Saturday Puzzle #20 – Ace Ventura, Card Detective

You and two of your friends (not facebook friends, real world friends, remember those?) are playing a game. The dealer holds three cards, which may contain any number of aces (0, 1, 2 or 3). Each player is dealt a card face down, and asked to hold their card up against their forehead so the value side is facing out.

At this point, no player knows which card he/she holds but all three players can see the other two players’ cards. The dealer asks you to raise your hand if you see one or more Aces (of any suit). You look around and see both of the other players are showing an Ace, so you raise your hand. The other two players also raise their hands. So, almost immediately, all three players have their hands in the air.

Next, the dealer says: “If you know whether your own card is an Ace or not, lower your hand”. A few long minutes go by, as all three players ponder this question. All three hands remain in the air. Here’s the challenge: given everything I’ve told you, can you determine whether you have an Ace or not?

As always, leave your guess in a comment. I now support Facebook comments, so feel free to use that option if you like.

Solution: The way to solve this puzzle is to put yourself in the other guy’s shoes, so to speak. Let’s call the three players A (you), B and C. Player B’s hand is in the air because he sees at least one ace – player C’s card. Now let’s imagine that you hold some card other an ace. Player B will reason as follows:

Player C’s hand is in the air so she sees an ace but it can’t be A’s card because I (player B) can see that player A’s card is not an ace. Therefore, player C must be looking at my ace.

Player C can make a symmetric argument (if A doesn’t have an ace, then B must be looking at my ace). Thus, if you don’t hold an ace, with a moment of thought it will be obvious to both B and C that they hold aces and their hands will go down in short order. The fact that they don’t reach that conclusion suggests that you must be holding an ace.

The solution to this puzzle involves indirect thinking in the sense that it requires you to reach a conclusion based on other people’s inability to reach a conclusion. To be fair, I probably should have stated something like this: the other players in the game are known to be perfectly logical people. Even without that clarification, some smart readers figured out the solution – congrats to Katy, Gareth and Muzaffer!

3 thoughts on “Saturday Puzzle #20 – Ace Ventura, Card Detective”

  1. Let’s try to solve it using ‘reductio ad absurdum’ (proof by contradiction); Suppose I do not have an Ace… Then, both of my clever friends would thought like this; ‘Muzaffer does not have an Ace but my other friend raised his/her hand so I must have an Ace’ and then lower his/her hand. This did not happened so my assumption must be wrong. If my not having an Ace is wrong then I must have an Ace :)

  2. Yes. I must have an ace, otherwise one of the other two players would have lowered their hand. If I don’t have an ace, player 1 can infer from the fact that player 2 doesn’t immediately lower his hand that he has an ace and would therefore lower his hand. Because this doesn’t happen and both the other players’ hands remain in the air, I must have an ace.

  3. Yes, (assuming players kept their hands up for the right reasons) I should be able to determine that my card is an Ace

    Statement 1: For 3 hands to be raised, the deal is either AAX or AAA.

    If the deal is AAX, two players must see only one Ace, and must conclude that they hold the second Ace (statement 1) so these two players hands should go down when asked if they can determine whether or not their card is an Ace. But all hands stayed up so the deal wasn’t AAX, which means the deal must be AAA.

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