Loved the unexpected shout out! Wife was going to give you the explanation, but someone beat her to it. Did give you a tip to visually see it though.. some students need to look at math differently. Like base ten blocks or other manipulatives to assist with their learning. Your manipulative is the other colored deck. Hope that helps and thanks for sharing and teaching Sean👍🤓👍
Looked to like a variation on the Automatic Placement Principle. I learned from a Bannon book Destination Zero but I think it may go back to Marlo. I also found this principle in Card College Lighter or Lightest.
It is a very nice card trick Sean! Love it. It is mathematical but not really all that tricky. Let me explain what I grasped from the video. It hardly matters whatever number the spectator cuts. What matters is ...how many cards you would deal. Example:1 Let us assume that the spectator cuts 7 cards and you dealt say 15 cards (spectator selects the 7th card among the 15 you dealt.). Proceed with rest of the trick as demonstred in the video by Sean. Now deal (15+1) cards into 3 or 4 piles. Let the last pile be of 3 or 4 cards. --- Example 2: Let us assume now that the spectator cuts 13 cards and you dealt say 16 cards (spectator selects the 13th card among the 16 you dealt.). Proceed with rest of the trick as demonstred in the video by Sean. Now deal (16+1) cards into 3 or 4 piles. Let the last pile be of 3 or 4 cards. ---- The only necessary condition in this trick is that number of cards you deal should always be more than what the spectator cuts initially. Let us say spectator cuts '☓' cards. You deal 'Y' cards. Y has to be always greater than X. Finally, when magician makes four piles number cards dealt are Y+1. I hope ...I could explain well.
You're an unending source of inspiration for my card effects! Continued success. P.S. What happens if the person cuts more than 12 cards in the beginning?
Hey Sean :) Great video similiar trick (using the same principle) is in Harry Loraynes - The Magic book - called Its impossible .... You can go more then 12 card - the selection always ends the number of cards you show +1 .... so if they cut 15 cards and you show 15 cards their card will be 16th :) and again great video :)
So clever! I love mathematical principles like this. If they chose 9 cards. You show 12. They remembered #9 but you keep dealing to 12 putting 3 cards on top of it. When you add the 9 originally counted cards it makes it 13.
Say they choose 8 cards. You then count down to the 8th card which is theirs. then 9, 10, 11, 12 (4 cards to 12), but now they are reversed with their card the 5th one down. Take the original 8, place it on the last 4 cards and theirs will be the 13th. Note that this works for any number. Say the limit is not 12 but n cards but they take k cards. (n - k + 1) + k = n + 1. It will always be n + 1 down assuming they didn't take more than n.
If i understood correctly, you don't have to have all 52 cards in the deck, I wrong? And another question, if I always stop at 12, does that mean they have to return cards to the pile if they cut more than 12?
Hi Sean. The basic mathematical idea here is similar to the premise upon which Dai Vernon’s original version on “Triumph,” first published in STARS OF MAGIC, is based. I like your trick but I think the Triumph plot and ending are superior. Just my opinion.
Let say they get x cards. When you show them the cards, they get reversed so their card ends up at the 13-xth position from the top of the pile. Then your replace their cards on top, so their card is a the position 13-x+x=13
My wife agrees. She also said Sean, for you demonstration purpose. Once you cut the cards for the spectator, change the deck visually. Grab any color deck other than that one. Count off the same amount of cards and continue as before. You'll be able to better visually see what happens. 👍🤓👍
@@valovanonym shoot.. I hadn't thought of it as a trick. For it will be very visual as to what is happening. Just worked it next to my wife so I could see what happens visually. Maybe a do as I do routine and have the decks set up. I wonder what kind of shuffle can be done though.. that a participant could do. Maybe.. instead of going back into the deck the piles are created.. then switch piles across magician forcing the necessary ones and continue the trick leaving the same card different colors as the reveal at the opposite sides.. 🤔 Sean.. you have some ideas for this?
Loved the unexpected shout out! Wife was going to give you the explanation, but someone beat her to it. Did give you a tip to visually see it though.. some students need to look at math differently. Like base ten blocks or other manipulatives to assist with their learning. Your manipulative is the other colored deck. Hope that helps and thanks for sharing and teaching Sean👍🤓👍
I love this. Thank you so much.👍
Sean, your tricks are always the best of the best and your tutorials never miss a thing thanks for being such a wonderful inspiration.
Thank you sean. You are by far my fave magic teacher!
Looked to like a variation on the Automatic Placement Principle. I learned from a Bannon book Destination Zero but I think it may go back to Marlo. I also found this principle in Card College Lighter or Lightest.
