Quantum games

Quantum Games and Quantum
Strategies
The Future of Decision Making ?

Quantum Games !!!
● Can principles of Quantum communication be
used to develop efficient and unbiased marketing
strategies ?

● Quantum communication and Quantum
cryptography can be regarded as Games
played between 2 legal players Alice and Bob and
the illegal players in between them trying to decrypt
the secret message.

What is a Game ?
● We have a set of players i

● a strategy set Si for each player i.

● Pay-off for each player Pi(s1,s2,....), where siSi.

Where does “Quantum” comes into
“play”?

● Game theory does not explicitly concern itself with
how the information is transmitted once a decision is
taken. Bearing in mind that a game is also about
the transfer of information, it becomes legitimate
to ask what happens if these carriers of
information are taken to be quantum systems,
quantum information being a fundamental notion of
information.

Strategies and Equilibriums
● A quantum strategy sA is called dominant strategy of
Alice if PA (sA,s'B) ≥ PA (s′A, s′B )
for all s′A ∈ SA , s′B ∈ SB

● A pair ( sA , sB ) is said to be an equilibrium in
dominant strategies if sA and sB are the players’
respective dominant strategies

Strategies and Equilibriums

● A combination of strategies ( sA , sB ) is called a
Nash equilibrium if PA (sA,sB) ≥ PA (s′A, sB ) and
PB (sA,sB) ≥ PB (sA, s′B )for all s′A ∈ SA , s′B ∈ SB .
● A pair of strategies ( sA , sB ) is called Pareto
optimal, if it is not possible to increase one player’s
pay-off without lessening the pay-off of the other
player.

Nash Equilibrium

● A Nash equilibrium implies that neither player has a
motivation to unilaterally alter his or her strategy
from this equilibrium solution, as this action will
lessen his or her pay-off. Given that the other player
will stick to the strategy corresponding to the
equilibrium, the best result is achieved by also
playing the equilibrium solution.

The Prisoner's Dilemma

● Two prisoners are being questioned by the police.
They are held in separate cells and cannot talk to
each other. The police make the following offer to
both the prisoners: if one confesses that both
committed the crime then the confessor will be set
free and the other will spend 5 years in jail; if both
confess then they will each get 4 yr jail term: if
neither confess, then they will each spend 2 yrs in
jail

● Table of Pay-offs :

P1 Cooperation Defection
P2
Cooperation (3,3) (0,5)

Defection (5,0) (1,1)


● If both the prisoners co-operate with each other then
they are each awarded 3 yrs of freedom, if none of
them co-operate with each other then each will be
awarded 1 yr of freedom and if one of them has
conscience and the other is defected then the 'good'
prisoner has to serve full 5 yr in jail while the 'bad'
one will be set free immediately i.e. Full 5 yrs of
freedom for the 'bad' prisoner. So what is the optimal
strategy that both of them apply here?

Nash Equilibrium
● In the Prisoner's dilemma problem if the dominant
strategy is (D,D) with pay-offs (1,1) for Alice and
Bob. Now if Alice changes her strategy while Bob
sticks to his strategy, i.e. Alice decides to become
'good' then she pays the price with 0 pay-off or in
this case a complete 5yr jail sentence. Same is true
for Bob, thus if either of them decides to become
'good' while the other stays 'bad', the his/her pay-off
decreases, hence (D,D) is in Nash equilibrium. But
(D,D) is not Pareto optimal because (C,C) has better
pay-offs than (D,D).

Quantum Prisoner's Dilemma
● In traditional 2 × 2 games where each player has just
a single move, creating a superposition by utilizing a
quantum strategy will give the same results as a
mixed classical strategy. In order to see non-
classical results it is necessary to produce
entanglement between the players’ moves

● Initial state is the maximally entangled state
|> = (|00> + i|11>)/√2

● The final state is represented as following :
|ψf> = J†(UA ⊗UB) J |CC>
where J=exp(iγD' ⊗ D'/2), γ ∈ [0, π/2] is a real
parameter, UA and UB are respective strategies for
Alice and Bob.

