<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">IB</journal-id><journal-title-group><journal-title>iBusiness</journal-title></journal-title-group><issn pub-type="epub">2150-4075</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/ib.2010.21004</article-id><article-id pub-id-type="publisher-id">IB-1438</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Business&amp;Economics</subject></subj-group></article-categories><title-group><article-title>
 
 
  A Study of Quantum Strategies for Newcomb's Paradox
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>akashi</surname><given-names>Mihara</given-names></name><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><author-notes><corresp id="cor1">* E-mail:<email>mihara@toyonet.toyo.ac.jp</email></corresp></author-notes><pub-date pub-type="epub"><day>26</day><month>03</month><year>2010</year></pub-date><volume>02</volume><issue>01</issue><fpage>42</fpage><lpage>50</lpage><history><date date-type="received"><day>August</day>	<month>24th,</month>	<year>2009</year></date><date date-type="rev-recd"><day>October</day>	<month>11th,</month>	<year>2009</year>	</date><date date-type="accepted"><day>November</day>	<month>23rd,</month>	<year>2009.</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  Newcomb’s problem is a game between two players, one of who has an ability to predict the future: let Bob have an ability to predict Alice’s will. Now, Bob prepares two boxes, Box1 and Box2, and Alice can select either Box2 or both boxes. Box1 contains $1. Box2 contains $1,000 only if Alice selects only Box2; otherwise Box2 is empty($0). Which is better for Alice? Since Alice cannot decide which one is better in general, this problem is called Newcomb’s paradox. In this paper, we propose quantum strategies for this paradox by Bob having quantum ability. Many other results including quantum strategies put emphasis on finding out equilibrium points. On the other hand, our results put emphasis on whether a player can predict another player’s will. Then, we show some positive solutions for this problem.
 
</p></abstract><kwd-group><kwd>Game Theory</kwd><kwd> Newcomb’s Paradox</kwd><kwd> Quantum Strategy</kwd><kwd> Meyer’s Strategy</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>Quantum mechanics has been incorporated into many fields called information. The most famous results are a quantum factoring algorithm by Shor [<xref ref-type="bibr" rid="scirp.1438-ref1">1</xref>] and a quantum database search algorithm by Grover [<xref ref-type="bibr" rid="scirp.1438-ref2">2</xref>]. Moreover, the studies on quantum information have succeeded in such as quantum computation, quantum circuit, quantum cryptography, quantum communication complexity, and so on. Recently, game theory based on quantum mechanics, quantum game theory, has been also proposed and it has been shown that quantum game theory is more powerful than classical one.</p><p>Game theory is one of the most famous decision making methods and has been used in many situations both theoretically and practically. There had existed the basic concept with respect to these games since early times but in corporation with Morgenstern, von Neumann [<xref ref-type="bibr" rid="scirp.1438-ref3">3</xref>] firstly constructed the theory systematically. However, the main principle of this theory was based on classical physics although he was familiar with quantum mechanics.</p><p>In 1998, for a coin flipping game, Meyer [<xref ref-type="bibr" rid="scirp.1438-ref4">4</xref>] proposed a quantum strategy for the first time and showed that the quantum strategy has an advantage over classical ones. This game is called PQ Penny Flip.