Self Working Card Trick FOOLS Penn and Teller | Chris Capehart Tutorial
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- Опубликовано: 4 мар 2024
- This self working card trick is a real fooler! Take some time to practise the first part and you will fool magicians with ease!
This was performed by Chris Capehart on Penn nd Teller's Fool Us 
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I also replicated the trick from the Penn and Teller show 3 days ago, but the TV direction cut out the sequence of cards at key moments, so I had to come up with my own routine to faithfully replicate the trick performed. I used 8 forced cards, which I cleverly arranged so that counting from the bottom deck there were always even numbers from the combination 1-4, and counting down from the top deck there were always odd numbers from the combination 1-4. It is out of 54 cards. This way means I can generate even position 12, and odd position 21. Not just 21, like in your example. I can generate any positions from a combination of 1-4. I only discard cards from the top of the deck, or the bottom of the deck, whether the combination is even or odd. The viewer can choose which combination they want, for example 13 or 31. All will be available from the combination 1-4. When discarding cards, a forced card can appear 2x, maximum 3x. But no one notices - tried - after all, no one knows what card is forced and the trick is only performed once in front of the same audience. Your move is correct because it shows a +1 card in number 23, so 24. Your move reduces the forced card to 2 pieces and can appear in the discard pile 1x, which no one will notice. The problem with this trick is the number 33. I wouldn't do any silly manipulations at all with the jokers you put in the middle of the deck, but I would simply say, there can't be two same numbers (33) because that would be very easy. The viewer is very happy to change his number to make it "hard". :) Good work. As for the second trick, it is performed by young children in my country. That's why he didn't fool Penn, Teller, or me. :)
You can reduce the force cards to 4x by using a 55 card deck (3 extra cards, presumably the force cards). Combinations like 23 and 24 are equivalent, either you discard 23, and the next card is the force, or the 24th card is the force. Having a 55 card deck lets you count from the bottom and in the same way 32 and 31 also reach the 24th card from the top. three other force cards would be required for the other twelve possible selections.
The original trick involves jokers, however, so I am inclined to to think that there are ways of narrowing it down even further. I strongly dislike "lets get rid of the jokers" because it draws suspicion to fixing the count.
Hi steven ! Thanks for sharing! 💯👍🙏🌟🌟🌟🌟🌟 For your information i used to perform with Christ and Penn and teller on the street of Philadelphia 44 years ago. Penn and teller are real good people, very nice guys in real life. I passed cluds with Penn. Christ was my partner sharing the spot. Anyway Penn was not fooled he said he wants to help him to get the trophy. In America with that trophy you can get more bookings. You was puzzled why Penn was fooled. Penn admired Christ from the past beginning of time. That was an era of street performing. No youtube no internet. Pure magic. But thanks to you bringing knowledge to mankind.🎉🎉🎉🎉
Chris Capehart made a wonderful performance for P&T :)
I think they were fooled by the first trick, even with the explanation it looks straight up impossible! :)
I got a lil bit "inspired" (it probably exists already, I'm not an inventor by any means) by the second part which totally fooled me as well by the way, but after I knew it was "just" 4 forced cards in a stacked deck, if you only have regular decks, I got a little idea:
- For the set up: if you wanna keep the jokers in, set up the 8s in 11th, 22nd, 33 and 44th position, save a pair of 4s (or any pair that makes 8) , place them wherever you like in the deck as long as you remember the cards and suits to be sure to take them out and not disturb the order.
if you don't keep the jokers in, set up the 6s in the exact same way and save a pair that equals 6.
- Do a bunch of false cut/shuffles while you start to lie... hum... tell your patter/present yourself, don't mention that you are cutting the deck if you did nothing yet cos, well, they can see you are cutting/shuffling the deck... but it's only my 2 cent, do whatever you like ^^
- Show the deck face up to your audience so they can see nothing is fishy about it (wink), insert your best patter/ bunch of magician lies to kinda explain the effect you'll do, take the deck and look for your prediction with whatever motivation linked to your presentation, hide them or show them what are those 2 cards is up to you again, then proceed to force the 4 cards to your victim(s), with the jokers in the deck, you'll automatically end up with 8 cards, just like the four of a kind "they" selected, and just like the prediction you made at the start.
and if you've taken out the jokers, you'll end up with 6 cards.
give them the deck to examine and as a souvenir while they are in shock, or runaway as fast as you can because they summoned the devil with all of those 6s...
