J. Software Engineering & Applications, 2010, 3: 240-244
doi:10.4236/jsea.2010.33029 Published Online March 2010 (http://www.SciRP.org/journal/jsea)
Quantum Number Tricks
Takashi Mihara
Department of Information Sciences and Arts, Toyo University, Kawagoe, Japan.
Received November 23rd, 2009; revised December 21st, 2009; accepted December 29th, 2009.
ABSTRACT
Some results indicate that quantum information based on quantum physics is more powerful than classical one. In this
paper, we propose new tricks based on quantum physics. Our tricks are methods inspired by the strategies of quantum
game theory. In these tricks, magicians have the ability of quantum physics, but spectators have only classical one. We
propose quantum tricks such that, by manipulating quantum coins and quantum cards, magicians guess spectators’
values.
Keywords: Quantum Trick, Entangled State, Game Theory
1. Introduction
The studies on quantum information have succeeded in
such as quantum computation, quantum cryptography,
quantum communication complexity, and so on. For
example, Shor’s quantum factoring algorithm is one of
representative results in these fields [1]. In addition,
quantum game theory has been also proposed and it has
been shown that quantum game theory is more powerful
than classical one.
In 1998, for a coin flipping game, Meyer proposed a
quantum strategy for the first time and showed that the
quantum strategy has an advantage over classical ones [2].
Moreover, he also showed the importance of a relation-
ship between quantum game theory and quantum algo-
rithms.
After that, other types of quantum strategies have been
also proposed. For example, Eisert et al. proposed a
quantum strategy with entangled states for a famous
two-player game called the Prisoner’s Dilemma [3] (also
see Du et al. [4,5], Eisert and Wilkens [6], and Iqbal and
Toor [7]). For another famous two-player game called the
Battle of the Sexes, Marinatto et al. also proposed a
quantum strategy with entangled states [8]. For these
games, they showed quantum Nash equilibriums different
from classical ones.
In this paper, we propose quantum tricks based on
methods inspired by the strategies of quantum game the-
ory. Magicians have the ability of quantum physics, but
spectators have only classical one. By manipulating
quantum coins and quantum cards, magicians guess
spectators’ values. For example, we propose tricks such
that by using entangled states, a magician transmits a
spectator’s value to another magician without communi-
cating between them.
The remainder of this paper has the following organi-
zation. In Section 2, we define notations and basic opera-
tions used in this paper. In Section 3, we propose quantum
coin tricks. In Section 4, we propose quantum card tricks.
Finally, in Section 5, we provide some concluding re-
marks.
2. Preliminaries
First, we denote some basic notations. Let


01

B,


0 1 ...1
nn

 Z
n a
b
ab
, and for a positive
integer . Let and be integers. We say that is
congruent to to modulus if is a divisor of

1 2 ...1
nn
Z
b
n n

a

and denote by (mod )ab n

, and we denote an
inner product modulo 2 of and b by . Finally,
let
aab

be an exclusive-OR operator, e.g., (1 100)

 
(1 0 1 0)(011 0)

 .
Next, we define some basic quantum notations. As
states of qubit, let and , where 0(10)
T
1(01)
T


 is Dirac notation and AT is the transposed matrix of
matrix A. Throughout this paper, we take


01
q


n
B
12
…bbb
as a computational basis and a measurement basis.
Moreover, we denote an -qubit basis state by n
12
bbb
12 nn
bb b

, where

is a tensor product and iq b


N
B1, 2,...,(). In
addition, we denote a basis in an -dimensional system,
in

Quantum Number Tricks241
a basis of qudit states, by , where
() is an integer. We call

q
NN
xxZZ
N2
x
 a quantum register.
Finally, we define some unitary matrices used for
quantum tricks in this paper. Let
I
be the 22 identity
matrix. This operation means no operation. A
Walsh-Hadamard operation
H
is
11
1
11
2
H 


(0(12)(01)H
and
1(12)(01)H ).
Note that H=H-1. This operation is used when we make
a superposition of states. An operation used when a coin is
flipped is X,
01
10
X 

( and ).
01X 10X
Moreover, we define an operation between two qubits.
A Controlled Not gate, CNOT, is
1000
0100
0001
0010
CNOT






(, where the first bit c is the
controlled bit and the second bit is the target bit). We
denote the operation by CNOT(ij) when the i-th bit is the
controlled bit and the
CNOTctc tc
t
j
-th bit is the target bit.
Entangled states can be made by using H and .
For example,
CNOT
1
00(01)0
2
1(001 1)
2
N
H
COT


 

Finally, we define two matrices for -state transition.
Let . A quantum Fourier transform [1], QFT, is
N
N
xZ
1
2
0
1
|
N
xy N
y
QFT xey
N



