Dick's problem is an odd one, partly because wine is a mixture containing water, and partly because one component of wine- ethanol- dissolves in water such that 1 pint water + 1 pint water results in not 2 pints of mixture, but about 8% less.

Although there is "water" at one end of the scale and "wine" at the other, there is no point at which you can say that it has ceased to be water, and become wine- and vice versa. And water is never pure H2O. So one one level, the water is still water and the wine is still wine.

But let's simplify- assume no significant change in volume on mixing, and assume that the "wine" is a component that can be labelled put in a pot marked A (for Alcohol). The water is of course marked W, and we assume unit volumes W and A.

Let the size of the spoon be S (0 < S < 1). Then a spoon of wine is SA, and putting this in the water results in W+SA.

The wine is now (1-S)A.

Now take a spoonful of the mixture S(W+SA) = SW + SSA, and add it to (1-S)A = (1-S)A+SW+SSA. The result is A-SA+SW+SSA.

The remainder in the water is (1-S)(W+SA) = W-SW+SA-SSA.

The quantity of wine in the pot marked W is SA-SSA.

The quantity of water in the pot marked A is SW.

Therefore there is more (absolute quantity) of water in the wine than there is wine in the water.

You might want to go on and work out which is the greater proportion of the other.