</p><p>PQ Penny Flip: The starship Enterprise is facing some immanent—and apparently inescapable—calamity when Q appears on the bridge and offers to help, provided Captain Picard can beat him at penny flipping: Picard is to place a penny head up in a box, whereupon they will take turns (Q, then Picard, then Q) flipping the penny (or not), without being able to see it. Q wins if the penny is head up when they open the box.</p><p>This game is a two-player zero-sum game and the probability that each player wins is at most 1/2 with classical strategies. Meyer showed a quantum strategy with which Q can always win by using a superposition of quantum states effectively. In this game, Picard is constrained to play classically. The quantum strategy is then executed in the following way. Let <img src="4-8601005\71200c75-628a-48c7-920e-b88cb4cf1458.jpg" /> and <img src="4-8601005\b7f8e709-912d-42f9-958c-117fa57fc19f.jpg" /> represent the head and tail of the penny, respectively. First, Picard prepares <img src="4-8601005\21b35755-2520-4e11-a9ed-c0f0a4fe6ad2.jpg" /> and Q applies a Walsh-Hadamard operation <img src="4-8601005\1bbfbaff-ad96-42d5-96cb-10074976f43b.jpg" /> defined in the next section to the state:</p><p><img src="4-8601005\d440a366-abbf-40ee-a510-c4672fc62bba.jpg" /></p><p>Next, Picard decides classically whether he flips it or not. However, the state does not change even if Picard flipped it. Finally, Q applies <img src="4-8601005\0eb2a656-4fb7-4890-b8c1-565880d3c898.jpg" /> to the state and always obtains<img src="4-8601005\0d456430-9bc4-4a55-a4a6-810245386f74.jpg" />. This means that Q always wins.</p><p>Moreover, he also showed the importance of a relationship between quantum game theory and quantum algorithms.</p><p>Later, other types of quantum strategies have been also proposed. In their strategies, all the players can use quantum operations. For example, Eisert et al. [<xref ref-type="bibr" rid="scirp.1438-ref5">5</xref>] proposed a quantum strategy with entanglement for a famous two-player game called the Prisoner’s Dilemma (also see Du et al. [6,7], Eisert and Wilkens [<xref ref-type="bibr" rid="scirp.1438-ref8">8</xref>], and Iqbal and Toor [<xref ref-type="bibr" rid="scirp.1438-ref9">9</xref>]). In their strategy, entanglement plays an important role. For another famous two-player game called the Battle of the Sexes, Marinatto et al. [<xref ref-type="bibr" rid="scirp.1438-ref10">10</xref>] also proposed a quantum strategy with entanglement. For these games, they showed quantum Nash equilibriums different from classical ones. Furthermore, there are many results being related to games such as the Monty Hall problem by D’Ariano et al. [<xref ref-type="bibr" rid="scirp.1438-ref11">11</xref>], Flitney and Abbott [<xref ref-type="bibr" rid="scirp.1438-ref12">12</xref>], and Li et al. [<xref ref-type="bibr" rid="scirp.1438-ref13">13</xref>], Parrondo’s game by Flitney et al. [<xref ref-type="bibr" rid="scirp.1438-ref14">14</xref>], games in economics by Piotrowski and Sładkowski [15–17], Newcomb’s paradox by Piotrowski and Sładkowski [<xref ref-type="bibr" rid="scirp.1438-ref18">18</xref>], and so on.</p><p>In this paper, we study Newcomb’s problem. Newcomb’s problem is a thought experiment between two players, Alice and Bob. Alice is a common human being. On the other hand, Bob may be a wizard having an ability to predict the future, or not. Bob can predict Alice’s will if he is a wizard. Then, the problem is as follows:</p><p>Newcomb’s problem: Bob prepares two boxes, Box<sub>1</sub><sub> </sub>and Box<sub>2</sub>, and Alice can select either Box<sub>2</sub> or both boxes. Box<sub>1</sub> contains $1. Box<sub>2</sub> contains $1,000 only if Alice selects only Box<sub>2</sub>; otherwise Box<sub>2</sub> is empty($0). Which is better for Alice?</p><p>No one knows the answer except Bob. Namely, there exists no best classical strategy. Therefore, this problem is called Newcomb’s paradox. We show some quantum strategies for Newcomb’s paradox by using entanglement.