 

and
1
12
0
1N
xy N
y
QFT yex
N






3. Quantum Coin Tricks
In this section, we show some quantum coin tricks using
quantum states. Throughout this paper, we use Alice and
Bob as names of magicians, and use Carol and Davis as
names of spectators participating in a magic show.
Moreover, Alice and Bob can cooperate but cannot
communicate with each other during each show.
First, we show a simple coin trick using a two-qubit
entangled state.
Coincidence: First, Alice prepares one coin and put it
in a box. The box is a container such that no one cannot
see the state of the coin but can operate it. Next, Carol
flips the coin or not. Then, Bob guesses the state of the
coin, i.e., either head (H) or tail (T).
Method of Coincidence
We denote H and T by 0

 and , respectively. 1
1) Beforehand, Alice and Bob share an entangled state
1(001 1)
2

 
where Alice has the first qubit and Bob has the second
qubit. Alice’s qubit is in a box.
2) Carol flips Alice’s coin or not. This means that Carol
applies
X
to Alice’s qubit if she wants to flip the coin;
otherwise she applies
I
to it. Then, if she flips it, the
state becomes
1(1 00 1)
2

  
3) Alice and Bob apply
H
to the state. Then it be-
comes
1(0011)
2

 
if Carol flipped the coin; otherwise the state does not
change, i.e.,
1(001 1)
2


4) Bob measures his qubit and announces the value(H
or T) to Carol.
5) Carol opens the box and confirms that her value is
same as Bob’s value.
This trick can be easily extended to multiple coins by
preparing the entangled states (12) (0011)


corresponding to the number of coins.
Next, we show a trick guessing the number of Carol
flipping coins.
Flip-Flop1: First, Alice prepares coins in all the
coins being head. Next, Carol flips some coins such that
the state of coins is . Alice flips some coins. Carol
flips some coins. Alice flips some coins. Then, Carol finds
that the state of final coins is m.
k
k
mB

Quantum Number Tricks 242
Method of Flip-Flop1
1) Alice prepares a state (all the coins are head),
exhibits it to Carol, and puts it in a box.
0k 
2) Carol flips some coins and the state becomes m

.
3) Alice applies k
H
 to it and the state becomes
21
0
1(1)
2
k
mx
kx
x




4) Carol flips some coins and the state becomes
21
0
1(1)
2
k
mx
kx
x
r




where .
k
rB
5) Alice applies k
H
 to it and the state becomes
2121
()
200
21 21
()
200
1(1) (1)
2
1(1) (1)
2
(1)
kk
kk
mxx r y
kyx
rymy x
kyx
rm
y
y
m
 





 

 

 
6) Carol opens the box and confirms . m
Finally, we show a trick modifying Flip-Flop1.
Flip-Flop2: First, Alice prepares k coins in all the
coins being head. Next, Carol flips some coins such that
the state of coins is . Alice flips some coins.
Carol flips some coins such that the added state of coins is
. Alice flips some coins. Carol flips some coins.
Alice flips some coins. Then, Alice guesses the value of
if Carol announces the value of ; otherwise, Alice
guesses the value of if Carol announces the value of
.
1
k
mB
2
m
2
k
mB
1
m
1
m
2
m
Method of Flip-Flop2
1) Alice prepares a state 0k

, exhibits it to Carol, and
puts it in a box.
2) Carol flips some coins and the state becomes 1 m

.
3) Alice does not flip them in her turn.
4) Carol flips some coins and the state becomes
.
12
mm
5) Alice applies k
H
 to it and the state becomes
12
21
()
0
1(1)
2
k
mmx
kx
x






6) Carol flips some coins and the state becomes
12
21
()
0
1(1)
2
k
mmx
kx
x
r






where k
r

B.
7) Alice applies k
H

to it and the state becomes
12
12
2121
() ()
200
21 21
()
200
12
1(1) (1)
2
1(1) (1)
2
(1)
kk
kk
mmx xry
kyx
mm yx
ry
kyx
ry
y
y
mm
  


 





 
 

