Hoist by my own petard

The expression in the title, which nowadays means “blown up by one’s own devices”, first occurred in Hamlet. It is interesting to note that Shakespeare dropped the “d” and used the word “petar” which apparently means “fart”. 

When solving problems, my students were sometimes blinkered by unacknowledged assumptions that really had little to do with the mathematics at hand. To highlight the situation, I used to distribute a sheet of puzzles that could be difficult to solve if you did not recognize your hidden assumption. Most of the puzzles were cribbed from Martin Gardner’s books. Here is one of my favourites:

I propped up four cards against the blackboard as follows.

The puzzle here is to determine what cards that you have to turn over to make sure that the following statement is true:

    If a card has a blue back then it has a 2 on the other side.

Note that we are not asking whether the statement is true. We are being asked which of the four cards must be turned over to enable us to determine if the statement is true. Here is how we analyzed it in class.

Blue card: We all agreed that you do have to turn over the blue card. 

Red card: We eventually agreed that you do not have to turn over the red card. Several students initially said that you had to turn it over because the statement would be false if there is a 2 on the other side. This is a familiar logical error whereby a statement and its converse are considered to be the same. The statement at issue does not say “If one side has a 2 on it then the card has a blue back.” It says “If a card has a blue back then the other side has a 2 on it.” It’s the same as the difference between the two statements “If your car is out of gas then it won't start,” and “If your car won't start then it's out of gas.”  So we agreed that we do not have to turn over the red card — it doesn’t matter what number is on the other side. 

The 3 card: All agreed that you do have to turn over the 3 card. You have to check that the other side is not blue, for if it were, you would have a card with a blue back that did not have a 2 on other side and the statement would be false. 

The 2 card: We agreed that you do not have to turn over the 2 card. It doesn’t matter what is on the other side. If the other side is blue, it’s fine; if it’s anything else it doesn’t matter. 

So there we have it. We only need to turn over the blue card and the 3 card to determine whether the statement is true.  

Here’s what happened next: I turned over the blue card and it had a 2 on the other side. I turned over the 3 card and the other side was red. At this point we agreed that we had enough information to conclude that the statement had to be true.

Well, of course, this was a flagrant set-up. When I turned over the remaining cards it revealed that the statement was false. How could this be? Is our logic not impeccable? Keep reading.

There’s actually nothing wrong with our logic. There is however the assumption that the four cards were standard playing cards. They were not. The red card was a two-backed card — red on the visible side and blue on the other. So it was a card with a blue back that did not have a 2 on the other side. So we do have to turn over the red card.

This a routine that I used for many years, and you can imagine my delight when I encountered Shecky Riemann’s recent post that links to the Wason selection task in Wikipedia. The Wason selection task goes like this:

You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. The visible faces of the cards show 3, 8, red and brown. Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?

In the Wason puzzle, the roles of the numbers and colours are reversed from what I did in class. The Wason task is about what colour is opposite the number rather than what number is opposite the colour. The statement we want to check in the Wason task is this:

If a card shows an even number on one face, then its opposite face is red.

The answer given in the Wikipedia article is that to check the veracity of this statement you only have to turn over the 8 card and the brown card. 

Except for the reversal of colours and numbers, the Wason task appears to be identical to  the Martin Gardner puzzle. I gleefully dashed off an email to SheckyR and pointed out that in the Wason task you would also need to turn over the 3 card, because there is a possibility that there is an even number on the other side. 

SheckyR quickly replied that this is not possible because the statement of the Wason task very clearly states that 

You are shown a set of four cards placed on a table, each of which has a number on one side and a colored patch on the other side. 

He has a point. It is pretty hard to interpret this as allowing numbers on both sides of a card.  
The Gardner puzzle, on the other hand, does not preclude this from being the case, and in fact it is deliberately set up to tempt you to assume that it is the case. My familiarity with the Gardner puzzle and my careless reading of the Wason puzzle caused me to bring unwarranted assumptions to the Wason puzzle and resulted in pretty hefty brain fart.

References:

Martin Gardner, Combinatorial Card Problems, in Time Travel and other mathematical bewilderments. 

The playing card version described above is a bit different than the one in Gardner's book. There, the puzzle involves five playing cards, and Gardner attributes it to Tom Ransom, a Canadian amateur magician. 

Shecky Riemann’s blogs are here and here. If you like things mathematical in the Martin Gardner vein, you should visit his blogs.