bigideas

The Wason Selection Task is a fairly well-known puzzle that tests whether or not you have an intuitive grasp of the thing logicians call the material conditional. It turns out that most people who aren’t logicians don’t. Here it is, along with some (so far as I know) original variants.

The Basic Wason Selection Test

In case you haven’t seen it before, here’s a classic version of the test. You’re presented with 4 cards lying on a table. You’re told that each card has a number on one side and a letter on the other — you can take that for granted. You’re asked to check only that all four cards obey the following additional rule:

If a card has an even number on one side then it must have a vowel on the other.

Of course, the visible face of each of the four cards on the table shows either a letter or a number. In this case, they’re 1, 2, A and B. Which cards must you turn over, at a minimum, to verify that the cards obey the rule? Think about this and come up with your own answer before reading on.

If you haven’t thought much about logic and implication before this question, though simple, is easy to answer incorrectly. Most people think you have to turn over the card marked 2 to make sure it has a vowel on the back. Many people think you have to turn over the card marked A to make sure it has an even number on the back. Many also think you should turn over the card marked 1 to see that it doesn’t have a vowel on the back.

The answer is to turn over the cards marked 2 and B. The first case is easy; you have to check it has a vowel on the other side. The second is trickier; you have to make sure it doesn’t have an even number. You needn’t look at the card marked 1 because the rule says nothing about what happens when the number is odd. You needn’t check the card marked A because, again, the rule doesn’t constrain which numbers can be paired with vowels.

Material Conditionals

This bit explains briefly why the answer is true. If you aren’t interested in logic, you can skip it, but you might find the variations in the next section more tricky as a result.

The material conditional is the logical connective that we usually call “if…then”. It’s written “→” and we write “if p then q” as “p→q”. In classical logics, this proposition is always true unless p is true and q is false. Our rule states:

A card has an even number on one side → it has a vowel on the other

So we need to find cases of an even number paired with a consonant; if we don’t find any such cases, the material conditional makes the whole thing true “by default”.

We don’t need to look at 1 because it’s not an even number; by the same token, we do have to look at 2. We don’t look at A because it isn’t a consonant, but we do look at B. That’s, formally speaking, how the answer is arrived at. You might have got the right answer by common sense, although it turns out that that’s actually uncommonly sensible.

Some Basic Variants

If you’re interested in the psychology of logical thinking you can try asking the question in different ways — abstractly, as we did, using numbers and letters; using concrete examples; using ethically or emotionally loaded examples and so on.

Instead, let’s see if we can make the puzzle itself puzzling enough to puzzle even some logicians who might be reading. In all these cases, as in the original version, we can assume that each card has a number on one side and a letter on the other. Try to solve each problem as soon as it’s stated before reading on.

Here’s a slightly different kind of rule:

If there’s an even number on one card then the card to its left has an odd number.

This time the cards are arranged slightly differently: from left to right, A, 1, 2, B. Which cards must be turned over to verify this rule?

We needn’t look at A, since it has no card to its left. We needn’t look at 1 since it doesn’t have an even number on it. We needn’t look at 2, since we can see that the card to its left satisfies the rule. We do however, need to turn over B, to make sure it doesn’t have an even number on the other side.

Let’s wind up the complexity a bit:

If there’s a vowel on one card then either it or the card to its left has an even number.

From left to right, the cards read B, A, 1, 2. Which cards must be turned over?

It seems at first that we needn’t turn over B, since it has no vowel. But A has a vowel; we must check that either it or B has an even number. 2 definitely obeys the rule whether it has a vowel on the other side or not. We can try first turning over A. If it has an even number then both A and 1 fit and we’re done. If not we have to turn over B and 1 as well. In the worst case we would have to turn over 3 of the cards in total — A, B and 1.

This one is a warm-up:

If the card has a vowel on one side then, if the number on the other side is even, that vowel has to be an A.

The cards you’re given are 1, 2, A, B. Which do you need to turn over? The 1 isn’t even, so whether it has a vowel on the other side or not it doesn’t matter whether or not it’s an A. The 2 is even, so we need to turn it over and check it doesn’t have a non-A vowel (E, I, O or U). The A is a vowel, but whether the number on the other side is even or odd, that vowel is an A, so the rule is satisfied. Finally the B isn’t a vowel, so the rule doesn’t apply at all. Therefore you only need to turn over the card showing the number 2.

Too easy? OK, here’s something closely related but more fiendish:

Either: if there’s a vowel on side then there’s an even number on the other, or: if the number on one side is divisible by 3 then the letter on the other is from the first half of the alphabet.

The visible faces of the cards are W, 8, B, 3. See if you can work out the answer — you need a very firm grasp of exactly when “if…then” statements are false. The “either…or” here is inclusive: “x or y” is only false if both x and y are false.

The logical structure of this statement is “(v→e) or (d→h)” where v, e, d and h are the following propositions:

• v = vowel on one side
• e = even number on one side
• d = number divisible by 3 on one side
• h = letter from the first half of the alphabet on one side

We’re using the strict logical meanings of the words “if…then” and “or”. Remember, “if p then q” is true as long as p isn’t true while q is false.

We needn’t turn over W; v→e is true (because v is false) and so v→e or anything is also true. Similarly we can leave 8 alone; its presence makes d→h true. B is from the first half of the alphabet, so you might have worried about that, but B makes v→e true just as much as W does, and that’s enough to satisfy the whole rule.

We do have to turn over 3, and check that we find a letter from the first half of the alphabet or any consonant; that is, that it doesn’t show either of the letters O or U. Even if you’ve studied some logic, that would be a tricky question to answer if you were put on the spot. Of course, by combining logical connectives you can create variants of arbitrary complexity.

Question-Answer Variations

Instead of cards, we can do something similar involving people; then instead of turning over cards we can ask questions and get answers in response, which is more fun. Here’s an example. You’re presented with three people: Anita, Brian and Cerys. You need to check that if someone is a member of the Smith family then they work at the family firm, Smith & Co. You already know that Anita is a Jones and that Cerys doesn’t work at Smith & Co. You’re allowed to ask each person at most one question. How to proceed?

Here’s one way. You first ask Dave whether he’s a Smith; if he is, the rule is broken. You don’t know anything about Brian, so you ask him whether he’s a Smith. If he is, you ask Anita whether Brian works at Smith & Co. This introduces another possibility for variation: the person I ask might lie. But that makes it much too easy to come up with really difficult problems.

I hope you enjoyed some of these variations, and maybe had fun coming up with some of your own.