 
Then, Alice measures it and obtains .
12
mm
8) Carol announces either or . Then, Alice
guesses if Carol announced ; otherwise she
guesses .
1
m2 m
12
m
1
m
m
Let be the number of coins. Then, the complexity of
these methods mentioned in this section is in time
because each operation of
k
()Ok
X
,
H
, and can be
executed in time.
CNOT
(1)O
4. Quantum Card Tricks
In this section, we show some quantum card tricks using
quantum states. Magicians Alice and Bob guesses the
numbers selected by spectators Carol and Davis.
Throughout this section, let arithmetic operations be
executed to modulus a prime integer . N
First, let (Alice, Carol) and (Bob, Davis) be two pairs.
Then, we show tricks such that Alice guesses Davis’s
number and Bob guesses Carol’s number.
Telepathy: First, Alice prepares a card written a
number, and puts in a box. The number of this card can be
rewritten. Next, Carol multiplies it by and adds a
random to it, where . Finally, Bob prepares
the
m
r
1
N
mr  Z
N

numbered cards. Carol opens the box and
obtains a number. By turning over Bob’s card corre-
sponding the number, Carol confirms that the reverse side
of the card is m.
Method of Telepathy
1) Beforehand, Alice and Bob share the following en-
tangled state.
1
0
1N
x
x
x
N




where Alice has the first register and Bob has the second
register. Alice’s register is put in a box.
2) Carol multiplies Alice’s register by , and adds
to it. Then, the state becomes
mr
1
0
1N
x
mx rx
N





3) Alice and Bob apply to it and the state be-
comes
QFT

Quantum Number Tricks243
12
12
112
12
1
12
111
2( )2
12
3000
111
22()
12
3000
2
12
0(mod )
1
1
1
NNN
mxr yNxyN
yy x
NNN
ryNmyyxN
yyx
ry N
my yN
eeyy
N
eey
N
eyy
N



  






y
















)
Then, Bob measures it and obtains satisfying
.
2
y
12
0(mod )my yN
4) Bob prepares a set of pairs satisfying
. That is, he writes to the sur-
face of a card and writes to the reverse side. He
makes cards corresponding to all the possible pairs of
. Then, he exhibits the set of the cards to Carol.
1
(my
1
y
12
0(mod )my yN
1
()my
m
5) Carol opens the box and knows . Then, she turns
over Bob’s card written and confirms that the value of
the reverse side is m.
1
y
1
y
Mutual Telepathy: Let Alice and Carol be one pair,
and Bob and Davis be another pair. First, Alice prepares a
card written a number, and puts in a box. Bob also pre-
pares a card written a number, and puts in another box.
Next, Carol multiplies it by , and adds a random to
it. Davis multiplies it by , and adds a random to it.
Here, . Finally, Bob prepares the
1
m
2
1
r
N
m2 r
1212 N
mmrr Z1

numbered cards. Carol opens the box and obtains a
number. By turning over Bob’s card corresponding the
number, Carol confirms that the reverse side of the card is
. In addition, Alice prepares the numbered
cards. Davis opens the box and obtains a number. By
turning over Alice’s card corresponding the number,
Davis confirms that the reverse side of the card is .
1
m1N
2
m
Method of Mutual Telepathy
1) Beforehand, Alice and Bob share the following en-
tangled state.
1
0
1N
x
x
x
N




where Alice has the first register and Bob has the second
register. Alice’s register is put in a box, and Bob’s register
is put another box.
2) Carol multiplies Alice’s register by and adds
to it. Davis multiplies Bob’s register by and adds
to it. Then, the state becomes
1
m
2
m
1
r
2
r
1
1122
0
1N
x
mx rmx r
N




In addition, Davis announces to Bob.
2
m
3) Alice and Bob apply to it and the state be-
comes
QFT
1122112 2
12
1122
112 2
11 1
2( )2()
12
300 0
2( )
12
0(mod )
1
1
NN N
ry ryNmy myxN
yy x
ry ryN
mym yN
ee y
N
eyy
N


 
 
 


y





 



Then, Alice and Bob measure it and obtain 1
y
and
2
y
, respectively, satisfying .
112 20(momym yd )N
4) Bob prepares a set of pairs satisfying
11
(my)
112 20(mod )mym yN


11
()my
. That is, he writes to the
surface of a card and writes to the reverse side. He
makes cards corresponding to all the possible pairs of
1
y
1
m

. Then, he exhibits the set of the cards to Carol.
5) Carol opens the box and knows y1. Then, she turns
over Bob’s card written y1 and confirms that the value of
the reverse side is m1. Note that Alice can also know m1
here.
6) Alice also prepares a set of pairs (m2,y2) satisfying
112 20(mod )mym yN

, and Davis can find the correct
pair (m2,y2).
Next, we show a card trick similar to Flip-Flop2.
Prediction: First, Alice prepares a card written 0, and
puts it in a box. Next, Carol adds to it. Alice
executes some operation. Carol multiplies it by m2 and
adds a random to it, where . Alice executes
some operation, opens the box, and obtains a number.
Finally, Alice prepares the N–1 numbered cards. By turn-
ing over Alice’s card corresponding m1, Carol confirms
that the reverse side of the card is m2.
1N
m Z
N

Zr2 mr
Method of Prediction
1) Alice prepares a state , exhibits it to Carol, and
puts it in a box.
0
2) Carol adds to it and the state becomes
1
m1 m