● Expected Pay-off for Alice :
<$> = ACC|<ψf |CC>|2 + ACD|<ψf |CD>|2 + ADC|
<ψf |DC>|2 + ADD|<ψf |DD>|2

● The matrix representation of operators
corresponding to quantum strategies from this set is
given by U(,) = {( eicos(/2), sin(/2) ),
( -sin(/2), e-icos(/2) )}

● The strategies are :
C={(1,0),(0,1)} and D={(0,1),(-1,0)}

● γ is a measure for the game’s entanglement

Pay-Offs is what matters !!

● Alice's expected pay-off with the quantum
strategies:
PA (θA , φA , θB , φB ) = 3 |cos(φA + φB ) cos(θA /2)
cos(θB /2)|2 + 5 |sin(φA ) cos(θA /2) sin(θB /2) −
cos(φB ) cos(θB /2) sin(θA /2)| 2 + |sin(φA + φB )
cos(θA /2) cos(θB /2) + sin(θA /2) sin(θB /2)|2 .


● Bob's expected pay-off with the quantum strategies:

PB (θA , φA , θB , φB ) = 3 |cos(φA + φB ) cos(θA /2)
cos(θB /2)|2 + 5 |sin(φB ) cos(θB /2) sin(θA /2) −
cos(φA ) cos(θA /2) sin(θB /2)| 2 + |sin(φA + φB )
cos(θA /2) cos(θB /2) + sin(θA /2) sin(θB /2)|2 .

● Assuming Bob chooses D'= U (π, 0), then
PA (θA , φA ,  , 0 ) = 5 |sin(φA ) cos(θA /2)| 2 +
|sin(θA /2)|2 <= 5 for θA = 0 and φA = /2
● Thus Alice’s best reply would be Q = U(0, /2) =
{(i,0),(0,-i)}
● While assuming Bob plays C'= U (0, 0) Alice’s best
strategy would be defection D'. Thus, there is no
dominant strategy left for Alice. The game being
symmetric, the same holds for Bob, i.e., D' ⊗ D' is
no longer an equilibrium in dominant strategies

Does the Prisoners escape the dilemma?

● It is interesting to see that Q ⊗ Q has the property to
be Pareto optimal , that is, by deviating from this
pair of strategies it is not possible to increase the
pay-off of one player without lessening the pay-off
of the other player. In the classical game only
mutual cooperation is Pareto optimal, but it is not an
equilibrium solution. One could say that by
allowing for quantum strategies the players
escape the dilemma .

Real Life Dilemmas !!
● Nuclear proliferation among nations: When a
nation tries to up its nuclear armory and sources
with the help of its more powerful allies, its rival
nations feels threatened and they too power up their
nuclear warfares and sources with powerful allies,
thus increasing tension between nations and
increasing possibility of nuclear war. Thus one
might speculate that Governments of nations using
Quantum strategies might even end the possibility of
a nuclear disaster in near future, but then
“everything” needs to be “Quantum” !!!

This one's from “Batman-Dark Knight” : Joker
planted bombs on the 2 ferries where the triggers for
the bombs were given to the opposite parties. The
classical Nash equilibrium strategy would lead the 2
parties to trigger the bombs simulteneously thus
killing everybody. But in the movie they might have
figured out the “Quantum strategy” to co-operate
thus saving all of them. One might think “Quantum
strategies” as the most “socially acceptable” or
“near conscience” but wait till you have seen the
opposite face of the “coin”.

● Breakdown of talks at the Climate change Summit at
Copenhagen: Rich nations having larger number of
industries are emmitting larger amount of CO2 in the
atmosphere. But only developing nations trying to build up
its commercial and industrial economy are asked to cut
down CO2 emmision. Such nations felt that the rich are
trying to suppress them down and they deny to decrease
CO2 emmision from factories etc. and thus talks broke
down. Defection( increased release of CO2 by all nations)
seemed the Nash equilibrium strategy compared to
Cooperation(decreased release of CO2 by all nations). Thus
a “Quantum World” should be the paradise or the heaven to
live in.

Should you exchange doors ?

● The answer is yes if you are a rational person !
● You win 2/3 of the time you exchange the doors and
lose 1/3 of the time !!
● From an information theoretic point of view, by
opening a door without a prize Monty has given
information about where the prize is. Let's see how?

Should you exchange doors ?

