A logic puzzle for your Friday enjoyment


What kind of person would I be if I didn’t leave you with a brain-twisting logic puzzle to help work your neurons and burn off all those calories from your Halloween candy? A bad person, that’s what. And I’m not a bad person.

So, let’s say you’ve heard of the Monty Hall problem. And you’re not like the uncultured masses, you know that the probability of getting the car if you switch is 2/3, not 1/2. And you feel so smart, don’t you, knowing the fundamentals of probability better than Joe Schmoe down the street. Well, try this one, tough guy:

Suppose you pick four cards out of a standard deck: the Ace of Spades, the Ace of Hears, the 2 of Spades, and the 2 of Hearts. You shuffle these four cards very well, and then hand two of them to me (we both know what the four cards are). So, now I’ve got two out of the four cards, and it’s your job to guess what the probability is that I have both aces.

At the moment, the probability is clear. There are six possible card combinations that I could have:

A♠A♡   A♠2♠   A♠2♡

A♡2♠   A♡2♡   2♠2♡

and since they occur with equal probability, I’ve got a 1/6 chance of having both aces.

But now I say, “I have at least one ace.” (and we will assume I’m not lying). Now we want to compute the probability of my having both aces, given that I have at least one. In math terms, this would be P( both aces | one ace ).

Well, there is one combination out of the six that does not have any aces (2♠2♡), and we can eliminate that possibility. So now my chance of having both aces rises to 1/5.

I don’t stop there, though. Now I say “I have the ace of spades”, and again, I’m not lying. What does this do to the probability? Clearly, it eliminates all of the possible combinations that don’t have the ace of spades, leaving

A♠A♡   A♠2♠   A♠2♡

A♡2♠ A♡2♡ 2♠2♡

and so the probability is 1/3. Notice, though, that we could apply exactly the same reasoning if I had said “I have the ace of hearts”:

A♠A♡   A♠2♠ A♠2♡

A♡2♠   A♡2♡   2♠2♡

and the probability would again be 1/3.

But wait a minute. We said before that if I announce, “I have one ace”, the probability that I have both aces is 1/5. But because we both know what the four cards are, if you know that I have one ace, you know it must be either the ace of spades or the ace of hearts, and either way, the probability of my having both aces is 1/3.

So which is it? If I say, “I have one ace”, do I have a one-in-five chance of having both aces, or a one-in-three chance? They can’t both be right. More importantly, explain why this is the case, and why the wrong answer this wrong (it’s very easy to be right for the completely wrong reason with this puzzle).

Now, this is not a trick question. It’s like the xkcd Blue Eyes puzzle: there’s no lying or guesswork involved, and the solution is not some stupid out-of-the-blue side-attack that will make you groan. It takes some careful deliberation to work out the right answer.

I can’t claim credit for this question: Cornell’s Professor Halpern used it in a recent Decision Theory lecture to help illuminate the dangers of improperly constructed state spaces for us. I did a little Googling and couldn’t find it on the internet, though, so I figured I’d share it for your enjoyment.

I’ve tried to make the question as clear and unambiguous as possible, but if you need clarification on something, use the comments or you can hit me up on Twitter. Have fun!


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