# Tag Info

36

I'm going to assume Alice looked at the cards and chose which one to give back to you. The key to the puzzle is then to encode a single card's suit and value in 4 cards without the luxury of choosing those 4 cards arbitrarily from the whole deck. The suit is easy. In 5 cards there must be a double of at least one suit. So the first (or last, but I'll choose ...

26

The magic trick is as follows. Source. Explanation:

25

Do these series work? If so, I know the answer. Example 1 "Am I thinking about a raisin?" "Nope" "Am I thinking about an ostrich?" "Nope" "Am I thinking about a finger?" "Nope" "Am I thinking about a moose?" "Nope" "Am I thinking about an astronaut? "Yes" Example 2 "Am I thinking about a snake?" "Nope" "Am I thinking about the Eiffel ...

25

Claim: If David Copperfield distributes $n\ge3$ cards with numbers $1,2,\ldots,n$ over the three top hats, then only the following two types of distributions make the trick work: Put $1,2,3$ into different hats; put every $k\ge4$ into the same hat as its residue modulo $3$. Put $1$ into the first hat; put $2,\ldots,n-1$ into the second hat; put $n$ into the ...

17

One way to do this If sum is $53$, it is only possible that one card is drawn from the red hat and one from the blue If sum $> 53$, the blue hat card and a yellow hat card has been drawn Else, a card from the red hat and one from the yellow hat has been drawn

17

If cards are not prearranged then it is impossible to do this trick. It is so simple arrangement that you wouldn't notice. Because Lan got your card by This way

16

Here is my answer: Key Item 1: Key Item 2: Key Item 3: Key Item 4: Key Item 5:

16

Presto! Changeo! Abracadabra! 3, 9, 1, 8, j, k, l, m, n and x are all different and none are 0.

13

Firstly arrange the dice into a 3x3 matrix. If the sum of all dice is even, choose the column of first cup according to the total sum mod 3, and the row according to the sum of its own column. Let's say the first row/column means mod 3 = 0 and the second means mod 3 = 1. If the first cup is O (its own column denoted by o), choose the second cup between A, B ...

11

This is a great card trick known as Fitch Cheney's Five Card trick. Alice selects the fifth card to be identified and arranges the remaining four cards in a specific sequence. Then Bob interprets the four card sequence and identifies the fifth card. Here are a few references that describe how to sequence the four cards: Batman on wordpress Gray Matters ...

10

My younger brother likes magic and so I recognize the boat. The boat + the tokens is already a set, everything else is from another trick. The trick is that you magically place the same sequence of images as a volunteer. There are two tokens of each image and there is a trick token which is black on both sides and looks like the bottom of the holes. I ...

10

He can put all cards 0 mod 3 in the red hat, all 1 mod 3 in the blue hat and all 2 mod 3 in the yellow hat. If the sum is 0 mod 3, it can only be obtained from 1+2 mod 3, so the unused hat is red. The same goes for 1 mod 3, it can be obtained from 0+1 mod 3, so the hat is yellow. For 2 mod 3 we have 0+2 mod 3, so the hat is blue.

9

"Egyption Prediction" YouTube video "Egyptian Prediction" & other magic tricks (How-To) PDF

9

Stealing from 2012rcampion's answer, we need to create a one-to-one mapping from five-card hands to four-card messages (four cards plus permutation). By symmetry, each four-card message can be mapped to an equal number of five-card hands, so by the regular version of Hall's Theorem there must exist such a one-to-one mapping (this is laid out in more detail ...

8

I'd ask for your mother, a T-Rex, Michael Jackson, a chair and a rainbow (which is the answer). Let me ask for a beach, a skyscraper, a castle, a park bench, world's fastest computer and a smartphone

8

She knows your card because The key moment is at

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My answer: For those who would like a clearer example, here is a tutorial:

7

A slightly different way to think about this problem is that the magician is trying to guess which five-card hand was initially drawn from the deck. Since he can see the other four cards, this is the same as guessing the volunteer's card. There are $\binom{N}{5}$ possible five-card hands, and $4!\binom{N}{4}$ four-card "messages" (including orderings) that ...

7

I think this is impossible, but I hope someone points out an error in my reasoning. There are $\binom{52}{3}\cdot 3=52\cdot 51\cdot 25$ situations the assistant can be handed. However, there are only $52\cdot 51\cdot 3$ possible arrangements that the magician can be handed by the assistant. This is because what he sees is completely specified by the two ...

6

From the point of view of Charlie (the one who holds the card to be identified and verifies Bob's answer), a trick when Charlie himself pulls the card from Alice's hands will look more showy. To make this possible-

6

There are more than one way to do this. For example; the number of hats mod which requires putting all the same numbers in "the number of hats" mod. For this one it is mod 3. (explained by @BianB BB's solution) Putting the smallest number in one hat, largest number in one hat, and the rest is the last hat. 1 is the smallest, 52 is the biggest. You cannot ...

5

In the first example, I notice that in the second example, However, this seems to fall apart in the third example: I don't know about the fourth one, but in the fifth (update) example:

5

The final permutation done by the volunteer is σ α = σ₁₀ σ₉ σ₈ σ₇ σ₆ σ₅ σ₄ σ₃ σ₂ σ₁ β = σ₂₀ σ₁₉ σ₁₈ σ₁₇ σ₁₆ σ₁₅ σ₁₄ σ₁₃ σ₁₂ σ₁₁ σ = β σₛ α where σᵢ are transpositions and σₛ is the silent transposition. Then, the magician can Each time the volunteer makes a "loud" transposition, the magician Finally, he gets Then, he checks In order to achieve this, ...

5

Based on Jaap's answer (which seems to be addressing a slightly different problem): (EDIT:)

5

4

Wanted to point out that the classic solution is just the most "human" way of doing this and while we humans tend to appreciate things that are easy for us, the classic technique is wasteful and there is another generic solution that came to me while observing the problem. Of course this is harder to do cause it is not human friendly but it is much more ...

3

Partial answer - only satisfies 1, 2 and 3 and assumes that the magician is permitted the luxury of using a pack of cards that has an assymetric back. I doubt that the audience would be suspicious if this was a otherwise plain back blatantly advertising the name of the show, the magician or the venue. It's not perfect, but will certainly work once and, for ...

3

Maybe it is not a A simple example: "Am I thinking of a unicorn?" "Nope" "Am I thinking of an angel?" "Nope" "Am I thinking of a pen?" "Nope" "Am I thinking of (whatever you want)?" "Yes" Another try: "Am I thinking of a flying unicorn?" "Nope" "Am I ...

3

Ask a volunteer/victim to pick a number between 1 and 63. Then they tell you the letter of any papers on which they see their number. From that, you tell them their number almost immediately. How it works Boom!

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