# Mind reading

*Mind reading* is a puzzle or a game in which the "mathe-magician" is able to figure out the correct answer, based on a small set of information that you provide. By tracing out the mathematical/logical steps involved, we can figure out the underlying trick.

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## Definition

Magicians read minds in the following ways:

- Clever mathematics (like very long math operation that gives a predictable result)
- Common psychological patterns (like using most common associations and trends when guessing what you think)
- Reading expectations from a human body (most react slightly seeing familiar things, hearing emotionally loaded words or things they want to hear)
- Faking or guessing

But in this section we will be dealing with the first way, which is used in famous card tricks and many more. "Mathe-maticians," as the name suggests, rely on the utter immutability and predictability, i.e. mathematics. Unlike other methods, this method gives a \(100\)% guarantee in its outcome.

Most types of tricks can be done with a little knowledge of high school algebra and a quick manipulation of basic arithmetic. In designing a problem like the one below, no matter what the original number is, you can make it return to some unchanging number. Consider the following example:

- Think of a number, any positive integer.
- Square it.
- Add the result to your original number.
- Divide by your original number.
- Add, oh, how about \(17\).
- Subtract your original number.
- Divide by \(6\).

The number you are thinking of now is \(3,\) right? Whoooah!!!

To understand how this trick really works, let the number you pick first be \(x\) and make it go through the above procedure:

\[\begin{array} &x &\Rightarrow & { x }^{ 2 }&\Rightarrow & x+{ x }^{ 2 }&\Rightarrow & \dfrac { x+{ x }^{ 2 } }{ x } =1+x&\Rightarrow & 18+x&\Rightarrow & 18&\Rightarrow & \dfrac { 18 }{ 6 } =3.\end{array}\]

Now we can clearly see what is happening. The trick is to obscure the variable \(x\) with a series of arithmetic such that the variable \(x\) vanishes. The same goes for card tricks, where the variable is the card you choose and the shuffle is equivalent to the binary operation we performed.

Also, don't forget to sell it, but do it like any magician would do!!

## Tricks

In Las Vegas, a street artist said that he could read your mind. He asked you to think of a certain two digit number, multiply it by \(25,\) add \(17\) to it, and then multiply it by \(4.\) As you rattle off your result, he is able to confidently state the number that you were thinking of. How did he do this?

Let's say that the ordinary random citizen, clueless of maths (you?), thought of a number \(\overline{ab}\) with \(a\neq 0\) and \(a,b\in\{0,1,2,\ldots,9\}\). Now, we proceed step-by-step to demonstrate you how he blew your (clueless?) mind. First of all, the number you chose in your mind can be written as

\[\textbf{Your number}=\overline{ab}=10a+b.\]

Let \(s\) be the number that you rattle off as your result to the mathe-magician. Then you can see how \(s\) is related to your number as follows:

\[\begin{align} 25\times \overline{ab} =&25(10a+b) &&(\text{multiply by }25)\\ =&250a+25b\\ \\ \Rightarrow \quad &250a+25b+17 &&(\text{add }17)\\ \\ \Rightarrow \quad &\left(25\times \overline{ab}+17\right)\times 4 &&(\text{multiply by }4)\\ =&(250a+25b+17)\times 4\\ =&1000a+100b+68\\ =&1000a+100b+60+8\\ =&\overline{ab68}. &&(\text{your final result}) \end{align}\]

So, if you had thought of \(69\), your result would be \(6968\). If you had thought of \(34\), your result would be \(3468\). See the pattern?

The first two digits from the left of the result you rattle off to the mathe-magician is the number you initially thought of! So, he isn't reading your mind. You're telling him what you thought of indirectly. Hence, he has "read" your mind. \(_\square\)

\[\textbf{"MIND=BLOWN"}\]

## \(27\) cards trick:

- You pick a card from a deck of \(52\) cards, remember it, and put it back in.
- The stack is split into 3 piles, and you say which pile the card appears in. Piles are stacked up.
- The stack is split into 3 piles, and you say which pile the card appears in. Piles are stacked up.
- The stack is split into 3 piles, and you say which pile the card appears in. Piles are stacked up.
- The stack is split into 3 piles, and you say which pile the card appears in. Piles are stacked up.
## Then, your card magically appears at the top of the stack. How is this done?

After you pick your card, remember it, and put it back in the deck of \(52\) cards, I shuffle the deck thoroughly.

