Mixology

by Kevin J. Lin

Neither. Don't be confused by the juggling of jiggers- we subtracted a jigger from the first glass and added it to the second. Then we subtracted a jigger from the second glass and added it to the first. Therefore both glasses have the same volume. Any amount of lager now residing in the stout must have been replaced by an equal amount of stout.

Update, 10/7/98:

We have certainly received a number of emails about this puzzle! There are two areas of contention on the part of the respondents. One camp was led astray by the wording of the puzzle. They felt that the question asked which glass contained a greater concentration of lager. The question actually asked if the concentration of lager in the stout glass was less than or greater than the concentration of stout in the lager glass.

The second camp disputed the results, based on the following logic: "If a jigger of 100% lager is transferred to the stout glass, then a jigger which is less than 100% stout is transferred back, there is an imbalance. Therefore there must be more lager hanging around in the stout glass than there is stout in the lager, no?"

No- this is a rather deceptive piece of specious logic. In the first transfer the lager glass lost a pure jigger of lager. But this was offset in the second transfer when it was given back some its original lager along with the stout.

Consider this gedankenexperiment (literally, "Thought experiment"): Take an equal number of black and white marbles (or Go chips or whatever). Let black be stout, and white be lager, and put them in separate bags. Move any number of white marbles to the black bag, and shake it. Now reach in and pull out the same number of marbles you put in, of random color, and place them in the white bag. You can see that no matter how the chips are distributed, there must be as many white marbles in one bag as there are black marbles in the other.

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