The two card problem

Suppose I have two cards in my hand and make the following statement:

If one of my cards is a Queen, then the other card is an Ace; or if one of my cards is not a Queen then the other card is an Ace

Although it appears that in both cases which are mutually exclusive one of the cards is always an Ace, the logical conclusion is:

I don’t have an Ace

The illusion arises from the difficulty of mentally processing conditional statements. Using Propositional Calculus and Boolean algebra we can derive the logical equivalent of the statement for analysis. Using simplified notation we can formulate the problem as follows:

(Q \Rightarrow A) \veebar (~Q \Rightarrow A)

Where

  • Q and A represent having a Queen and having an Ace respectively;
  • ~ is the negation operator. if the proposition P is True, then ~P is False;
  • \Rightarrow is the material conditional operator “if … then…”. This operator joins two propositions named antecedent and consequent. The material conditional takes the value of the consequent when the antecedent is True, and True when the antecedent is False; and
  • \veebar is the exclusive disjunction. The operator joins two propositions and its value is True when the value of only one of the propositions is True, otherwise it is False.

The material conditional is a secondary operator and can be replaced with primitive Boolean operators noting that X \Rightarrow Y \Leftrightarrow ~X v Y (where v represents the simple disjunction ‘or’ and the double negation cancels out) we arrive at the following substitution:

(~Q v A) \veebar (Q v A)

The exclusive disjunction \veebar is also a secondary operator and can be substituted using the following equivalence X \veebar Y \Leftrightarrow (X v Y) ^ (~X v ~Y) where ^ represents the conjunction operator ‘and’:

((~Q v A) v (Q v A)) ^ (~(~Q v A) v ~(Q v A))

The statement on the left of the conjunction is a tautology (always True) due to the following laws: ~Q v Q \Leftrightarrow True; A v A \Leftrightarrow A; and A v True \Leftrightarrow True. Noting that True ^ X \Leftrightarrow X we simplify the statement and rearrange the order of the propositions:

~(A v ~Q) v ~(A v Q)

Applying deMorgan’s law: ~(X v Y) \Leftrightarrow ~X ^ ~Y we obtain:

(~A ^ Q) v (~A ^ ~Q)

This statement is the result of applying the distributivity of ^ over v. The reversal yields:

~A ^ (Q v ~Q)

Noting that X v ~X is a tautology, and that Y ^ True \Leftrightarrow Y the statements resolves to

~A

In other words, I don’t have an Ace.


Johnson-Laird, P. N. (2001). Mental models and deduction. Trends in Cognitive Sciences, 5(10), 434-442.

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