The quantum Monty Hall problem

Author: Marc Ventura Olivella. Facultat de F´ısica, Universitat de Barcelona, Diagonal 645, 08028 Barcelona, Spain.

Advisor: Miquel Montero Torralbo (Dated: January 16, 2019) Abstract: We consider a variation of the Monty Hall problem and we present a quantum version of it in which the quiz show master does not know where the prize is and can accidentally reveal it. We prove that if the quiz show master is allowed to play quantum and the player is not the game becomes a fair game in contrast with the classical version of the game.

I. INTRODUCTION who knows where the prize is hiding, opens a door in accordance with the following rules: Inspired by the work of von-Neumann and Oskar I Alice must not open the door picked by Bob. Morgesten [10] many mathematicians and economists have used simple games to understand concepts of game II Alice must not open the door containing the prize. theory, well-known examples are the prisoner’s dilemma [9] and “the battle of sexes” [7], among others. Moreover III If Alice can open more than a door (i.e., Bob made the study of games of chance has been used for classical the right choice) without violating rules I and II, information theorists since 1950s. Meyer [6] in 1999 in- then she opens one of these doors randomly. troduces a way of generalizing the classical game theory Given this extra information, Bob is now asked if he into the quantum world, creating quantum games. In wants to stay with his initial choice or otherwise change quantum games we assume some of the parts (or all of his choice to the unopened door. It can be shown that them) of the classic game have been quantized, we can 1 Bob has 3 probability of winning the car if he sticks to his quantize the players’ strategies or the objects used in the 2 initial choice and 3 probability of winning if he switches game. to the unopened door. For further information see [8]. Some games have already been brought into the quantum We want to present a variation of this game when some world, in relation to prisoner’s dilemma, Eisert, Wilkens, of the rules concerning Alice’s choice of a door have been and Lewenstein [2] show that this game ceases to pose relaxed. a dilemma if quantum strategies are allowed. They also construct a particular quantum strategy which always gives reward if played against any classical strategy. An- A. A variation of the classic problem other well-known dilemma in classical information is the Monty Hall problem and the revision of this problem into Suppose that the prize has been hidden by Clarice, the quantum world can be of interest in the study of whose only function in this game is to prepare the game, quantum strategies of quantum measurements. therefore the quiz show master, Alice, does not know Our paper is organized as follows, we will explain the clas- where the prize is hidden. In this case after Bob’s initial sical Monty Hall problem and a variation of it in section choice, Alice will open one of the two remaining doors II. In section III we expound a genuine way to quantize randomly, therefore there is a chance of her opening the variation of the problem previously explained and we the door containing the car. In that way we eliminated solve the problem giving the best choice for the player condition II. in any case. Finally in section IV we discuss some other versions of the quantum Monty Hall problem which have Notice the main differences between the original been already published. game and this version, Alice can open the door with the prize and Bob has no more the two strategies after Alice opens a door (stay, switch) and has three II. THE CLASSIC MONTY HALL PROBLEM strategies (pick door one, pick door two, pick door three).

The Monty Hall problem [8] is a well-known and a As the reader can check, this is a simple problem 2 frustrating brainteaser in all of mathematics. On the of probabilities, and Bob expects to win in 3 of the surface this problem appears to be simple, however it times no matter what strategy he follows as long as can be challenging. This problem is due to a TV quiz he is a rational player, in other words this means Bob show where are shown three doors, two of them hiding is not switching to the opened door if Alice reveals no a goat and the other one hiding a car. The player, lets prize in it, the result is not sensible to Bob’s strategy call him Bob, makes a guess and picks one of the three in this case. And if Alice reveals the prize when she doors. Then, the quiz show master, let’s call her Alice, opens a door, then Bob is picking that door as his final The quantum Monty Hall problem Marc Ventura Olivella choice. Thus this variation of the game does not modify We want to study how Alice can decide whether help the classical probabilities even though impoverishes the Bob to win the car or make a fair game for her with her game. choice of a “door”, which in this case will be a projection on H.

