Martin Gardner died last month, on 22 May 2010. He was 95 years old, and in the words of Elwyn Berlekamp, John H. Conway, and Richard Guy, he “brought more math to more millions than anyone else.” I know he had a lot of influence on me when I was growing up, and having read the last few weeks’ worth of blogs, columns, and articles, it appears many others felt the same. Gardner was also, in my opinion, a first-rate scientist and arguably the original skeptic, with a very interesting and thoughtful perspective on the relationship between reason and religion.
But most importantly, he loved mathematics. He particularly loved mathematics “for fun”– what no one but other mathematicians could manage to refer to with a straight face as “recreational mathematics.” This includes the mathematics involved in playing games, magic tricks, gambling… and mathematical puzzles in particular.
For someone who enjoys mathematical puzzles, and has enjoyed them for some time, finding new and interesting ones can be difficult. (E.g., Alice and Bob are locked in a room with three doors and two light switches on the wall. They are each wearing a hat that is red or blue. What color is the bear? That should cover most of the classics.) A new puzzle, or even an interesting variant of an old puzzle, is a wonderful thing.
I am not a very eloquent writer, so I will not attempt to eulogize Martin Gardner here. Instead, I want to share something that I found this past week while reading what others had to say about him: another interesting puzzle that I have not seen before. It would certainly be in the spirit of Mr. Gardner’s influence if just a reader or two found it interesting as well.
The problem comes from an MAA column from just last year, dedicated to Martin Gardner for his 95th birthday. It caught my eye because it reminded me of one of my favorite puzzles, a card trick with some very interesting mathematics in it. Before getting to the new puzzle, let me describe the old one:
A magician asks a spectator to choose five cards from a standard 52-card poker deck. The magician takes the chosen cards, then hands four of them, one at a time, back to the spectator who announces each card in turn to the magician’s assistant who is blindfolded. After hearing the four cards, the assistant immediately announces the fifth chosen card. How do they do it?
The trick is not terribly difficult to learn, but is a remarkable effect when executed well. More importantly, the trick and its generalizations are a great vehicle for many different mathematical ideas, from the pigeonhole principle to saturating matchings in graphs. (For the mathematically inclined, consider the magician handing back to the spectator k of m cards selected from a deck of n, and the assistant naming the remaining m-k cards. For what values of (n,m,k) can the trick be done?)
Now on to the new puzzle. A magician hands a spectator a standard 52-card poker deck and asks him to select five cards. The spectator hands the five cards to the magician, who inspects them and places them all face up in a row on a table. The magician then asks the spectator to select one of the five cards, and remove and replace it with another card selected from the remainder of the deck… with the only requirement being that the selected card and its replacement be different colors (i.e., replace red with black or vice versa).
At this point the spectator is asked to call the magician’s assistant on the phone, and to read aloud the five cards as they appear on the table, in his own words and with no communication from the magician. After hearing the five cards, the assistant is able to name which of the five cards is the replacement. How does he do it?
Reference: Colm Mulcahy, “Poker-Faced Over the Phone,” the MAA column where I found the description of the puzzle. I am tacking this at the end of the post because I am pretty sure that it includes a solution to the problem, and I don’t want to spoil it for readers.