</p><p>It is thought that entanglement is essential as the main power of quantum information and many results mentioned above also have used entanglement effectively. First, we show some basic quantum strategies with entanglement. In the other related studies mentioned above, each player operates only each assigned qubit although the states are entangled. On the other hand, our proposed strategies operate not only one qubit but also states between two qubits. Consequently, we show that our quantum strategies with entanglement are more powerful than classical ones.</p><p>Finally, we show some quantum strategies for Newcomb’s paradox. Piotrowski and Sładkowski showed a quantum solution for Newcomb’s paradox by using Meyer’s strategy [<xref ref-type="bibr" rid="scirp.1438-ref18">18</xref>]. Newcomb’s paradox is whether a player can predict another player’s will. We also study this problem by applying our strategies. Then, we obtain positive results. That is, in some case, a player can predict another player’s will.</p><p>The remainder of this paper has the following organization. In Section 2, first, we define notations and basic operations used in this paper. Moreover, as the tools of our quantum strategies, we show two fundamental lemmas with relation to entanglement. In Section 3, we denote two types of two-player zero-sum games. We then show that in these games, each player cannot win with certainty with classical strategies but one side player can win with certainty with quantum ones. In Section 4, we study Newcomb’s paradox. We modify this problem and show some quantum strategies to it by using the results in Section 3. Finally, in Section 5, we provide some concluding remarks.</p></sec><sec id="s2"><title>2. Preliminaries</title><p>In this section, first, we define some notations used in this paper. Let <img src="4-8601005\23e89fd4-2871-44d1-a895-5eaa0d205f06.jpg" /> be a bitwise exclusive-or operator, i.e.,<img src="4-8601005\c05374d0-1eae-42f2-a4ac-429acef0e0f4.jpg" />. Let <img src="4-8601005\2e5d0f16-e147-4cde-b560-cd113f091fc2.jpg" /> be the negation of a bit <img src="4-8601005\dbcf68c5-dfd1-4182-a5a0-4d6169cafdd9.jpg" /> for<img src="4-8601005\02a33b23-078e-47a6-a394-6ca038087d41.jpg" />, i.e.,<img src="4-8601005\e4c46cba-6e8d-499a-80cc-83fb9b3d45b2.jpg" />. Moreover, let <img src="4-8601005\4e9292b1-12ca-44de-b16e-7f4441045a0b.jpg" /> be the inner product of <img src="4-8601005\c4ab4d18-9f1a-45a5-9532-1fec75e07c12.jpg" /> and<img src="4-8601005\f15afce9-21b2-4a63-92a4-00a906e2fe0b.jpg" />, where <img src="4-8601005\e8217858-fdbe-4846-a478-16531eac33ba.jpg" /> and <img src="4-8601005\9c46183f-e1b8-4286-84f5-d5d6832a61b1.jpg" /> for <img src="4-8601005\f447db1e-f7ca-4fc6-9d0a-bd0d89af5f9f.jpg" /> (<img src="4-8601005\ddbad9c9-ff65-41a6-a0e5-7aba5f574330.jpg" />).</p><p>Next, we define some basic operations. Let <img src="4-8601005\a794652c-553c-4f71-bca9-44eb80876f00.jpg" /> and<img src="4-8601005\17b35239-be4e-436a-ba38-272b63b9ad94.jpg" />, where <img src="4-8601005\3bdd9439-b3b6-4fcf-8ca1-4c7c7278e313.jpg" /> is Dirac notation and <img src="4-8601005\6286c501-8014-43b3-8ab3-926b9344b3c0.jpg" /> is the transposed matrix of a matrix<img src="4-8601005\b0dee83e-c20f-4b20-a1e2-c334bea3177e.jpg" />. Let <img src="4-8601005\67f575d9-4eeb-413b-b010-db1194e748aa.jpg" /> be the <img src="4-8601005\8a734d96-6751-447a-9365-44184bffb904.jpg" /> identity matrix. This operation means no operation. A Walsh-Hadamard operation <img src="4-8601005\1d5c53da-2933-431f-8171-e2ca7cec4e88.jpg" /> is</p><p><img src="4-8601005\f1ad30db-f573-4797-81c2-3705b73d666b.jpg" /></p><p>(<img src="4-8601005\f1e74135-0e28-4fcc-b42d-b7b5660ae51c.jpg" />and<img src="4-8601005\0ac4ec56-eb01-4da5-b4b0-b8ca181b9b8a.jpg" />). Note that<img src="4-8601005\6e58e41f-80bc-494a-87ac-7c6e93265b3a.jpg" />. This operation is used when we make a superposition of states. As an operation used