III. THE QUANTUM MONTY HALL PROBLEM Let the initial state of the prize be as equation (1) and without loss of generality |φ0i = |2i the initial A. Analysis of the game under limited information choice of Bob. Alice must pick a direction orthogonal to Bob’s choice so she picks any direction in the plane {|0i, |1i}: In this section we will set forth the quantum version of the variation of the classic Monty hall problem |χi = sin θ1|0i + cos θ1|1i. (3) described in the previous section. The main quantum variable will be the position of the prize, which lies in π With θ1 ∈ [0, 2 ]. Alice can control the θ1 parameter a 3-dimensional Hilbert space H. We will also quan- and it is public information so Bob knows exactly Alice’s tize the Alice’s choice of a box in contrast with Bob’s move. We can compute now the expected value of the first and second choice of a Box that will remain classical. probability of Alice hitting the prize: Z Z Before the game starts the system is prepared quantum 2 1 2 |hψ0|χi| P (θ0, ϕ0)dθ0dϕ0 = |hψ0|χi| sin θ0dθ0dϕ0 mechanically (by Clarice) so that a quantum particle is S 2π S in a superposition of the states |0i, |1i, |2i; representing the doors of the classic problem. Let {|0i, |1i, |2i} be π π Z 2 Z 2 the orthonormal basis of H. The prize will be in any 1 2 3 1 2 3 2 = sin θ1 sin θ0dθ0 + cos θ1 sin θ0dθ0 point of the upper hemisphere of the S as don’t take 2 0 2 0 into account the imaginary parts, in this work we will stay real. Then, the initial state can be described as: 1 2 1 2 1 = · sin2 θ + · cos2 θ = . (4) 2 3 1 2 3 1 3 |ψ0i = sin θ0 cos ϕ0|0i + sin θ0 sin ϕ0|1i + cos θ0|2i. (1) So we expect that in one third of the games Alice re- π Where ϕ0 ∈ [0, 2π] and θ0 ∈ [0, 2 ]. We consider veals the prize, however this does not ensure Bob a easy only the upper hemisphere since the states (θ0, ϕ0) and win. In this case the position of the prize will have col- (π − θ , ϕ + π) are the same, with this restriction, from 0 0 0 lapsed in the state |ψ i = |χi = sin θ1|0i + cos θ1|1i, but a quantum point of view we remove any redundancy. We Bob is still restricted to pick a direction along one of the can use different probability distributions for the param- axis, so if he picks the door |0i his probability of hitting eters θ and ϕ . We will assume that the prize is any 2 2 0 0 the prize is sin θ1, and cos θ1 if he picks the door |1i. point of the upper hemisphere with equal probability so Therefore if Bob is a rational player his expected payoff the probability distribution we will use: is:

1 2 2 max {sin θ1, cos θ1}. (5) P (θ0, ϕ0)dθ0dϕ0 = sin θ0dθ0dϕ0. (2) θ ∈[0, π ] 2π 1 2 The game proceeds in the following stages: In the other two thirds of the games Alice won’t hit the prize so after her measurement the position of the prize I Bob chooses a projection p = |φihφ| along one of will have collapsed to the space (1 − q)H. Now the posi- the axis on H, i.e., Bob is restricted to choose either tion of the prize can be given by: |φi = |0i, |1i, |2i. |ψ0i = (1 − |χihχ|)|χi II Alice’s choice of a door is q = |χihχ|, by rule I, it must be another door so q ⊥ p. During this   step Alice could show the “location” of the prize = g(θ0, ϕ0) cos θ1|0i + sin θ1|1i + cos θ0|2i. (6) as we do not assume q ⊥ |ψ0ihψ0|, that is due to

the fact that Alice does not know where the car is. Where g(θ0, ϕ0) = sin θ0 cos(ϕ0 + θ1). As the reader may After Alice performs a measurement the quantum see |ψ0i is not normalized so we have to normalize it. Let 2 2 system will have collapsed in the two-dimensional N = cos θ0 + g (θ0, ϕ0) be the normalizing constant, space (1 − q)H if Alice don’t hit the prize and will then the state of the prize is: collapse in qH if she hits the prize.