It is a very nice card trick Sean! Love it.
It is mathematical but not really all that tricky. Let me explain what I grasped from the video.
It hardly matters whatever number the spectator cuts. What matters is ...how many cards you would deal.
Example:1 Let us assume that the spectator cuts 7 cards and you dealt say 15 cards (spectator selects the 7th card among the 15 you dealt.). Proceed with rest of the trick as demonstred in the video by Sean. Now deal (15+1) cards into 3 or 4 piles. Let the last pile be of 3 or 4 cards.
---
Example 2:
Let us assume now that the spectator cuts 13 cards and you dealt say 16 cards (spectator selects the 13th card among the 16 you dealt.). Proceed with rest of the trick as demonstred in the video by Sean. Now deal (16+1) cards into 3 or 4 piles. Let the last pile be of 3 or 4 cards.
----
The only necessary condition in this trick is that number of cards you deal should always be more than what the spectator cuts initially. Let us say spectator cuts '☓' cards. You deal 'Y' cards. Y has to be always greater than X.
Finally, when magician makes four piles number cards dealt are Y+1.
I hope ...I could explain well.
Great trick Sean, thanks for sharing. 👍👍
This is so cool Sean! Your math trcks are always so good! I am going to practice this a lot and then give it a go!
Congrats on 9K!
#Roadto9K!
Fantastic job thanks 😊👌
Nice and easy,great Sean
Thanks so much. That was a super cool trick.🤪🔥🔥🔥🔥🔥🔥
You're an unending source of inspiration for my card effects! Continued success. P.S. What happens if the person cuts more than 12 cards in the beginning?
I’d simply say if you have more than 12 add the digits to get a more random number. As long as it’s less than 20 it will be 1-10.
Man so good😍😍😍
Hey Sean :) Great video similiar trick (using the same principle) is in Harry Loraynes - The Magic book - called Its impossible .... You can go more then 12 card - the selection always ends the number of cards you show +1 .... so if they cut 15 cards and you show 15 cards their card will be 16th :) and again great video :)
Oh, that's fantastic! Thanks for the heads up!!
So clever! I love mathematical principles like this.
If they chose 9 cards. You show 12. They remembered #9 but you keep dealing to 12 putting 3 cards on top of it.
When you add the 9 originally counted cards it makes it 13.
Awesome trick 🤜🤛👍😊👍😊
Now you know, why its call mind- boggler. 'Do you gets it.'👀
Say they choose 8 cards. You then count down to the 8th card which is theirs. then 9, 10, 11, 12 (4 cards to 12), but now they are reversed with their card the 5th one down. Take the original 8, place it on the last 4 cards and theirs will be the 13th. Note that this works for any number. Say the limit is not 12 but n cards but they take k cards. (n - k + 1) + k = n + 1. It will always be n + 1 down assuming they didn't take more than n.
The math is quite simple, you are puting always exactly 12 cards on top of their cards. If they pick only 1 card you will put 12 cards in top.
If i understood correctly, you don't have to have all 52 cards in the deck, I wrong?
And another question, if I always stop at 12, does that mean they have to return cards to the pile if they cut more than 12?
Hi Sean. The basic mathematical idea here is similar to the premise upon which Dai Vernon’s original version on “Triumph,” first published in STARS OF MAGIC, is based. I like your trick but I think the Triumph plot and ending are superior. Just my opinion.
Let say they get x cards. When you show them the cards, they get reversed so their card ends up at the 13-xth position from the top of the pile. Then your replace their cards on top, so their card is a the position 13-x+x=13
My wife agrees. She also said Sean, for you demonstration purpose. Once you cut the cards for the spectator, change the deck visually. Grab any color deck other than that one. Count off the same amount of cards and continue as before. You'll be able to better visually see what happens. 👍🤓👍
@@davidmurphy9049 that's a very good idea! I don't know if it would make sense but I will try to apply it and see how my audience react to that change
@@valovanonym shoot.. I hadn't thought of it as a trick. For it will be very visual as to what is happening. Just worked it next to my wife so I could see what happens visually.
Maybe a do as I do routine and have the decks set up. I wonder what kind of shuffle can be done though.. that a participant could do.
Maybe.. instead of going back into the deck the piles are created.. then switch piles across magician forcing the necessary ones and continue the trick leaving the same card different colors as the reveal at the opposite sides.. 🤔 Sean.. you have some ideas for this?
Mindblown. An automatic placement do as I do method. I'm going to work on this!
🙏🏻🤟
Thanks
Your biggest fan, your very first comment.
I must have a go😂
Prittttteee cool