when they flip a coin classically, players use an operation<img src="4-8601005\4bd49904-9e2f-4766-a252-02cccfa1c204.jpg" />,</p><p><img src="4-8601005\15fea09d-fe8f-4653-877c-19f2545d9eb2.jpg" /></p><p>(<img src="4-8601005\277a8d52-b58e-4a49-8ba6-73783783c9f7.jpg" />and<img src="4-8601005\f79f2966-8828-4502-8897-0ffd494295f6.jpg" />). We also define a phase-shift operation <img src="4-8601005\ab599f52-0392-4edb-a116-19b333dbe761.jpg" /> by</p><p><img src="4-8601005\a25ba683-4cb4-4b7b-b760-fdb8606f4024.jpg" /></p><p>(<img src="4-8601005\39b505dc-3599-4961-82d4-b814acc7afaf.jpg" />and<img src="4-8601005\b0e40ef9-23dc-4f31-b764-d526d71bced6.jpg" />), where<img src="4-8601005\dd73405d-8ffa-4893-9caf-e6500f85ba17.jpg" />.</p><p>Moreover, we define an operation between two qubits. Here, we denote an <img src="4-8601005\06cbd2b8-3f3e-47a2-8722-fa382701b408.jpg" />-qubit state by<img src="4-8601005\68ce7c50-7d51-4ebb-a540-84ff53df0c64.jpg" />, where <img src="4-8601005\8ba1a3b0-6d3f-44d1-a2f7-9f8a0a1fc2e9.jpg" /> is a tensor product. Then, let <img src="4-8601005\0cebba8d-f5bb-4d69-a54e-cce071ab7d22.jpg" /> be a Controlled Not gate,</p><p><img src="4-8601005\b17c7138-9881-4f98-8379-cb7149cb1410.jpg" /></p><p>(<img src="4-8601005\3505aaa2-2aa7-4d01-b043-555a302dbcac.jpg" />, where the first bit <img src="4-8601005\58a88a0b-ad20-4996-83f2-80e995947b6b.jpg" /> is the controlled bit and the second bit <img src="4-8601005\8ad7ada2-5f4c-46f9-9506-dcffc574b6c9.jpg" /> is the target bit). We denote the operation by <img src="4-8601005\eedac724-cb09-4071-a15a-9cae6db2135a.jpg" /> when the <img src="4-8601005\d72c9d64-2077-46cb-8e60-f360569150f3.jpg" />-th bit is the controlled bit and the <img src="4-8601005\079e966d-bf51-4bcb-be10-251715a97361.jpg" />-th bit is the target bit.</p><p>Finally, we show two basic results operating entangled states. These results can be used as tools of making a specific state in order that one side player always wins games by using our quantum strategies mentioned in the following sections.</p><p>Lemma 2.1 Let</p><p><img src="4-8601005\fff8ca18-a8cd-46b3-bc5b-6900a4fb9ba5.jpg" /></p><p>be a <img src="4-8601005\957f2e5c-5e07-46d7-82c7-fed243c1e4be.jpg" />-qubit entangled state, where <img src="4-8601005\7ed8687c-2115-463e-892c-1459f4661d0a.jpg" /> (<img src="4-8601005\db0f8faa-8e14-4af8-994d-aa625f98b394.jpg" />) and<img src="4-8601005\742f16b5-648e-4aa1-ac9f-0c7942d8413f.jpg" />. Then,</p><p><img src="4-8601005\63a8de7a-ad16-4340-a1ca-ab8368c72488.jpg" /></p><p>when we apply the Walsh-Hadamard operation <img src="4-8601005\7080b6f5-20df-443e-94e7-945842027c97.jpg" /> to all the qubits of<img src="4-8601005\7240b5fc-4aa3-4d2f-b213-4d04961e3d7a.jpg" />.</p><p>Proof. When we apply <img src="4-8601005\1e01a115-0e33-41d0-bc45-cc288b054406.jpg" /> to all the qubits of<img src="4-8601005\c6513e09-610a-4e64-a7ca-53db1e8ce294.jpg" />,</p><p><img src="4-8601005\591cf252-e0ea-4f0a-8f50-3d4623ca1580.jpg" /></p><p>where<img src="4-8601005\35eed8eb-136d-4e86-a151-e9c8a57f91a8.jpg" />. Then, the statement of this lemma is then satisfied.</p><p>This lemma means that a player can obtain a state <img src="4-8601005\9ff49f3b-562a-41e0-9abd-15d8f954c802.jpg" /> satisfying <img src="4-8601005\59e3162b-102c-4f14-8ccc-f6bd0efd155f.jpg" /> if he makes the state <img src="4-8601005\42e27589-0475-47bc-8eef-2d5058ddbd67.jpg" /> and that he can obtain a state <img src="4-8601005\857b2f51-fcf3-4623-9b63-7047ebfd7262.jpg" /> satisfying <img src="4-8601005\dacf8163-dc30-4658-9090-3c36f393f5c9.jpg" /> if he makes the state<img src="4-8601005\ef82e04c-e6c0-4c13-adc6-bf066f4b6b3e.jpg" />.</p><p>Next, we show a result used as a player’s quantum strategy when another player flips coins classically.