III After seeing Alice’s move and with all the informa- 1 h   i |ψ1i = √ g(θ0, ϕ0) cos θ1|0i + sin θ1|1i + cos θ0|2i . tion but the parameters θ0 and ϕ0, Bob can pick a N direction again as in step I. (7)

Treball de Fi de Grau 2 Barcelona, January 2019 The quantum Monty Hall problem Marc Ventura Olivella

At this moment, Bob knows that the state of the prize at least, if Bob is a rational player, his expected payoff 1 is the one described in equation (7). Now he is asked to equals 2 by picking |φ1i = |2i as his final choice in this decide whether stick to his initial choice |φ0i or switch to case. Now we can compute the total expected payoff for another door. He can compute the quantum probabilities Bob in this quantum version of the game, which will take i 2 P (θ0, ϕ0; θ1) = |hψ1|ii| of hitting the prize when the into account the probabilities of Bob winning the game state (θ0, ϕ0) is known if Bob picks the door |ii. Then if when Alice hits the prize when she picks a door and when he picks the |0i: she doesn’t hit the prize.

2 1  2 2 2 1 0 cos θ1 h$i = max sin θ1, cos θ1 + · P (θ , ϕ ; θ ) = . (8) π 0 0 1 2 2 3 θ1∈[0, ] 3 2 1 + cot θ0 sec (ϕ0 + θ1) 2 If he picks |1i:   1  2 2 = 1 + max sin θ1, cos θ1 . (14) 3 θ ∈[0, π ] sin2 θ 1 2 P 1(θ , ϕ ; θ ) = 1 . (9) 0 0 1 2 2 1 2 1 + cot θ0 sec (ϕ0 + θ1) It is fulfilled that 2 ≤ h$i ≤ 3 . The lower bound is π reached when Alice plays θ1 = 4 , which corresponds with And if he picks |2i: a move |χi = √1 |0i+ √1 |0i. Playing like this, Alice plays 2 2 the “the most quantum possible”, taking advantage of 2 1 P (θ0, ϕ0; θ1) = . (10) the fact that Bob is restricted to playing classical, and 1 + tan2 θ cos2(ϕ + θ ) 0 0 1 makes a fair game for her. On the other hand the upper π Now we can compute Bob’s expected value of the prob- bound is reached when Alice plays θ1 = 0 or θ1 = 2 , ability of winning when he picks any of the doors by in- which corresponds with the moves |χi = |0i or |χi = |1i. tegrating these probabilities on the upper hemisphere of These plays do not take profit of the possible quantum 2 because (θ0, ϕ0) are unknown to Bob. Remember that moves which Alice can make as she remains classical, so S 2 the distribution of the probability is given by equation Bob can reach his expected payoff of h$i = 3 like in the (2). Then: classical game. Z 0 h$0i = P (θ0, ϕ0; θ1)P (θ0, ϕ0)dθ0dϕ0 B. The influence of complete information S

