Since after the two transfers there is still a equal volume in each glass, an equal amount of water must now be in the wine as there is wine in the water! Mathematically: Let V be the volume of water in the first glass. V is also the volume of wine in the second glass. Let S be the volume of one spoonful. After the first transfer the volumes are: In the 1st glass: Water = V-S, Wine=0 In the 2nd glass: Water = S, Wine=V The fraction of water in the second glass is S/(S+V) Therefore the second transfer consists of: Water = SxS/(S+V), and Wine= S - SxS/(S+V) And so after the second transfer: In the 1st glass: Water = V - S + SxS/(S+V), Wine = S - SxS/(S+V) In the 2nd glass: Water = S - SxS/(S+V), Wine = V - S + SxS/(S+V) There is the same amount of wine in the water as there is water in the wine. Try it with V=90 and S=10: After first transfer, first glass is (80 water, 0 wine), second glass is (10 water, 90 wine) which is 10% water. So the second transfer is of (1 water, 9 wine) giving a final position: First glass (81 water, 9 wine), second glass (9 water, 81 wine).