</p><p>Lemma 2.2 Let <img src="4-8601005\28533e0a-24c5-49af-afa1-c5358ed267d4.jpg" /> be the <img src="4-8601005\5110c62e-2075-4114-a0e5-5a87f8b0cf2a.jpg" />-qubit entangled state of Lemma&#160;2.1, and let</p><p><img src="4-8601005\b9e35bac-d2ed-4b53-a87a-4bc44a539a18.jpg" /></p><p>Now, let the operation <img src="4-8601005\c59970f2-973d-4294-be00-5966a965623f.jpg" /> be applied to some qubits of<img src="4-8601005\d3dff824-c6a5-492c-abe6-0e0a10c68641.jpg" />, i.e.,</p><p><img src="4-8601005\2ee9e0bb-b855-496c-9fb4-f5c32b8fafe6.jpg" /></p><p>where<img src="4-8601005\b227f87d-6087-4b1c-aa2b-d91777bca8c0.jpg" />. Note that <img src="4-8601005\53aee51c-a8e7-4f70-b6bc-c3f0f52dc3a6.jpg" /> if <img src="4-8601005\4eeb6609-09e8-445f-b9ac-9f72fb87fafd.jpg" /> has been executed. Then,</p><p><img src="4-8601005\8196d2df-a0f8-47d4-b0d0-9cd5b15ae4fe.jpg" /></p><p>Proof. When we apply <img src="4-8601005\2acb444e-529d-407d-af89-db54755ffcbd.jpg" /> to all the qubits of<img src="4-8601005\bb480464-9ce6-4c79-bc43-c949386b935b.jpg" />,</p><p><img src="4-8601005\91881617-ffd6-47a6-a355-cd9147d4f34d.jpg" /></p><p>where the last expression is obtained by noting that except for the states corresponding to <img src="4-8601005\af5c1a54-1846-4da4-8f45-dd49ad3f6c67.jpg" /> and<img src="4-8601005\7236620e-af5a-4f9d-b3d1-80e58d18484d.jpg" />, the states vanish. The statement of this lemma is then satisfied.</p></sec><sec id="s3"><title>3. Quantum Strategies Using Entangled States</title><sec id="s3_1"><title>3.1 Strategic Games</title><p>We denote a strategic game <img src="4-8601005\4bcb8a2a-d475-4c74-8c59-992cded2587c.jpg" /> as<img src="4-8601005\f8991b6a-1ee5-49e2-8626-41af45cee866.jpg" />, where</p><p><img src="4-8601005\213adec3-2086-4da8-8538-efd01e55056a.jpg" />is the set of players,</p><p><img src="4-8601005\1d9a6dfc-a3ed-4bae-9853-26a2e67cd80d.jpg" />is the set of strategies of player<img src="4-8601005\9ee12807-39fa-461e-9c83-dccde2db7eaf.jpg" />, and</p><p><img src="4-8601005\5250b399-aecf-43e0-aead-04425108f9ca.jpg" />is the payoff function of player<img src="4-8601005\5518b5a4-cadf-43bf-a747-4653428d8aaf.jpg" />, i.e., <img src="4-8601005\dfaed68e-a327-473d-9bbd-baede51ca29f.jpg" />(the set of real numbers).&#160;</p><p>The game <img src="4-8601005\0f79bbbe-ee34-4500-a93e-53d9ef6c9098.jpg" /> can be given also by a matrix shown in <xref ref-type="fig" rid="fig1">Figure 1</xref>. For more details, we shall refer the reader to the books by, e.g., references [22,23]. For two players, <img src="4-8601005\f17a57ea-01d4-4a69-83ed-76fd18d0f3fe.jpg" />={Alice, Bob}, the set of Alice’s strategies is<img src="4-8601005\08b420be-bab8-4585-a356-d40b8cb0a9c4.jpg" />, and the set of Bob’s strategies is<img src="4-8601005\cc916a9c-54f7-4146-970c-ca463808d84d.jpg" />. Then, each value of Alice’s payoff function is<img src="4-8601005\29a0f4d2-64a8-4634-b8f6-a15ea535923d.jpg" />, and each value of Bob’s payoff function is<img src="4-8601005\dd8085df-62bd-4622-b943-abedf9761471.jpg" />, where<img src="4-8601005\df1b4382-edf8-4293-b9ef-4132e7fad78e.jpg" />. If <img src="4-8601005\4142ecc2-94eb-4acc-b36e-d8fdb5735891.jpg" /> for any<img src="4-8601005\8d7297c9-a2d2-4c69-8058-43dd83c232f4.jpg" />, <img src="4-8601005\88dab7fc-d17f-4ac2-a71a-ca363698d761.jpg" />can be omitted. In certain circumstances, we represent a value of payoff function not as a real number but as some players’ win or loss, and so on. In addition, quantum strategies are permitted any unitary matrices, i.e., we can make a super</p><p>position of some fundamental strategies by quantum strategies.</p><p>Now, let us formalize PQ Penny Flip. The set of players is<img src="4-8601005\e49071e1-0148-451d-9fd3-ede175522efd.jpg" />, the set of Piard’s strategies is<img src="4-8601005\b4aa0efc-d130-45bd-a6af-8be369b7f8a1.jpg" />, the set of Q’s strategies is<img src="4-8601005\159f6a28-adcf-4726-9c55-f95dd06493a6.jpg" />. Operation <img src="4-8601005\ebf077f5-3e34-4dbb-af13-cffc11b96567.jpg" /> means no coin flip, and operation <img src="4-8601005\596fc30b-4c5b-422b-b60d-b52d144bd530.jpg" /> means a coin flip. The payoff matrix is shown in <xref ref-type="fig" rid="fig2">Figure 2</xref>. “Unchanged”/”Changed” means that the state of the coin is finally unchanged/changed. Meyer showed a quantum strategy that Q always wins [<xref ref-type="bibr" rid="scirp.1438-ref4">4</xref>].