^^
Nicely done, great explanation!
Both tricks are amazing. This second trick l have done a number of times but without the blank deck. I always reveal the Aces last it adds a nice kicker. Both tutorials were awesome 👏😊👏😊😊👍👍
You can get a single spectator to holdup a number on each hand, then chose to read the number from either your or their perspective. Also it's better with two force cards say a duplicate of the 24th card in your left shoe and the 44th card in your right shoe.
Great tutorial I'm sure Penn and Teller weren't fooled by the second trick and only the first.
Love the second trick 👍
Steven this is absolutely brilliant! 👏🏻😁🏆♣️♥️♠️♦️
Cheers dude 👌🏽
Great Job
If you have 2 jokers at the bottom of the card deck and take 53 cards, excluding the 4 jokers (2 at the top and 2 at the bottom of deck), all numbers work.
Bravo ❤❤❤
Thank you!!!!!
Is the Chris Capehart's own trick or has it been published elsewhere? Anyone know the name? Would like to track it down.
What happens if they tell you to stop at 1? You can only put ONE card down (the queen) the ace is still in your hands. What possible excuse could you use to put another card down under the king when you have not done it that way all along? That is why if you watch Chris Capehart on Fool Us and he gets to Tellers last count down, he gets very worried when he is down to 4,3, he gives Teller a look, 2, finally Teller signals stop and Chris said "I thought HE was gonna be the difficult one" pointing to Penn. because if Teller would have made him count down to one, Chris could only put the queen down. He would now have 5 cards remaining in his hand and one of them would be the last ace. How do you explain that?
What is this trick called? Is it in a book?
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I assume that Chris Capehart did a deck switch when he moved on to the second part of his routine.
Nope. No deck switch at all. You can see both decks in view during his performance.
@@steventaputoro I just rewatched it. He just smoothly moved into the second routine with the second deck. I assumed he was using the same deck throughout. 🤷🏻♂️
Hi there miss you man 💘
The screenshot didn't account for 12 13 and 14. I guess at this point magician force the number.
I think there is a detail that's been missed for this trick that makes this less complicated than its presented here. You even mention it in your introduction of selecting the numbers, but then procced to interpret it differently. The selection is between 1 and 4 not from 1 and 4, which limits the number combinations severely.
The only numbers selectable between 1 and 4 is 2 and 3, which creates combinations 22, 23, 32, 33 when rearranged in their possible arrangements. This means that you only have to place the prediction cards at locations 23 and 33. If 22 and 32 are picked then it's just a simple matter of counting to the required number (22 or 32) then turn over the next card. No extra cards, no multiple outs to be memorized.
How in hell did you puzzle all of this together? 😄 Honestly, please
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Hi Steven,
I think the Number 32 is not possible. When you count face up, the force card is on position 31.
Rewatch the tutorial
In your System stand read the Number 1 and 3 as 31 and then count from the bottom to 31. On Number 32( 3 and 2) stand also count from the bottom, but to 32. The Force Card can not be at the same time on both positions. On bottom you have no jokers to compensate that.
Rewatch.
I agree with @joinmylife. There's no way the force card will be at position 32, counting from the face. It's only at position 31. The only way possible, is to count off two at a time face up, then the 7H will be on top of the face up pile.
I think it's better to have 55 cards - a few little changes: 11 and 31 would count from bottom, then next card reveal, 33 (if you allowed that combo) would only require 1 joker inserted, or else just a subtle move of one card top to bottom.
Not if done right. Follow my tutorial and it works perfect.
what if they stop at 1
Then you put 1 card down 🤷🏽♂️
@@steventaputoro
Then how do you put the ace down?
I genuinely think that Penn & Teller were fooled by the first trick, since they have a free choice and there can be many possible outcomes.
1 and 2 is 21, 1 and 3 is 31, and 1 and 4 is 41? What if the spectator says he wants 12, 13 or 14? I think you did not mention those numbers.
Rewatch the tutorial
you just ask for numbers, you dont ask them to name the combination. You do that yourself. It's your magic trick meaning youre leading it
penn and teller are getting really stupid - i'm not even going to shuffle the deck - doh! so I have a preset deck already set-up for the trick so don't ruin it and btw you can't even look at the deck before I start my trick - OMG how the f* did they not work this one out? - just giving away trophies for the sake of it now
just being able to say its a selfl working trick doesnt count buddy. They have to add more info