one orientation, and the two other glasses that are diagonal from each other are in the other orientation.
ADJACENTS) Two glasses that are next to each other are in one orientation, and the two other glasses are in the opposite orientation.
SINGLETON) One glass is in one orientation and three glasses are in the opposite orientation.
We can therefore describe the game as in one of three states: state DIAGONALS, state ADJACENTS, or state SINGLETON.
By turning over one (random) glass, the bartender will....
... if it's in state DIAGONALS, switch it to state SINGLETON.
... if it's in state ADJACENTS, switch it to state SINGLETON.
... if it's in state SINGLETON, switch it to state DIAGONALS or state ADJACENTS or win.
By turning over two glasses that are adjacent to each other, the bartender will...
... if it's in state DIAGONALS, switch it to state ADJACENTS.
... if it's in state ADJACENTS, switch it to state DIAGONALS or win.
... if it's in state SINGLETON, keep it in state SINGLETON.
By turning over two glasses that are diagonal from each other, the bartender will ...
... if it's in state DIAGONALS, definitely win.
... if it's in state ADJACENTS, keep it in state ADJACENTS.
... if it's in state SINGLETON, keep in in state SINGLETON.
Now for the solution
1) Turn over two diagonal glasses.
If you were in state DIAGONALS before this move, you will win.
If you were in state ADJACENTS before this move, you will stay in state ADJACENTS.
If you were in state SINGLETON before this move, you will stay in state SINGLETON.
2) Turn over two adjacent glasses
You could not be in state DIAGONALS before this move (all prior scenarios would leave you in state ADJACENTS or SINGLETON).
If you were in state ADJACENTS before this move, you will go to state DIAGONALS or win.
If you were in state SINGLETON before this move, you will stay in state SINGLETON.
3) Turn over two diagonal glasses.
If you were in state DIAGONALS before this move, you will win.
You could not be in state ADJACENTS before this move.
If you were in state SINGLETON before this move, you will stay in state SINGLETON.
4) Turn over one glass.
You could not be in state DIAGONALS before this move.
You could not be in state ADJACENTS before this move.
Since you must be in state SINGLETON before this move, you will go to state DIAGONALS or ADJACENTS.
5) Turn over two diagonal glasses.
If you were in state DIAGONALS before this move, you will win.
If you were in state ADJACENTS before this move, you will stay in state ADJACENTS.
You could not be in state SINGLETON before this move.
6) Turn over two adjacent glasses
You could not be in state DIAGONALS before this move.
If you were in state ADJACENTS before this move, you will go to state DIAGONALS or win.
You could not be in state SINGLETON before this move.
7) Turn over two diagonal glasses.
If you were in state DIAGONALS before this move, you will win.
You could not be in state ADJACENTS before this move.
You could not be in state SINGLETON before this move.
Edit: renamed states for clarity
Hmmm... does your "checkmate in 7" work if all four glasses are initially in the same state, or would that make it a "checkmate in 8", or ??
Simple money probability
in New Logic/Math Puzzles
Posted
It's not a complete answer, as the 15 coins could also be 9 dimes, 1 nickel, and 5 pennies.
I think you're on the right track, but the final solution must factor in all the possible combinations of coins that meet the 19 and 15-coin criteria.