</p><p>Next, we show a simple solution for the Battle of the Sexes using an entangled state. This idea leads to the results of the following subsections. The payoff matrix is shown in <xref ref-type="fig" rid="fig3">Figure 3</xref>. Alice prefers movie to soccer, and Bob prefers soccer to movie. However, both prefer having a date.</p><p>The strategy is as follows. Alice and Bob share the following entangled state:</p><p><img src="4-8601005\22086a8d-d33d-41ad-8bc3-11ece216b60d.jpg" /></p><p>where Alice has a first qubit and Bob has a second qubit. In deciding either soccer or movie, both measure the state. Alice/Bob selects soccer when the outcome of the bit is 0, otherwise she/he selects movie. Note that if Alice’s outcome is 0, Bob’s outcome is also 0, and vice versa. Namely, the probability of selecting (Soccer, Soccer)/(Movie,Movie) is 1/2, and the probability of selecting (Soccer,Movie)/(Movie,Soccer) is 0. Therefore, they can have a date with certainty. Moreover, for example, if the payoff of (Soccer,Soccer) is (4,6) instead of (3,5), the probability of selecting (Soccer,Soccer) becomes greater than that of (Movie,Movie) by preparing</p><p><img src="4-8601005\f293ce2b-2998-4b96-b837-4d597d1f1dca.jpg" /></p><p>as the entangled state, where <img src="4-8601005\3cc6407b-335a-4819-9e21-954406d731c2.jpg" /> is complex numbers satisfying <img src="4-8601005\3f410f0f-2c3e-4566-a6db-47afe3ea35ba.jpg" /> and<img src="4-8601005\32306bbc-1de9-46f1-851a-6687a056532f.jpg" />.</p></sec><sec id="s3_2"><title>3.2 <img src="4-8601005\7abb12ef-b9b6-41f4-9fa2-c2d72ce13ea8.jpg" />-Coin Even-Odd Games</title><p>First, we denote a zero-sum game using <img src="4-8601005\48562a71-eef6-4f0c-9478-6da3d84f2291.jpg" /> coins between two players, <img src="4-8601005\19c37dc3-abd1-4610-a281-28de34b22b5a.jpg" />={Alice, Bob}. Throughout this paper, suppose that Alice is constrained to play classically, i.e.,<img src="4-8601005\3ed409bd-6a21-4139-a220-f7d7f96ea507.jpg" />. Then, we show that neither Alice nor Bob can win the game with certainty.</p><p>With only classical strategies but Bob can win it with certainty with our quantum strategy. Here, let H and T</p><p>represent the head and tail of a coin, respectively.</p><p><img src="4-8601005\c68febae-ba69-4316-add9-cb51249b8ab8.jpg" />-Coin Even-Odd Check: First, Alice prepares <img src="4-8601005\4f274c22-b6a5-4abf-a9da-132afe1d6143.jpg" /> coins and puts them into a box in the state of all the coins being heads, i.e., (H, H,…, H). Suppose that any player cannot see the inside of the box. Next, Bob flips some coins(or not), Alice flips some coins(or not), and Bob flips some coins(or not). Finally, they open the box. Alice wins if the number of H is odd; otherwise Bob wins if the number of H is even.</p><p>Because we can regard this problem as whether Bob can predict Alice’s strategy, the payoff matrix can be also shown as <xref ref-type="fig" rid="fig4">Figure 4</xref>. It is obvious that if they use only classical strategies, neither Alice nor Bob can win the game with certainty. However, if Bob uses a quantum strategy, he can win the game with certainty. Now, we show a quantum strategy for this game with which Bob wins with certainty.</p><p>Theorem 3.1 For <img src="4-8601005\a7872def-1937-4d9c-9e86-9e27d4463677.jpg" />-Coin Even-Odd Check, there exists a quantum strategy with which Bob wins with certainty.