Great card trick...but why does it work? (mathematically)

[dOrKuS]

This is one of my favorite card tricks, but I'm not sure why it works, lmk if you have any ideas...
Kind of lengthy explanation for the trick, but it's worth it.

Shuffle deck
Flip over the first 26 cards (half the deck), face up, in a pile (must keep them in order)
As you're doing this remember the 7th card
Take those first 26 cards and put them on the bottom of the deck
Now flip over the top three cards (next to each other)
Count to 10 with each one. (If you flip a 3, K, 6, you place 7 cards from the deck on the three, 0 on the king, and 4 on the six. All face cards count as 10, and the Ace counts as 1)
Add up the three cards you just flipped face up. (In the example it would equal 19)
From the deck you are holding, the 7th card from the initial flipping of 26 will be the 19th card (or whatever the sum of the 3 cards you count to ten with is).

Give it a try, and if you know why this works lmk.
Hopefully the directions are clear enough.

~Brian

edit: can't add 10 + 6 + 3, thanks conjur

 

[conjur]

Wouldn't your example = 19, not 20? 3 + 6 + 10 (K)?

 

[dOrKuS]

Yep, sorry

 

[conjur]

DOH!!

 

[rh71]

Whoever found that out in the first place needs better things to do with their time.

 

[Skyclad1uhm1]

You should have been able to think of the error in that even sooner if you had taken 3 aces as example draw. 9 cards extra for each, and you'd be stuck with finding the 30th card in the original 26 card draw.

 

[Ness]

The magic lies in the fact that you are always counting the same number of cards out, no matter what the numbers are.

If you have the 7th card from the top of 26 cards, and you add 26 more on top of it, it will ALWAYS be the 33rd card.

Now, when you pull the 3 cards down, you bump that card up to 30th card, right?

Let's say for instance, you get a 4, 5, and 6. You put 6 cards on top of the 4, 5 on top of the 5, and 4 on top of the 6. That's 15 cards in total. Then, you remove the total of the original 3 from the top, which is also 15. Which means that the 30th card is now on top.

Now, to illustrate why this trick works, rather than removing a number of card equal to the total of the original 3 AFTER you place the cards on top of it, let's just remove the same number as the number on the card in addition to it. (rather than counting down at the end, let's just go ahead and count the cards at the end with the cards that are placed on the other cards.) So, on top of the four, you place 6 cards. Then, because it is a four, you place an additional 4 cards on it.
What you have to realize is that each of the original 3 cards will require you to draw a total of 10 cards, because no matter when you pull them off, or what pile you place them in, you will get 10.

As for an equation, try this:

let x equal the original amount of one of the 3 cards. Since we know that we must bring the total to 10 when counting out the cards to place on top, we can say "10 - x = y" , where y is the number of cards counted out the first time. Now, we know that in the end, we must count out the a number of cards equal to the original card amount (we'll call it z). So in this case, so z can also be defined as x. ( z = x ).
Therefore, if x is the same as z, we could also say that 10 - z = y, furthermore, we could say that 10 = z + y. Because of that, we know that we will always count out 10 cards, because z is the number we could later, while y is the number we count out first. and clearly z + y is equal to 10.

So, since there are 3 stacks, we know that there will be 30 cards counted out, and then you add the original 3.. which means that 33 cards will be counted from the top, ALWAYS. As stated above, the seventh card will ALWAYS be 33rd from the top. Picking the seventh card in the stack is NO coicidence, and the confusion of the trick lies in the fact that you count out the cards at different times, to give the appearance that the number of cards being counted out is changing.

 

[dOrKuS]

say you draw three aces...
49 cards left in the deck
49-27 (cards pulled to make 10 w/ each) = 22 left
3 aces = 3
22 - 3 =19
19 would be the 7th card deep into the initial 26, which you placed on the bottom of the deck

I see why that works, but flipping the 3 cards flipped are soo random... thats the part I don't understand.

 

[dOrKuS]

ness1469,

Your explanation was perfect, thank you.
Gotta admit though, it's still an interesting little trick (even though its not a trick at all).

Have a good weekend all,

~Brian

 

[Ness]

Originally posted by: dOrKuS

I see why that works, but flipping the 3 cards flipped are soo random... thats the part I don't understand.

the numbers on the cards are the only thing that is random. The fact is no matter what the card is, you end up counting out 10 cards for each card, plus the card itself. It will always be 33 cards drawn from the top. If you really need to, open up a notepad document and write out as many hypothetical situations as you have to realize it.