The EASIEST Card Trick Ever - You Can't Screw Up!

This week's trick requires no sleight of hand, elaborate math-magicry or fancy moves. It's so easy, it even fools the performer! Watch Brian and learn.

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ABOUT SCAM SCHOOL
If Harvard offered a PhD in deceit, this would be it. Award-winning magician Brian Brushwood takes viewers on an inside tour of bar tricks, street cons, and scams. If you watch carefully, you'll never have to pay for a drink again!

[asdf]

Opens up a scam school on the internet
*Gets scamed by his barber..*

[MichaelSimmons.]

If anyone is looking for a reason to deal out the 10 cards, I tell people since there are 3 piles face down, we subtract the 3 from 13, which equals 10. Therefore, we place 10 cards face down, and continue with the trick.

[HaydenLau.]

This show makes me feel good about my hair

[StViers]

When I do tricks like this, I spell out "Hocus Pocus" instead of saying I'm just putting 10 cards down. Slight muddling!

[TyberiusWilson]

To all of you who are wondering, "Why didn't it work? What did I do wrong?" There's a few things you may have missed:

1. There has to be 52 cards in the deck, and no Jokers. No more, no less.

2. You have to make sure that you count up from the first card dealt up, no matter what the other cards are, until you reach up to the "13th" card. For example, if the first card you dealt down was a 4, then you would pretend that there were already four cards down and count up from 4 to 13, adding cards and counting "5... 6... 7... 8..." until you reach 13.

3. Aces = 1. Jacks = 11. Queens = 12. Kings = 13, don't deal down any more in that stack if it started with a King. All the other cards are simply what the number says. Tens = 10. Sixes = 6. etc.

4. *IMPORTANT* : Put the leftover cards you had from dealing out the decks with the cards from the decks the spectator did not choose while you were looking away.

5. The spectator must pick 3 decks from the number of decks you dealt. No more, no less.

6. When you get to the last 3 decks and you're about to use the "Discard pile" for finding out the top card of the 3rd deck, FIRST deal off 10 cards, then the number of the other two cards, and the number of the remaining cards is the value of the top card on the 3rd deck.

[robertbrandywine]

First, you must be able to intuit that each stack contains 14-N cards, where N = the number of the initial card.  If you can't get that far, then the below won't make sense.

Number of cards = 52 = R + (14-x + 14-y + 14-z), where R = the number of cards in the collected face up stacks plus the discards.

14-x + 14-y +14-z = 42 -x -y -z

so substituting:

52 = R + 42 -x -y -z  

R = 10 + x  + y + z 

Say y is the unknown card

y  = (R-10 - x -z)
 

[vaclavpernicka]

Ok here is the math:
Assume you picked three cards (thats how it ends)... their values are X, Y, Z
Their piles have 14-X, 14-Y, 14-Z cards, because you count from that card to 13 inclusive.
Lets say piles with Y and Z have been revealed. So you count down 10 + Y + Z cards.
There is total 52 cards.
Now the final math how many cards you have in hand:
You take all cards (52) subtract the cards that you just put on the table (10 + Y + Z) and also subtract the piles (14-X + 14-Y + 14-Z) and the final result is:
52 - (10 + Y + Z) - (14-X + 14-Y + 14-Z) = (52 - 10 - 42) + (Y - Y) + (Z - Z) + X = X
So its X cards what remains in your hand.

[theflyingpie]

"The EASIEST Card Trick Ever - You Can't Screw Up!"
*Screws up*
.-.

[Colm]

The EASIEST Card Trick Ever - You Can't Screw Up! i'm just thinking "ooo i'll find a way"

[mohammadparhizkar]

Mathematical perspective: 

I assumed this situation: we have 6 piles, the value of Xi shows the number of first card at pile i and the value of Yi shows the number of cards that should add to this pile.