</p><p>Proof. We denote H and T by <img src="4-8601005\f6aa39a2-0470-4917-82d5-02592b7bf89c.jpg" /> and<img src="4-8601005\e769f3f4-b28d-4d03-aa5b-f409531c42b1.jpg" />, respectively. This means that Alice prepares<img src="4-8601005\e169d00e-bc81-43ab-a124-53413fee6de2.jpg" />. First, Bob executes the following operation.</p><p><img src="4-8601005\7ec999aa-49a6-4b61-981b-cfa364dce94e.jpg" /></p><p>Next, Alice flips the coins using the operations <img src="4-8601005\a940c2ed-f996-4263-a831-b2a98dd66b92.jpg" /> and<img src="4-8601005\21b2a5c7-c2a9-4549-95d0-11d053adbe23.jpg" />, because she can only execute classical strategies. Then, the state becomes</p><p><img src="4-8601005\3cc853b1-712d-49e2-a9a8-8e0e7672f08b.jpg" /></p><p>where <img src="4-8601005\51e5df8c-2d38-4645-9cc3-2e8c734205f0.jpg" /> (<img src="4-8601005\7c45f783-e577-4aad-b269-ed28f98f85b5.jpg" />).</p><p>Finally, by using Lemma 2.1, Bob obtains</p><p><img src="4-8601005\23ee87ee-7e06-47a9-9493-3e1a899471a0.jpg" /></p><p>Thus, Bob can obtain the bits <img src="4-8601005\ddbb0d86-8435-4f8f-8005-a2edf1bb58c5.jpg" /> satisfying<img src="4-8601005\1a278a43-d7e0-4bbc-a103-6eec89daa2c0.jpg" />. This means that the number of H is even and he can win the game with certainty.</p><p>We can prove the same theorem by using the Meyer’s quantum strategy [<xref ref-type="bibr" rid="scirp.1438-ref4">4</xref>] for <img src="4-8601005\4d53d8b8-8eb2-44cc-b20c-59ac0f2427e2.jpg" /> coins:</p><p><img src="4-8601005\c9e57ff7-a794-4782-9ea1-0914568a95d9.jpg" /></p><p>Therefore, we next denote a generalized even-odd game such that Bob cannot win with certainty by using only Meyer’s strategy but can win with certainty with our strategy.</p><p><img src="4-8601005\f94a71be-3bb8-46e1-9b86-c8e5bc152f3c.jpg" />-Coin Even-Odd Check(G): First, Alice prepares <img src="4-8601005\21eeba4a-bb97-47d8-8fb7-e866f711621c.jpg" /> coins and puts them into a box in the state of (H,D<img src="4-8601005\926168bf-c72f-4ea3-92ba-4b5b2da49b6e.jpg" />,…, D<img src="4-8601005\2c52ef9a-81c5-4127-b1e0-7bbd72f2ded7.jpg" />), where <img src="4-8601005\0cc51031-60ea-4a3f-af67-9ab86633b5cf.jpg" /> (<img src="4-8601005\986b98c9-85db-4a9d-9665-615065b59afe.jpg" />). Suppose that any player cannot see the inside of the box and that Bob does not know D<img src="4-8601005\50903406-8914-49a7-ac36-ee3328bee6fe.jpg" />. Next, Bob flips some coins(or not), Alice flips some coins(or not), and Bob flips some coins(or not). Finally, they open the box. Alice wins if the number of H is odd; otherwise Bob wins if the number of H is even.</p><p>The payoff matrix of this problem can be shown as same as <xref ref-type="fig" rid="fig4">Figure 4</xref>. Also in this case, it is obvious that if they use classical strategies, neither Alice nor Bob can win the game with certainty. Moreover, because Meyer’s strategy uses the property of:</p><p><img src="4-8601005\de432b21-afe4-49d3-a3c0-fda12b0c315e.jpg" />Bob must know the initial state of all the coins in order to win the game. However, if he uses our quantum strategy, Bob can win the game with certainty. Now, we show a quantum strategy for this game with which Bob wins with certainty.</p><p>Theorem 3.2 For <img src="4-8601005\abba8a06-c4ee-47e9-972f-9130bc682e4a.jpg" />-Coin Even-Odd Check(G), there exists a quantum strategy with which Bob wins with certainty.</p><p>Proof. Also in this case, we denote H and T by <img src="4-8601005\4ece4471-b4b9-4874-abff-9b89b789417d.jpg" /> and<img src="4-8601005\43b8f4c5-2dce-48b2-9658-16a4af9b80e3.jpg" />, respectively. Therefore, Alice prepares<img src="4-8601005\702ad242-f836-4aca-bfb5-5a9a944f0c04.jpg" />, where <img src="4-8601005\5c9d25af-8075-456e-b253-c812ac4c1633.jpg" /> (<img src="4-8601005\5daa662a-451f-4317-87a2-0041d389d680.jpg" />). First, Bob executes the following operation:</p><p><img src="4-8601005\d5f82a1a-f9da-4b80-a4bb-8c6d273cf3a9.jpg" /></p><p>Next, Alice flips the coins using the operations <img src="4-8601005\aa48c1e5-2a90-4996-8326-62d722153bcd.jpg" /> and<img src="4-8601005\4aa36ebe-a2ae-42f1-8b40-c09c3f3a1786.jpg" />. Then, the state becomes</p><p><img src="4-8601005\595e8d59-7ccd-41c2-8f87-30dce2140e6a.jpg" /></p><p>Finally, by using Lemma 2.1, Bob obtains</p><p><img src="4-8601005\97a0894e-50f7-46c9-9594-da774cbbb3c0.jpg" /></p><p>Thus, Bob can obtain the bits <img src="4-8601005\03405a50-af62-48a0-a9d0-39bf58fd3049.jpg" /> satisfying<img src="4-8601005\193213b7-cca8-4dd9-a602-cb6bdf7e418f.jpg" />. This means that the number of H is even and he can win the game with certainty.