Conditional reasoning - When if - then's get muddled

I was having a conversation the other day on conditional reasoning and how bad we IT people can be at using logic in our programming. I was reminded of an example that was put to me when I was at University. This is relatively famous in Maths geeks circles and was put forward by Wason and Johnson-Laird in 1972 to show how poor we really are at reasoning.

Four cards are laid out as below (UPDATE Letter on one side number on the other - Thanks Richard!)

E

K

4

7

The conditional statement is then given

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

The question is now to decide what is the minimum number of cards that need to be turned over to prove that the conditional statement is true, and what are those cards?

If you said "E" well done that is "affirming the antecedent", "E" is a vowel and thus should have an even number on the other side. If there was an odd number on the other side, the statement would be false, so E must always be turned over to check the validity of the "if - then"

But what other card(s) do you need to turn over?

If like many people you choose "E" and "4" then with "4" you are affirming the consequent. Even though 4 is even, it may have a vowel or consonant on the other side and the statement is not falsified by either so that was wrong

If you said E and 7. Then 7 could deny the consequent and hence must be checked. If there was a vowel on the other side, the statement would be false.

So if you said E and 7 well done! If you said E and 4 well be careful how you use your IF / THEN arguments you might get bitten on the bum!

** Update **
The E / 4 answer shows how easily we fall into what is known as confirmation bias.. When we attempt to confirm the claim, we can forget that it is also important to try and falsify it. Finding a pairing of vowel and even number does nothing to support the claim; it's the failure to find a vowel with an odd number that confirms it.