Face-Up Card Puzzle Explained

When I taught a 3-week intensive course for 6th-8th graders called The Beauty of Mathematics several years ago, I used many mathematical card tricks. Some of these mathematical card tricks are quite amazing when embellished for an audience. They were quite fun to perform, instruct how to perform, and then to break down and analyze. This is one of the many great things the kids took away from that exerience.

The Riddler Express for last Friday isn’t a card trick so much as a card puzzle. It has a rather elegant solution and explanation that I hope you will enjoy and understand. The puzzle is as follows.

You are in a pitch black room and cannot use your sense of sight. Sitting at a table, you feel a standard 52-card deck of cards in front of you. Scattered randomly throughout that deck of cards are 13 face-up cards.

Your job is to separate this deck of cards into two piles of cards that both have the same number of face-up cards. Both piles must contain at least one card. How can you do this?

It helps to have some sort of visualizatoin, so here is a possible arrangement of the 52-cards spread out so you can spot the 13 face-up cards.

Example 52-card deck with 13 face-up cards

The Solution

The first thing I remember thinking was to simply cut the deck in half by counting 26 cards. I immediately found the issues with this, as you cannot split 13 cards evenly. The best I could hope for would be 6 face-up cards in one pile and 7 in another. This means that there must be some cards you have to flip over.

Next, I thought about flipping cards over one at a time and wondered when I should stop. I then asked myself what if I stopped at 13?

It turns out that this is a solution to the problem! Indeed, you could count out any 13 cards in the deck, flip them over to form a pile, and you would solve the problem.

Why does this work?

Here is the same deck as before, except with 13 of the 52 cards in the bottom pile, while the remaining 39 are in the top pile.

Separate Piles prior to turning any over

From the photograph, this particular example shows us that there are 3 cards in the bottom pile that are face-up leaving 10 cards in the top pile that are face-up. But let’s analyze every possible case. Some of you are afraid of variables, but hang with me, there is only one.

Yes, there are 3 face-up cards in the example picture, but since we wouldn’t know in general, we must say there are x face-up cards in the bottom pile of 13 cards. What do we know about x?

The number of face-up cards in the bottom pile could be 0, 1, 2, etc. all the way up to 13. Thus, x is a number between 0 and 13. In the top pile, our example shows 10 face-up cards. We didn’t need to see the pile to deduce this. Since we know there are 3 face-up cards in the bottom pile, and 13 face-up cards in total, there are 13-3=10 face-up cards in the top pile.

Using our variable instead, we know there are 13-x face-up cards in the top pile.

Now, flip all of the cards in the bottom pile of 13 over. Here is the picture of our example case for clarity.

After flipping our bottom pile of 13 over

What happens? There were once 3 face-up cards and now there are 10 face-up cards…. matching the number of face-up cards in the other pile!

There are 13 cards in the bottom pile, x of which are face-up (meaning 13-x of them are face-down). So, when we flip them all over, we will have 13-x cards that are face-up and x of them face-down.

For whatever x is, 13-x (the number of face-up cards in our bottom pile now) will always be equal to 13-x (the number of face-up cards in the top pile). So, we’ll have the same number of face-up cards in each pile every time we do this no matter what.

By using one simple variable and thinking a little bit mathematically, we have demystified this fascinating riddle.

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