● The probability of originally choosing a goat is 2/3
and the probability of originally choosing the car is
1/3. Once Monty Hall has removed a "goat door,"
the contestant who chose the door with a goat
behind it will necessarily win the car, and the
contestant who originally chose the car will
necessarily "win" the goat. Because the chances are
2/3 of being a contestant who originally chose a
goat, probability will always favor switching
choices.

Quantum Monty Hall !!
● In our quantum version of the game let us call the
host as Alice and player as Bob.
● There is one quantum particle and three boxes |0 > ,
|1> , and |2>
● The state of the system can be expressed as
|> = |oba>
where a = Alice’s choice of box( prize door ),
b =Bob’s choice of box, and o = the box that has
been opened


● The final state of the system is
|f> = (S' cos  + N' sin )O'(I'  B'  A')|i>
● Where A' Alice’s choice operator or strategy, B'
Bob’s initial choice operator or initial strategy, O'
the opening box operator, S' Bob’s switching
operator, N' Bob’s not switching operator, I' the
identity operator, and  = [0,/2]

The open box operator is a unitary operator that can
be written as :
O' = ∑ijkl |∈ijk| |njk><ljk| + ∑jk |mjj><ljj|
● where |∈ijk| = 1 if i,j,k are different else 0.
m = (j+l+1) (mod 3) and n = (i+l)(mod 3).
● The first term considers the cases where Bob
chooses the door not having the car behind it and the
second term considers the cases where Bob initially
chooses the door having the car inside it

● The possible cases are :
|010> ⇒|210>, |020> ⇒|120>, |001> ⇒|201>, |
021> ⇒|021>, |002> ⇒|102>, |012> ⇒|012>

● Thus the operator should be |210>< 010| +
|120>< 020| + |201>< 001| + |021>< 021| +
|102>< 002| + |012><012| to introduce unitarity we
use the complete operator ∑ijkl |∈ijk| |njk><ljk|


● Bob's switch operator can be written as :
S' = ∑ijkl |∈ijk| |ikk><ijk| + ∑ji |imj><ijj| m=(3-i-j)
● The possible transformations we consider as
probable :
|120> ⇒|100>, |100> ⇒|120>, |011> ⇒|021>, |201>
⇒|211>, |021> ⇒|011>, |122> ⇒|102>, |
022> ⇒|012>, |012> ⇒|022>, |102> ⇒|122>, |200>
⇒|210>, |210> ⇒|200>, |211> ⇒|201>.

“Expectation”

● Bob wins if his choice of door is same as the door
Alice chooses to keep the car. Thus expectation
value of Bob's win = <$B> = ∑ij |<ijj|f>|2

● Alice wins if Bob is incorrect, so
<$A> = 1 - ∑ij |<ijj|f>|2


● Without entanglement and one of Alice or Bob
applying the identity operator we get back the
classical case where Bob wins with 2/3 probability if
he switches doors.
● Consider the case with maximum entanglement
The initial state is
|i> = |0>  (1/3)(|00> + |11> + |22>)


● O'(IB'A')|i> = (1/3) ∑ijkl |∈ijk| bljalk|ijk> +
(1/3) ∑jl bljalj|mjj>
● S'O'(IB'A')|i> = (1/3) ∑ijkl |∈ijk| bljalk|ikk> +
(1/3) ∑jkl |∈jkm| bljalj|mkj>
● If B'=I then <$B> = (1/3)sin2 (|a00|2 + |a11|2 + |a22|2) +
(1/3)cos2 (|a01|2 + |a02|2 + |a10|2 + |a12|2 + |a20|2 + |a21|2)


● If now Alice chooses an unitary operator whose
diagonal element all have absolute value (1/ 2) and
off-diagonal elements have absolute value (1⁄2)
then :
<$B> = (1⁄2) sin2 + (1⁄2) cos2 hence payoff for
Bob if he switches = 1⁄2 .

Is the Game Fair Now ?
● For the Monty Hall game where both participants
have access to quantum strategies, maximal
entanglement of the initial states produces the same
payoffs as the classical game. That is, for the Nash
equilibrium strategy the player, Bob, wins two-thirds
of the time by switching boxes. If the host, Alice,
has access to a quantum strategy while Bob does
not, the game is fair, since Alice can adopt a strategy
with an expected payoff of 1/2 for each person,
while if Bob has access to a quantum strategy and
Alice does not he can win all the time.