In this section we will analyze how often does Clarice π  s  2 Z 2 2 cos θ1  cot θ0  agree with Bob’s decision. We will do this by plotting the i = sin θ0 2π 1 − 2  dθ0 probabilities P (θ0, ϕ0; θ1) (which Clarice knows because 2π 0 1 + cot θ0 she knows the state (θ0, ϕ0)) as a functions of θ0 and ϕ0 ϕ0 which are unknown to Bob. Let u = 2π and ω = cos θ0. π 2 π As θ0 ∈ [0, ] and ϕ0 ∈ [0, 2π] then (u, ω ) ∈ [0, 1]×[0, 1]. Z 2   1 2 = cos2 θ sin θ − sin θ cos θ dθ = cos2 θ . Therefore we can rewrite equations (8) to (10) as: 1 0 0 0 0 2 1 0 1 (11) P 0(u, ω; θ ) = cos2 θ (1 − ω2) cos2(2πu + θ ). (15) In the same way we can find: 1 N 1 1 Z 1 1 2 1 1 2 2 2 h$1i = P (θ0, ϕ0; θ1)P (θ0, ϕ0)dθ0dϕ0 = sin θ1. P (u, ω; θ1) = sin θ1(1 − ω ) cos (2πu + θ1). (16) S 2 N (12) And finally: 1 P 2(u, ω; θ ) = ω2. (17) Z 1 N 2 h$2i = P (θ0, ϕ0; θ1)P (θ0, ϕ0)dθ0dϕ0 0 1 S It is obvious that max{P (u, ω; θ1),P (u, ω; θ1)} = 2 2 max π {sin θ , cos θ }, so we want to study when θ1∈[0, 2 ] 1 1 the probability of “door” 2 is higher or lower than 1 Z dθ dϕ = sin θ 0 0 “doors” 0 and 1. i.e., we want to study when it is prefer- 0 2 2 2π S 1 + tan θ0 cos (ϕ0 + θ1) able for Bob to change his initial choice or to stick to his 2 2 initial choice. Let β = max{sin θ1, cos θ1}, combining π equations (15),(16) and (17), we reach the following con- Z 2 sin θ0 1 = dθ0 = . (13) clusion: as Bob, picking as final decision |2i has higher p 2 0 1 + tan θ0 2 probability than picking |0i or |1i when:

2 2 Whatever Alice’s choice of θ1, Bob expects more or 2 β cos (2πu + θ1) 1 1 1 ω > 2 2 , (18) equal if he picks |2i because 2 ≥ 2 sin θ1, 2 cos θ1, so 1 + β cos (2πu + θ1)

Treball de Fi de Grau 3 Barcelona, January 2019 The quantum Monty Hall problem Marc Ventura Olivella and it is worthier to pick |0i or |1i when: 1 θ1 = 0.79 2 2 β cos (2πu + θ ) θ1 = 0.94 ω2 < 1 . (19) 2 2 0.8 θ1 = 1.10 1 + β cos (2πu + θ1) θ1 = 1.26 In this last case Bob would have to pick |0i or |1i θ1 = 1.41 0.6 depending on the parameter θ1 chosen by Alice which θ1 = 1.57 2 he already knows. ω 0.4 We can plot these probabilities to further compre- hension of Bob’s better choice and what does Clarice think about Bob’s decision. First of all we present a 0.2 graph showing the behavior of inequation (19) when π θ1 = 0 and θ1 = : 0 4 0 0.2 0.4 0.6 0.8 1 u 1

π π FIG. 3: Curves of probability for 4 ≤ θ1 ≤ 2 . 0.8 The conclusion is that picking |0i or |1i than |2i, i.e., it 0.6 would be preferable switching than sticking to your initial

2 choice has maximum worth when Alice plays θ1 = 0 or ω π θ1 = 2 , and it has minimum worth when Alice plays 0.4 π θ1 = 4 .

