Ah okay, it's a version of the same trick, but instead of using base systems to get the card in the same position as the audience's number, it's just always the eleventh card.
The order you pick the piles of cards up in may be the same, but you change the position of the pile containing the card you want relative to the other piles when picking them up. In the 27 card trick you break down the number given to you into powers of 3 (i.e. 10 is 1*3^2 + 0*3^1 + 1*3^0) which translates to where you need to put the pile containing the important card relative to the other piles when you pick them up (i.e. top/middle/bottom). This allows you to move any card to any position by dealing the cards 3 times as 3^3 is 27. The 49 card trick relies on a similar principle except it only requires the cards to be dealt twice as 7^2 is, as Matt said, 49. Matt Parker is brilliant.
I just spent about an hour(maybe less, I didn't time myself) figuring out how this could be done without removing any cards from the deck, ie: using a deck of *52 cards.* *Here's the solution I came up with:* Comparing with the video: the 27 card trick has 3 deals, and the cards are dealt into 3 piles each time. My solution for 52 cards has 3 deals of 4 piles each time. (there may be a simple solution where you deal out 13 and 4 piles but 1. 13 piles takes up a lot of space and 2. only has 4 in each pile so isn't as impressive) When the 'audience' picks the pile with their card in then that is the 'active pile' You then stack the piles back together to remake the pack, to encode a 'digit' in this recompiling stage you must put the 'active pile' in a position from the top, say you are encoding the digit '1' you put the 'active pile' at the top of the deck, for a '2' you place one other pile above it, two above it for '3' and put it at the bottom of the deck for a '4'. Now you just need to know the 'digit' sequences for each desired number the 'audience' could challenge you with. The 27 or 49 card version were simple for this, you would just input the number in base 3 or base 7 (although what I've called 1,2,3,4 would be 0,1,2,3 if you were doing that) For my version, to figure out the sequences, I drew a 'tree' for all 64 sequences you can do with 3 iterations of 4 digits, and annotated it with where a card that is picked by the 'audience' would end up. I figured out some pattern to make it easier to remember. (also included in comment is the full list in order, scroll down a bit) The pattern is: (lets call 'audience's number N for convenience) _3rd 'digit':_ if N in range *1-13:* 3rd digit is *1* if N in range *14-26:* 3rd digit is *2* if N in range *27-39:* 3rd digit is *3* if N in range *40-52:* 3rd digit is *4* _2nd 'digit'_ now minus 1, 14, 27 or 40 from N, lets call this O, if O in range *1-3:* 2nd digit is *1* if O in range *4-6:* 2nd digit is *2* if O in range *7-9:* 2nd digit is *3* if O in range *10-13:* 2nd digit is *4* _1st 'digit'_ now minus 1, 4, 7 or 10 from O, lets call this P, This one's kinda hard to explain linearly so I'll present it as a matrix. (there are also some number that have two sequences, you can do either and still get the same result, ie if in the matrix the element says '1or2' then you can have either 1 or 2 as your first 'digit') Get the element that is in (column='P' across from left to right, row='2nd digit' down from top) and use that as 1st digit. *1or2, 3, 4, -* *1, 2or3, 4, -* *1, 2, 3or4, -* *1, 2, 3, 4* There is another step to this. Half of the time the card's final position will be the top of the stack of N cards, the other half it will be the next card in your deck. You could try winging it with a 'second guess' but I think knowing to predict when this will happen will be more impressive. What you take into account is the first two 'digits' of the sequence and which pile is chosen as the 'active pile' the second time the 'audience' chooses. If the 'active pile' chosen is one of the 'problem piles', then the card will appear at the top of your deck rather than the top of N cards. I could try and explain the pattern, but it's easy enough to see, here: *first two digits:* problem piles *1.1;* none *1.2;* 1 *1.3;* 1 and 2 *1.4;* 1, 2 and 3 *2.1;* 1, 2 and 3 *2.2;* none *2.3;* 1 *2.4;* 1 and 2 *3.1;* 1 and 2 *3.2;* 1, 2 and 3 *3.3;* none *3.4;* 1 *4.1;* 1 *4.2;* 1 and 2 *4.3;* 1, 2 and 3 *4.4;* none As promised, the full list of sequences: *desired number:* sequence *1:* 1.1.1 or 2.1.1 *2:* 3.1.1 *3:* 4.1.1 *4:* 1.2.1 *5:* 2.2.1 or 3.2.1 *6:* 4.2.1 *7:* 1.3.1 *8:* 2.3.1 *9:* 3.3.1 or 4.3.1 *10:* 1.4.1 *11:* 2.4.1 *12:* 3.4.1 *13:* 4.4.1 *14:* 1.1.2 or 2.1.2 *15:* 3.1.2 *16:* 4.1.2 *17:* 1.2.2 *18:* 2.2.2 or 3.2.2 *19:* 4.2.2 *20:* 1.3.2 *21:* 2.3.2 *22:* 3.3.2 or 4.3.2 *23:* 1.4.2 *24:* 2.4.2 *25:* 3.4.2 *26:* 4.4.2 *27:* 1.1.3 or 2.1.3 *28:* 3.1.3 *29:* 4.1.3 *30:* 1.2.3 *31:* 2.2.3 or 3.2.3 *32:* 4.2.3 *33:* 1.3.3 *34:* 2.3.3 *35:* 3.3.3 or 4.3.3 *36:* 1.4.3 *37:* 2.4.3 *38:* 3.4.3 *39:* 4.4.3 *40:* 1.1.4 or 2.1.4 *41:* 3.1.4 *42:* 4.1.4 *43:* 1.2.4 *44:* 2.2.4 or 3.2.4 *45:* 4.2.4 *46:* 1.3.4 *47:* 2.3.4 *48:* 3.3.4 or 4.3.4 *49:* 1.4.4 *50:* 2.4.4 *51:* 3.4.4 *52:* 4.4.4
