Could someone come up with a formula explaining the following?

Question

This is a card "trick" that I was shown over the weekend, and I'm sure there must be a concrete mathematical way to explain why this always works:

Card values are:

• numbered cards - face value
• jack - 11
• queen - 12
• king - 13

Start with a pack of cards containing no jokers (so 52 cards) - shuffle the pack well.

1) Start dealing the cards into packs. Take the top card and place it face up. Starting at the card's value, continue placing cards on that pack (regardless of their value) until the pack reaches 13.

e.g. You place a 4 - place another 9 cards on top of it (5, 6, 7, 8, 9, 10, 11, 12,13). The values of the cards that follow the first card don't matter, you're just counting up to 13. This pack is now done. You place another card. It's a 7, you place another 6 cards on top of that to complete that pack. The next card is a queen - you place 1 more card on that to complete that pack.

2) Continue until you either use all the cards or until you are unable to complete a pack.

3) Flip over any 3 of the packs you created so that they are all face down, pick up the other cards and hold them in hand.

4) Flip over 2 of the cards on the packs that you flipped face down in the previous step.

5) Count 10 cards from the ones in your hand and set them aside, they are no longer part of the trick.

6) Add the values of the 2 face up cards from turn 4 - let's say they were a 4 and a queen. That means that the total is 4 + 12: 16.

7) Remove another 16 cards from the ones you are holding.

8) Count how many cards are left in your hand.

9) The amount of cards you have left in your hand will be the value of the card on the 3rd pack from stepped 3, which wasn't flipped face up.

This works every single time, provided you didn't slip up in counting somewhere.

I would love if someone could give me a formula for this!

Please feel free to edit this post and add any relevant tags, I am not sure which ones I should add.

Answer 1

Presumably the last pack (which might be incomplete due to step 2) is not one of the three special packs.

The most important cards are the three that "start" the three packs, which also happen to be the ones you see in step 4.

Let $x$, $y$, and $z$ be the value of the three cards that start these three packs. Then the number of cards in these packs is $14-x$, $14-y$, and $14-z$, due to step 1. The number of cards in your hand in step 3 is then $52 - (14-x)-(14-y)-(14-z)=10+x+y+z$.

In step 5, you then have $x+y+z$ cards in your hand. In step 6, you see a pair of the three original cards, say $x$ and $y$, and after discarding $x+y$ cards you have $z$ cards left in your hand, which is precisely the value of the remaining card.

Answer 2

It is not that hard in fact :)

Start with dividing cards in packs. Let $x_1,x_2,\dots,x_n$ represent the values of the cards you pick.

Then, number of cards in each pack is:

$\bullet \hspace{0.5cm}13-x_1+1$

$\bullet \hspace{0.5cm}13-x_2+1$

$\vdots$

$\bullet \hspace{0.5cm}13-x_n+1$

You flip three packs, lets say $13-x_i+1$, $13-x_j+1$ and $13-x_k+1$ over and pick cards from the rest in your hands.

Then, in your hands you have:

$52-(13-x_i+1)-(13-x_j+1)-(13-x_k+1)$ cards which simplifies to $10+x_i+x_j+x_k$.

Now, you remove 10 cards leaving you with $x_i+x_j+x_k$ cards in your hands.

Then, we flip any two cards from the packs facing down - suppose that we pick packs with $x_i$ and $x_j$ as top cards - remember that now the top card is $x_i,x_j$ and subtract $x_i+x_j$ cards from the hands.

This leaves you with $x_k$ which is exactly the value of the top card in the last pack.

Answer 3

The equation is: $52 -(14-a_i) - (14-a_j) - (14-a_k) - 10 - (a_i + a_j) = a_k$.

The three packs have cards $a_i,a_j,a_k$ on the top. They have $13-a_i$, $13-a_j$, and $13-a_k$ more cards in each pack. So the packs have a total of $14-a_i$, $14-a_j$, and $14-a_k$ cards each. The top cards of the packs are $a_i, a_j,$ and $a_k$.

There are $52 -(14-a_i) - (14-a_j) - (14-a_k)= 10 + a_i + a_j + a_k$ cards left over that you put in your hand. You remove 10 of them and you are left with $ a_i + a_j + a_k$ cards in your hand.

You turn over two $a_i$ and $a_j$. You remove $a_i + a_j$ from the $a_i + a_j + a_k$ cards remaining. That leaves $a_k$ cards remaining which is the top card of the last pack.