0.2 IV. OTHER VERSIONS OF THE PROBLEM

0 0 0.2 0.4 0.6 0.8 1 In this section we want to summarize three other ver- u sions of the problem that have been published. This three other versions are based in the original Monty Hall prob- π FIG. 1: Inquation (19) for θ1 = 0 (green) and θ1 = 4 (red). lem instead of a variation of the problem as we do. Nev- ertheless, these examples show that the quantization of a classical game is non-unique. The filled area shows us all the possible pairs (θ0, ϕ0) where is would be preferable to choose |0i or |1i as fi- • The quantization proposed in [4] is completely dif- nal decision, so it shows when Clarice and Bob would ferent from our approach. The parts that have been disagree (remember that Bob best strategy is to stick to quantized from the original problem are not mea- his initial choice |2i as we mentioned in the previous sec- surements as we do, Flitney and Abbott suggest tion). The following graphs put on display how the curve that the moves Alice and Bob can do are oper- changes when the parameter θ1 vary. ations on a tripartite system. Alice’s and Bob’s choices are represented by qutrits and they start 1 in some initial state ψ ∈ HA ⊗ HB ⊗ HO. These θ1 = 0.00 Hilbert spaces represent the door where Alice hides θ = 0.16 1 the prize, the door that Bob chooses and the door 0.8 θ1 = 0.31 that has been opened by Alice respectively. If the θ1 = 0.47 θ1 = 0.63 initial state is |ψii, the final state of the system is: 0.6 θ1 = 0.79     2 ˆ ˆ ˆ 1 ˆ ˆ ω |ψf i = S cos γ + N sin γ O ⊗ B ⊗ A |ψii. (20) 0.4 Where Aˆ is Alice’s choice of operator, Bˆ is Bob’s choice of operator, Oˆ the opening box operator, 0.2 Sˆ and Nˆ are Bob’s switching and not switching operators. Bob also controls the parameter γ to 0 play a mixed strategy of switching or not switching. 0 0.2 0.4 0.6 0.8 1 They reach the following result: if Alice has access u to quantum strategies and Bob doesn’t, she can make the game fair with a expected payoff of 1 FIG. 2: Curves of probability for 0 ≤ θ ≤ π . 2 1 4 for each player. Otherwise, if Bob has access to

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2 quantum strategies and Alice does not, then he can player is rational, he expects a payoff of 3 , see section win the game all the times. II A.

• A completely different different way to implement In the quantum version of our problem the prize is quantum strategies is proposed in [5]. In this quantum particle which is in a random superposition (1) version, Alice puts the quantum particle in a 2 on the upper hemisphere of the S . Alice is allowed to box, maybe in a superposition state like |ψip = perform quantum measurements and Bob is restricted √1 (|0i + |1i + |2i). After Bob picks one box, for 3 to classical moves. We prove that the expected payoff instance |0i, and Alice reveals no particle in |2i, 1 2 for Bob is 2 ≤ h$i ≤ 3 . Playing classical, Bob can never the state of the particle may be described by the overcome his expected payoff of the classical game. In density matrix: fact, Alice decides the value of Bob’s payoff when she makes her move (3). That’s due to the fact that Alice 1 2 ρp = |0ih0| + |1ih1|. (21) is allowed to play quantum and Bob isn’t, and this is a 3 3 clear advantage for Alice who can make the game fair Now Alice is allowed to perform a von-Neumann for both players by playing quantum. measurement on the particle, shuffling it around and the game is reduced to a coin tossing game. This game offers more research than we have done, it may be interesting to study how the players’ payoff is • The most closely related published version to ours affected when the probability distribution of the prize is [1]. Although the idea for this paper came up does not verify equation (2), or how it is affected when in a bar during a workshop as the authors confess, Bob has to pay or cede part of the prize if he wants to they came up with some interesting ideas to face change his initial choice after Alice’s move. the quantum Monty Hall problem. The most im- Acknowledgments portant one and the one we have exploited in this work is to see the position of the prize and all the strategies of the players as directions in the space. I would like to express my gratitude to my supervi- sor Dr. Miquel Montero Torralbo for his constant advice V. CONCLUSIONS and assistance, which have been vital for achieving this work, additionally he made me really interested in quan- We have presented a variation of the classical Monty tum games which, from my point of view are entertain- Hall problem where the quiz show master does not know ing , funny and very instructive in quantum mechanics. where the prize is. In contrast with the original version I would like to thank as well my family and friends who 1 where the player expected a payoff of 3 if sticking to the have been supporting me throughout the time of this 2 initial choice and 3 if switching, in our version, if the work.

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