You're looking too deeply into this (but your solution is great). The shuffling and asking for the pile in this video is all unnecessary. If you pile shuffle a deck any number of times where the number of piles used during each shuffle (say we shuffle 3 times, a b and c) multiply to equal the number of cards in the deck (D), we get the equation abc=D If you shuffle twice ab=D So on and so forth. As long the number of piles in each shuffle are factors of the number of cards in the deck and they all multiply to equal the number of cards in the deck (this video, 3x3x3=27, 27 card deck), you can either reverse the deck or keep it in the same order. If you use an even number of piles, the deck is reversed. An odd number, the deck is kept the same
NoBudgetDuelist What do you mean unnecessary? how does reversing the deck or keeping it the same put their desired card into their desired position? Also regarding factors. I could have done it the same way as the video, but 52 has the factors 2^2*13. Which means i'd need to deal out the deck into 13 piles (or 26) and at that point each pile only has 4 (or 2) cards in it so it's not much of a trick(almost on par with dealing into 52 separate piles of 1 and asking what pile it's in, it's ridiculous). Also, who has space to deal out 13 piles of cards (or 26)? Since I needed to avoid this I decided I'd deal into 4 each time. (which also meant working out all the weird stuff about where all the cards go for each combination since 52 is not a power of 4)
JNCressey You're right. I realized WELL after I wrote this comment that these magic tricks and what I was thinking were completely different. That's a mistake on my part. The work you did is great.
I think I'd just go for this: OK. Think of a card, but don't tell me what it is. This trick's a bit difficult, so you gotta give me four guesses. Is THIS your card? (Show the first card of the deck.) - repeat for your next two "guesses." If any of those work, it's a miracle, you did it, move on. If not, when you've done three cards say "Oh, OK. This is my last guess, and it doesn't seem to be going that well, so I'm going to try something a bit different…" and then do the 49-card trick.
im pretty sure the 27 card trick is on numberphile, and if the 49 card trick works in the same way then that must mean hes counting in base 7, awesome!
You want the ace of spades to be the 13th card. Hence you need 12 cards on top of it at the end of the trick. By converting 12 to base 7 (7 piles of cards during the trick) you get 15. After dealing out the cards for first time, when you pick them back up; you place the pile, with the AoS in it, 6 piles from the top (5 piles on top of it). On dealing them into piles again you do the same thing, but make sure the AoS pile is second from the top when you pick them up again (1 pile on top of it). So placing the AoS pile in a certain depth in the deck (correspondent to first: the units of your base 7 number, 15 in this case; and then the '7s' in the second pick up) you can position the AoS anywhere you want in the 49 card deck. Thus the card is buried 12 cards deep in the pile, so it will be the 13th card at the end of the trick.
This works for any power a^b. Personally I prefer the 1 card trick, it's simple and you don't need to worry about tedious card placing and base conversion. However not everyone is impressed when I show it.
with the first deal you narrow the choice to 7 cards. since you arrange the ordering of how the piles are stacked, when you deal the second time you can pinpoint the exact card (if its in the second pile you put on top after the first deal, then the card will come up in the second row in the second deal). since you know the exact position you need to put the card in (the number from calculator) you just calculate modulo 7 of it and after the first deal just put the choosen pile as the n'th pile, according to the number you got from modulo function. then, depending on what the original number from the calculator was, you just add required number of packs of 7 on top of it. Voila :D
This trick has to do with a property of card shuffling. Basically, if you have, say, 40 cards and you pile shuffle them (take the cards one by one and place them in piles in a pattern) multiple times where the number of piles you use multiply to the number of cards in the deck (the number of piles in each instance of pile shuffling is a factor of the number of cards in the deck), you will either reverse the order of the deck (if you pile shuffled an even number of times) or you will keep it the same (if you pile shuffle it an odd number of times). Knowing this property allows one to shuffle a deck in a semi-manipulated way. If you know the order of the cards, more or less, you can change the order by NOT using this method or you can keep it the same by using this method.