Simple Penny Flipover
● There are 2 players involved in the penny flip game.
Let's name them P and Q. P is to place a penny, head
up, in a box, whereupon they will take turns (Q, then
P, then Q) flipping the penny over (or not), without
being able to see it. Q wins if the penny is head up
when they open the box
● Q
---- NN NF FN FF
P

N -1 1 1 -1

F 1 -1 -1 1


● Suppose P doesn’t flip the penny over. Then if Q
flips it over an even number of times, P loses.
Similarly, if P flips the penny over, then if Q flips it
over only once, P loses. Thus PQ penny flipover has
no deterministic solution , no deterministic Nash
equilibrium : there is no pair of pure strategies, one
for each player, such that neither player improves
his result by changing his strategy while the other
player does not.

● Since this is a two-person zero-sum strategic game
with only a finite number of strategies, there is a
probabilistic solution: It is easy to check that the
pair of mixed strategies consisting of P flipping the
penny over with probability 1⁄2 and Q playing each
of his four strategies with probability 1⁄4 is a
probabilistic Nash equilibrium: Neither player can
improve his expected payoff (which is 0 in this case)
by changing the probabilities with which he plays
each of his pure strategies while the other player
does not.

● In the classical case both P and Q has a 1⁄2
probability of winning the game but if Q adopts a
quantum strategy then she will win with a
probability of 1
● |0> = |H> then
Q: |1> = H|0> = (1/2)(|H> +|T>) then
P(mixed classical strategy) : U= pF + (1-p)N,
|2> = U|1> = (1/2)(|H> + |T>), then again
Q: |3> = H|2> = |H>

Penny Flipover Applications

● This strategy is same as that of the game between
the oracle and the Grover's algorithm:
● In the problem of searching a database of size N, the
locations in the database correspond to pure
strategies. The oracle can be thought of as the player
P in the game who uses a mixed strategy to frustate
our search for an item at some specified location.
While the Grover's algorithm represents the player
Q who uses quantum strategy.

Penny Flipover Applications

● Even the Shor's algorithm can be interpreted as a
penny flipover game between a Quantum player and
a classical mixed strategist. After we apply the
inverse Quantum Fourier Transform, we measure a
certain value as the output measurement, then we
use classical continued fraction to find out the
period accurately with some probability. The
continued fraction implementation can be thought of
as the player P who uses mixed strategy to produce a
correct result with some probability.

Penny Flipover

● One might ask what if player P also uses a Quantum
strategy?
● In the penny flipover game P using the Quantum
operator U = √p X + ie- (1-p) Z where X is the
Pauli NOT gate and  is the phase, would produce
the same result as before with Q winning with
probability of 1.

Penny Flipover
● Theorem : A two person zero sum game need not
have a (quantum, quantum) equilibrium.
● Proof : Let us assume that the quantum strategies
[(U2),(U1,U3)] is a (quantum, quantum) equilibrium.
Suppose that U3 U2 U1 |H>  |H>. Then Q can
improve his expected payoff to 1 by changing his
strategy U3 to U1-1U2-1 which is unitary. Also suppose
U3 U2 U1 |H>  |T> then P can improve his
expected payoff to 1 by changing his strategy U 2 to
U3-1XU1-1 which is also unitary.

Concluding Remarks

● One might speculate that game theoretic perspective
will suggest new possibilities for efficient quantum
algorithms.

Concluding Remarks

● It may be too soon to say that Quantum game theory
and quantum strategies are going to be the future of
decision making in fields like Finance, Marketing,
Adminstrative Policies, Quantum Algorithms,
Quantum communication(cryptography) and
everyday life decision processess, given that
efficient hardware implementation of quantum
computation is still an uphill task with the noise and
decoherence effects playing their parts

Still a Hope !!

● The day when the Quantum error correction codes
becomes so efficient that all noises and
decoherences can be cancelled out, “Quantum” will
be the order of the day, and the world will take the
next leap forward changing gears from the
“Classical” to “Quantum”. I strongly believe that the
day is not far away and it will change the whole
world, “Logic and reasoning will dominate over
emotions and feelings”.

Quantum games

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