I formulized the problem as below:

X1+Y1=13
X2+Y2=13
X3+Y3=13

X4+Y4=13
X5+Y5=13
X6+Y6=13
Y7

And we know, Y1+Y2+Y3+Y4+Y5+Y6+Y7+6 (The number of piles)=52
If we flopped the first three piles and gather the others, we have this equivalent: 

Y4+Y5+Y6+Y7+3-X1-X3-10=? (It may be X2)

I put (-13+Y1) instead of -X1 and (-13+Y3) instead of -X3. Thus we have:


Also we know, Y1+Y3+Y4+Y5+Y6+Y7=46-Y2 finally we have: 46-Y2-33 that is equal with 13-Y2.

Thus the answer is always X2.

[businessgirlyt]

You have no idea how happy I am to have found this video! I saw it a few years ago as a recommended video and did the trick with my friends. Unfortunately, I ended up forgetting how to do the trick and I couldn't find the video to relearn it...

Joy of joys to see it again. :D

[vilhjalmurtheviking]

I learned this trick 40 years ago, good to know it's still going strong. I just count all the cards out that remain in my hand in front of everyone and subtract 10 from that, then just tell them what the last card is. Different way of doing it, still pretty cool.

[blairmunro]

Y'see, you say you can't screw up, and you're right. I managed to, because I somehow lost one of the cards in the deck before performing this trick.

However, that led me to an interesting variation. If you stash any card in the box for the deck, you're left with 51. This lets you count up to the reveal, and then turn over the final card as the last number in your count.

From memory, the magic card was a five, and I was left with four cards in hand. I mumbled something about magical wires being crossed, when I should have turned over the five and counted it as the fifth. Subsequent performances of this trick use this variation, and I've found that it neatens things up a little bit.

[Mowreets]

What I like to do with this trick is have someone pick the top card of the last deck (the card you would be guessing) and let them look at it without showing. In the meantime you count the remaining amount of cards under the table or if there are very few cards a quick glance at your deck. Then have them stare into your eyes and think about the number or picture their card was and you can guess their card correctly. Just to create a magical vibe.

[PKPK-rrrs]

Barber: Which haircut do you want today?
Brian: I want children to scream at me.
Barber: Say no more.

[jadon]

9:25 i love how everyone turns and looks at you guys like you all just let out the gnarliest fart they've ever witnessed

[hooperga]

Each pile contain (14 - number_of_the_card) cards as it goes from the number until 13. if its a King(=13) there must be 1 card, if its an ace (=1) ther must be 13 cards and etc.

There are 52 cards in the deck, that must be equal to:
the 10 cards left over + the number of cards on pile1 + the number of cards on pile2 + the number of cards on pile3 + the number_of_the_card1(dismissed cards from the top of pile 1) + the number_of_the_card2(dismissed cards from the top of pile 2) + the misterious number, so:

52 = 10 + (14-number_of_the_card1)+ (14-number_of_the_card2)+ (14-number_of_the_card3) + number_of_the_card1 + number_of_the_card2 + misterious_number
adding everithing: 52 = 52 + number_of_the_card3 - misterious_number
misterious_number = number_of_the_card3.

[paullelekis]

Glad you like this effect, Brian!  I created it about 40 years ago and was the most popular trick in years published in the Linking Ring magazine.  I was also asked by Stephen Tucker of the UK if he could include it in his magicians' e-zine, THE BUDGET in August of 2012.  I am Paul A. Lelekis if you're interested and you presented it well...however, I dropped using the face cards in the counting procedure, years ago because spectators have trouble with associating the values of 11, 12 and 13 to the Jacks, Queens and Kings, respectively, and it also leads them to the possibility of being a mathematical trick - which it is.  If you just use the spot cards and "burn" any face cards that show up, it becomes truly inexplicable!

[BradenBest]

"You Can't Screw Up!"

*screws up*
*gets kicked out of casino for card counting*

[bmaroney]

Has worked every time for me...once I ended up with one card after counting 10, then the appropriate number for the two turned up cards. The final flipped revealed an ace -- and the final count out card was also an ace -- just added to the "wow" factor!