.
3) Alice applies to it and the state becomes QFT
1
1
2
0
1N
mxN
x
ex
N




 
4) Carol multiplies it by and adds to it. Then,
the state becomes
2
m r
1
1
2
2
0
1N
mx N
x
emx
N




 r
 5) Alice applies to it and the state becomes
QFT

Quantum Number Tricks 244
5) Carol opens the box, obtains w, and confirms that the
value of the reverse side is m.
12
12
11
22()
200
11
2( ) 2
200
2
1
1
NN
mxNm xryN
yx
NN
mmyxN ry N
yx
ry N
ee y
N
ee
N
ey
 


 


y










 


 

Let be the time complexity of arithmetic opera-
tions, where is the size of the input. In addition, let
be the time complexity of QF . It is know that
both and are within polynomial of . Then,
the complexity of their methods mentioned in this section
is in
()cn
()cn
((loOc
n
gN
()qn T
()qn
) (lq
n
og ))N

time.
where . Then, Alice measures it
and obtains .
12 0(mod )mmyN


y
6) Alice prepares a set of pairs satisfying
. That is, she writes to the
surface of a card and writes to the reverse side. she
makes cards corresponding to all the possible pairs of
. Then, she exhibits the set of the cards to Carol.
12
(mm)
12 0(mod )mmyN


12
()mm
1
m
2
m
5. Conclusions
In this paper, we proposed new coin tricks and card tricks
based on quantum physics. In these tricks, magicians had
the ability of quantum physics, but spectators had only
classical one. Therefore, magicians could manipulate
coins and cards as quantum states. Moreover, by sharing
entangled states, they could transmit spectators’ values
without communicating between them.
7) Carol turns over Alice’s card written and con-
firms that the value of the reverse side is .
1
m
2
m
Finally, we show a trick such that Alice guesses the
number selected by Carol in a situation that Alice prepares
a set of cards beforehand.
Since our tricks are simple and straightforward ones
using quantum states, they are somewhat clumsy. There-
fore, it is a future work to construct polished tricks.
Moreover, in our tricks, spectators had only classical
power. Therefore, it is an interesting problem that we
construct quantum tricks when spectators also have
quantum power.
Mindreading: Beforehand, Alice prepares a set of
cards. She writes each to each card and
writes
1NN y Z
()y

to the reverse side, where ()y

is a ran-
dom permutation of
y
. First, Carol selects N m


N

Z
Z and
announces it to Alice. Alice prepares a card, and puts in a
box. Next, Carol adds a random to it. Alice
executes some operation. Finally, Carol opens the box,
and obtains a number. By turning over Alice’s card cor-
responding to the number, Carol confirms that the reverse
side of the card is m.
rREFERENCES
[1] P. W. Shor, “Polynomial-time algorithms for prime
factorization and discrete logarithms on a quantum
computer,” SIAM Journal on Computing, Vol. 26, pp.
1484–1509, 1997.
[2] D. A. Meyer, “Quantum strategies,” Physical Review
Letters, Vol. 82, pp. 1052–1055, 1999.
Method of Mindreading
1) Carol selects and announces it to Alice. m[3] J. Eisert, M. Wilkens, and M. Lewenstein, “Quantum
games and quantum strategies,” Physical Review Letters,
Vol. 83, pp. 3077–3080, 1999.
2) Alice prepares a state
1
2
0
1N
wxN
x
ex
N





  [4] J. Du, H. Li, X. Xu, M. Shi, J. Wu, X. Zhou, and R. Han,
“Experimental realization of quantum games on a quantum
computer,” Physical Review Letters, Vol. 88, 2002. where let ()wm

. This is put in a box.
3) Carol adds a random to it, and the state becomes r[5] J. Du, H. Li, X. Xu, X. Zhou, and R. Han, “Entanglement
enhanced multiplayer quantum games,” Physical Review
Letters, Vol. 302, pp. 229–233, 2002.
1
2
0
1N
wxN
x
ex
N




 r
[6] J. Eisert and M. Wilkens, “Quantum games,” Journal of
Modern Optics, Vol. 47, pp. 2543–2556, 2000.
4) Alice applies to it and the state becomes
1
QFT 
[7] A. Iqbal and A. H. Toor, “Evolutionarily stable strategies
in quantum games,” Physics Letters A, Vol. 280, pp.
249–256, 2001.
11
22()
200
11
22()
200
2
1
1
NN
wx Nxry N
yx
NN
ry NwyxN
yx
rw N
ee
y
N
ee
y
N
ew
  


 













 
[8] L. Marinatto and T. Weber, “A quantum approach to static
games of complete information,” Physics Letters A, Vol.
272, pp. 291–303, 2000.