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The newsletter of ASKE, the Skeptical Adversaria, is issued quarterly to members, and I have a little piece each time called ‘Logic and Intuition’ which features one or more puzzles that require the application of a little logic or the very simplest mathematics to solve.  This is the first one and it appeared in Issue 1 of 2004.

The ‘Random Cards’ Paradox

The above represents the best title I can come up with at the moment for this nice little puzzle.  Cut a full pack of playing cards (or any even number) into two exact halves.  Turn the top half over, re- combine it with the bottom half and shuffle thoroughly.  Thus we have a pack of 26 face-up cards and 26 face-down cards, randomly mixed.  Again cut the pack into two exact halves, turn the top half over, but this time place it next to the bottom half.  What are the chances that the two half-packs have the same number of face-up cards and the same number of face-down cards?

The answer is given below, so do not look until you are ready.

Answer

The answer is 100%.  One of the beauties of this puzzle is that the answer defies one’s immediate intuition.  No matter how thoroughly you shuffle the pack and ‘randomise’ the face-up and face-down cards, you always end up with the same number in the two halves of the pack!

Even readers with only a passing acquaintance with statistical theory will acknowledge that the calculation of probability is often more an exercise in logic than mathematics.  Yet one’s immediate reaction to the puzzle may be that the answer requires the construction of a complex equation that is best left to statisticians.  In fact, another beauty of this puzzle is that the logic or, if one chooses, algebra required to solve it could hardly be simpler.

Although we say that the order of the face-up and face-down cards is ‘random’, we must still remember that their original numbers are identical and equal to half the pack which is also the number of cards in the two piles that we are left with.  Likewise the number of original face-up and face-down cards remains the same.  The twist (literally) in the puzzle is that half their total is reversed and constitutes the turned-over half of the pack.  (This gibberish should make sense if or when the reader has given his or her thought to the problem.). 

Readers may be more familiar with a related puzzle.  From a glass of wine transfer a teaspoonful to a glass of water, mix, and transfer a teaspoonful of the mixture back to the glass of wine.  Is there more wine in the water or water in the wine?  Again, the answer is that the two quantities are identical.  Many years ago I introduced this puzzle to a fellow psychologist who pronounced it to be ‘a striking example of conservation’.  He was referring to Jean Piaget’s concept of conservation: in the earliest years of life we learn that certain properties of the material world, such as number, volume and weight, are conserved even when other properties, such as height, width and length, change.  Maybe ‘conservation of a high order’ is a more apt description.

I have not yet used this puzzle for demonstration purposes but it has merit, I think, as a way of demonstrating aspects of the scientific method.  For example, you might first produce several packs of, say, 20 cards each and ask some members of the audience to perform the experiment a number of times.  Hence the ‘law’ is derived by the simple process of ‘induction by enumeration’.  The drawback of this method can then be discussed: how many observations do you need to be sure the law is correct?  It may be for example that the number of face-up and face down cards is the same on 999 times out of 1,000 occasions.  Now you prove the law by deduction from the rules of logic or mathematics.  Or, you could deduce the law first and then demonstrate it empirically.

I derived the ‘random cards paradox’ from the description of a party trick that readers may like to try.  Put on a blindfold and have someone spread on a table an even number of coins (say 12), half heads-up and half (naturally) tails-up.  The challenge is to make two piles of coins with the same number of heads and the same number of tails in each.  Simply draw half the coins to one side, turning each one over.

P.S.  Since completing this little piece I came across the puzzle, in slightly altered form, while flicking through Martin Gardener’s Mathematical Puzzles and Diversions, Penguin Books, 1965 (Chapter 10).