I've been impressing family members and friends who reluctantly sit through the 27 card trick when I do it for them (people just don't appreciate magic - or math I suppose), and I just recently figured out how to do it quickly. Now I have to start practicing the 49 card trick, but it may take me a while to get it right. Thanks for this video Matt. Love your stuff. More numberphile calculator unboxings! Those are hilarious.
Matt, instead of using a giving number from 1 to 49, you could use the sum of the values of 3 unused cards. Of course the range is reduced to 3 to 39, but no one van suspect the number is made up to help you somehow.
@@L4Vo5 You could make the person pick four cards, then show you three of them and put the other back in the deck. Then at the end you count using the three you were shown
Theoretically, this same trick can work with any number of cards that is a number raised to the power of another number. Since 27=3^3, the trick deals out 3 decks 3 times. This same trick can be used with 49 cards. 49=7^2, which means to deal out 7 decks 2 times. You could even use 32. 32=2^5. Deal out 2 decks 5 times. I just felt that people can use this concept to change up the trick to have a little bit more fun
It's not just limited to powers. If the numbers of piles you use each time you shuffle are factors that multiply to equal the number of cards in a deck (52 card deck, 4 piles x 13 piles) and you shuffle the same way both times, you can reverse the order of the deck. If you pile shuffle the deck an odd number of times (52, 2 x 2 x 13) you will maintain the order of the deck. It's used to stack decks in competitive card games (sort of, it's almost impossible to completely stack a deck since both players shuffle each other's decks).
L4Vo5 I never said it was. Also, you're replying to a comment almost 2 years later. Woo. I also don't intend to discredit the original content of this video or claim that I know more, I just happen to know something similar to what was addressed in the video which was not addressed in the video. So... I guess it doesn't matter?
"I can't believe that worked" Of course you can ! Maths does work. ;) The number (less than 52) I have the most trouble with is 32 : too hard to know in advance what the chosen card when you deal 5 times !
I doubt you'll read this, but I've made a variation of this trick where you show them the cards and place them face down without ever looking at them. it is a bit slower however
I know this is an oooooold video - but doesn't this basically work out to mapping the location of the selected card in 3 dimensions in the 27 card trick (because it's a cubed number) and 2 dimensions in the 49 card trick (because it's a squared number) and then selectively placing each pile on the top or bottom of the deck as you collect them, to manipulate the position of the card in dealt order? I've probably explained that very poorly... but it makes sense in my own head. And when I use words, they mean precisely what I intend them to mean.
+Tzisorey Tigerwuf they explained in on numberphile at one point I think ( I also bug mathematicians so I learned about it from them you can generalize the idea it's basically about bases I think.
13=6*(7^0)+1*(7^1) the first digit (1) => 6 => the first draft, the pile where there's your card, you put it last the second digit (7) => 1 => the second draft, the pile where there's your card, you put it second. (0 is first 1 is second and so on) I guess that's the logic behind it right? It's funny, but I find the 49 trick easier to learn than 27 since 49 in only square, and 27 is cube...
+Zork you can generalize it I only know because I bug mathematicians with n^x cards you can use n piles and x dealings to put it anywhere I think is the key if I remember correctly they did a numberphile video on the 27 card version.
i still don't get it why the card is on the 10th position (27 card trick) wich was choosen randomly.... what if the calculator gave him the number 13 or 25 before? what did he had to do then? (i cant understand alasdair lowes's explanation...)
+DesmondAltairEzio thanks, now i understand, ha. Very easy. Also you can change the choosen number to +- 3/6/9/12... After the first and +- 9/18 after the second round for a special effect ;)
The trick deals in base 3. Whatever number is chosen, you convert to base three minus one. Let 0 = top, 1 = middle, and 2 = bottom. The first time you pick up the piles represents the 1s place, so put the pile with the selected card in it at the proper position. The second time represents the 3s place, and the last time represents the 9s place. After the third collecting of cards it should be in the right place. As for the 49 trick, it is in base 7, which means it is the same exact thing as the 27 trick, except you convert the chosen number to base 7, and only collect the cards twice since there are only 7^2 cards.