Then I deal out the \(52\) cards

face down into \(3\) piles just as a casino dealer would do to \(3\) players in front of him/her. Since \(52\div 3=17\) with remainder \(1,\) each pile will have \(17-17-18\) cards in the end. Then I show you each pile face up, asking if your card is in there. Upon hearing "yes", I grab that pile face down, and then stack up the other two piles on top of it also face down. (Remember your card is now one of the \(18\) cards at the bottom of the \(52\) card deck with a \(100\)% chance.)one by oneThen just as I did above, I deal out the \(52\) cards one by one face down into \(3\) piles again. Once the deal is done, I again show you each pile and ask if your card is in there. This time, upon hearing "yes", I grab the other two piles first and then put that pile containing your card at the top, of course, everything face down. (Your card is now one of the top \(6\) cards. Agreed?)

For the third time I deal the \(52\) cards again, and upon hearing from you "Yes, my card is in here.", I grab that pile and then stack up the other two piles on top of it just as in the first deal. (Where is your card now? It is one of the bottom two cards, right?)

For the last time, I deal the \(52\) cards again and when the deal is done, one of the two piles where the last two of the \(52\) cards went will have your card, at the top. Even though I know this, I pretend not to know it and ask you again which pile has your card. As soon as you pick the pile for me, I grab the other two piles first and then put that pile with your card on top of them, everything face down. Then I say to you, "Pick the top card and see if it is your card." \(_\square\)

## Birthday Card Trick:

The below 5 cards can be used to find the Birthday of friends/relatives. Give the five cards to Friends/Relatives and ask them to tell which of the cards have their birthday. Add the smallest number on the top of the respective cards will give their Birthday.

It is also known as

Binary Magic Card trick. It worked based on the binary number to decimal number conversion. The color code for red=1 which is equal to \({2}^{0}\), yellow=2 which is equal to \({2}^{1}\), green=4 which is equal to \({2}^{2}\), blue =8 which is equal to \({2}^{3}\) and purple=16 which is equal to \({2}^{4}\). We need to read the presence of the color in the following order purple->blue->green->yellow->red if the number exist place ‘1’ and if not place ‘0’ we get the 5 bit binary representation. For example, if the birthday exists in purple, blue and red cards it can be represented as \({11001}_{2}\). Now convert the binary number to its equivalent decimal as follows:\({11001}_{2}\) = (1 x \({2}^{4}\)) + (1 x \({2}^{3}\)) + (0 x \({2}^{2}\)) + (0 x \({2}^{1}\)) + (1 x \({2}^{0}\)) = 16 + 8 + 0 + 0 + 1 = 16 + 8 + 1 = purple + blue + red = \({25}_{10}\) = 25

The below table shows how the number from 1 to 31 are derived using presence of the respective color codes.

Color Code and Birthday:

A magician (read mathematician) selects five spectators by some random process (say throwing five balls blindfolded). Then he gives the first spectator a deck of cards and asks him/her to cut the deck and take a card from the top. Then the second one is asked to pick up the next card, the third one the next, the fourth one the next, and the last one the next card from top. The five spectators are asked not to show their cards to the magician but to concentrate on their own cards. The magician waves his hands a bit and then asks them a few questions including who hold red cards. He again waves his hands a bit, closes his eyes, and then correctly declares the names of the cards that each spectator is holding, and in correct order. How does he do this?

Now we shall see how the trick works. Clearly, the key to the trick lies in the question on people having red cards. Further, notice that 5 cards have been distributed, which means 5 bits of information, each bit being either "Red"(1), or "Black"(0). Knowing the "Red" (or "Black") bits, one can completely frame the 5-bit sequence of information. Clearly, there are \({ 2 }^{ 5 }=32\) such possible sequences, which indicates that the deck has 32 cards instead of 52.

Also, notice that if these 32 cards are given an order (i.e. a permutation), the order remains invariant under the operation of cutting the deck (note that the deck is being cut, not shuffled). So, the cutting does nothing to the order in which the magician has arranged the deck.

Finally observe that the magician tells the 5 cards only knowing their color sequence. This indicates something sneaky. Replace the 32 cards by a 32-bit sequence of information in 1's (Reds) and 0's (Zero's). So, the above feat can only be accomplished if every 5 consecutive bits of this sequence is a unique sub-sequence. Suppose the 5-bit sub-sequence is 11010. Knowing this sub-sequence, the magician can tell what the cards are, since he knows that the sub-sequence occurs only once in the 32-bit sequence. And that shows how the trick works. \(_\square\)