# Flippin' Cards

Here’s a riddle for you:

A friend brings you in a dark room and hands you a shuffled deck of 40 cards, 10 of which are facing up and the other 30 are facing down. Your task, she tells you, is to come out of the room holding two decks that have the same number of cards facing up. As it is dark in the room, you cannot see anything, and the cards cannot be distinguished by touch. How can you do it?

No tricks are necessary, it can be done with logic alone. You might want to take some time to think it through, before reading my solution.

Here are some things that you might have noticed:

- As the deck is already shuffled, there is no point in re-arranging the cards somehow.
- Flipping a card twice brings it back to how it was at the beginning, hence the problem actually consists in choosing which cards to flip.
- Since the cards cannot be told apart, the essence of the problem is to choose how many cards to flip.

Now this looks like a much simpler problem. The main idea is to split the deck in two smaller decks, and flip the cards in one of them. The question is, how many cards should be flipped?

Let’s introduce some notation. Say we take the first $k$ cards and flip them, and leave the remaining $40-k$ untouched (in light of point 1 above, it does not matter which cards are flipped). Now, assign a binary variable to each card, so that $x_i=1$ if and only if the $i$-th card is facing up. This allows us to express the operation of “flipping the $i$-th card” as $1-x_i$.

Call $m$ and $n$ the number of cards facing up in each deck:

\[\begin{equation} \tag{1} \sum_{i=1}^k x_i=m \end{equation}\] \[\begin{equation} \tag{2} \sum_{i=k+1}^{40} x_i=n \end{equation}\]From the statement of the problem, we know that there are 10 cards facing up in total, so that $m+n=10$. Now we flip the cards in the first deck, so that the number of cards facing up in this deck becomes:

\[\begin{equation} \tag{3} \sum_{i=1}^k (1-x_i)=k-\sum_{i=1}^k x_i=k-m \end{equation}\]We want the decks to have the same number of cards facing up, which means:

\[\begin{equation} \tag{4} k-m=n \end{equation}\]or, by rearranging:

\[\begin{equation} \tag{5} m+n=k \end{equation}\]But we know that $m+n=10$, so $k$ must be 10, too! Note how it also works if you flip the larger deck instead: Equation 3 would result in $40-k-n$ cards flipped, and $40-k-n=m$ tells us that $k$ would need to be $30$ (i.e., the number of cards in the larger deck would have to be 30, exactly as we found earlier).

Hence the solution is:

Create a deck by taking any 10 cards, and a deck with the remaining 30. Flip either deck.