Water & Wine Glasses

Start with two glasses, one filled with water, the other filled with wine.
wine glasses
A spoonful of wine is taken from the wine-glass and stirred into the water-glass, then a spoonful is taken out of the water-glass and returned to the wine-glass. The operations are repeated.

Is there more water in the wine-glass than there is wine in the water-glass or the other way round?


10 thoughts on “Water & Wine Glasses

  1. The problem is extremely tedious to work out if one goes through it stage by stage, assigning volumes to the spoon and glasses and finding the different concentration of wine and water at each stage. If, however, the point of view is shifted so that the situation at the end is considered, instead of its development, the solution is very simple.

    To be continued …

    from New Think by Edward DeBono

  2. Another way to phrase this question is, “given an even number of alternating turns, which liquid will contaminate the other more?” Since the wine get the first turn, it has the not-to-be-repeated step of contaminating the water with a spoonful of 100% pure wine. The return trip will always contain some percentage of wine returning to from the water glass. No matter how many times these two steps are repeated, the water will always be a little more contaminated with wine than the other way around.

    If you were to plot this on graph paper, with (wine volume – water volume) of either glass in the Y axis and (# of round trips) on the X, you’d get an “asymptote,” a kind of curve that approaches but never quite touches the X axis. Y will never be zero, i.e., the mixtures will never be *exactly* 50/50.

  3. Remember, each glass had two spoonfuls taken out, and two returned. So, the volume in each at the end must be the same as at the beginning.

    Therefore, any wine in the water glass must have displaced an exactly equal volume of water.

    And, any water in the wine glass must have displaced an exactly equal volume of wine.

  4. That’s in interesting and perfectly reasonable application of limit theory. I’ll have to think about that!

    It would make sense if you used two spoons and simulaneously took a spoonful of each liquid, traded glasses, and mixed.

    Perhaps the “gotcha” lies in the fact that the water glass temporarily has 2 spoonfuls more volume than the wine. Still, to maintain equation the wine would need to receive a spoonful of pure water on the first step, which is impossible.

    Where’d I put that thinking cap?


  5. The first spoonful put into the water glass is pure wine. Everything after that is some percentage less than pure – so more wine in the water glass.

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