A glass of wine. An equal glass of water. Take one spoonful of water and pour into wine. Mix thoroughly. Take one spoonful of mixture and pour into water. By volume, is there more water in the wine or more wine in the water?

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).