</p><sec id="s3_2_1"><title>3.3 <img src="4-8601005\e41b99d5-20ce-445a-8afe-40a443daaebd.jpg" />-Coin Flipping Games</title><p>Next, we denote zero-sum games between two players modifying the games in the previous subsection and show that neither Alice nor Bob can win the games with certainty with classical strategies but Bob can win them with certainty with our quantum strategy.</p><p><img src="4-8601005\09bc5a5e-262c-44d4-82c5-c399884d3b9e.jpg" />-Coin Flip: First, Alice prepares <img src="4-8601005\67e9c9b6-d8b6-47fa-92d6-bfc2a6fd399b.jpg" /> coins and puts them into a box in the state of all the coins being heads, i.e., (H, H,…, H). Suppose that any player cannot see the inside of the box. Next, Bob flips some coins(or not), Alice flips <img src="4-8601005\748950ac-126e-4881-b17f-5491a9136ccc.jpg" /> coins(<img src="4-8601005\3591688d-3f2c-43f5-bede-52c5b26f6181.jpg" />) under keeping the value of <img src="4-8601005\ed1bc252-10dd-4b0a-a1f7-7aabf5e48ebf.jpg" /> secret from Bob, and Bob flips some coins(or not). Finally, they open the box. Then, Bob wins if all the coins are in the following state: the state of the coins is (H, H,…, H) if <img src="4-8601005\54501002-9817-4413-85a0-c2f0842520ae.jpg" /> is even, or the state of the coins is (T,H,…, H) if <img src="4-8601005\2a070d01-40cf-4838-932e-cc51f73122e0.jpg" /> is odd. Otherwise Alice wins.</p><p>We show the payoff matrix in <xref ref-type="fig" rid="fig5">Figure 5</xref>. We can regard also this problem as whether Bob can predict Alice’s strategy, It is obvious that if they use classical strategies, neither Alice nor Bob can win the game with certainty. Moreover, by the same reason in the previous subsection, Bob cannot also win the game with certainty even if Meyer’s strategy is used. However, if Bob uses our quantum strategy, he can win the game with certainty. Now, we show a quantum strategy for this game with which Bob wins with certainty.</p><p>Theorem 3.3 For <img src="4-8601005\13006a50-ded2-47e9-8dbb-c3f69d2feb12.jpg" />-Coin Flip, there exists a quantum strategy with which Bob wins with certainty.</p><p>Proof. Alice prepares<img src="4-8601005\7432287a-ec91-4822-8f1c-2d35755f7c8b.jpg" />. First, Bob executes the following operation.</p><p><img src="4-8601005\b38e2e30-4ca8-40ee-a143-c8ec12110ca1.jpg" /></p><p>Next, Alice flips <img src="4-8601005\dd8dd239-5fc2-481a-938c-fae2effa51c3.jpg" /> coins using the operation<img src="4-8601005\16786c9a-6c69-4ce9-88f6-803f5f12b952.jpg" />.</p><p>Then, the state becomes</p><p><img src="4-8601005\24e02b31-be03-45e6-b1a9-aed540f9d1d2.jpg" /></p><p>where <img src="4-8601005\c12acf45-71ad-4026-a6e8-a45808138a33.jpg" /> (<img src="4-8601005\074d1e62-568c-4b3b-8adb-c169a4370bd5.jpg" />) and<img src="4-8601005\74d84814-1d55-4aa7-8d9a-fc61a7303854.jpg" />.</p><p>By using Lemma 2.2, Bob obtains</p><p><img src="4-8601005\fc0baec8-30bb-4c47-b18d-405676ca1fe7.jpg" /></p><p>Moreover, he executes the following operation.</p><p><img src="4-8601005\ec3f9d74-cc5c-4e5a-b02d-201b26c8da1f.jpg" /></p><p>Then, <img src="4-8601005\823e45ff-3601-4956-97af-c816eb766793.jpg" />if <img src="4-8601005\4dd5364e-c6b6-4954-a591-323546ddf1b4.jpg" /> is even; otherwise <img src="4-8601005\3c3bd766-e273-4e2f-b65f-16e9f1c983c3.jpg" /> if <img src="4-8601005\7616c7d3-b087-4d57-a352-3e3ffb9e7ae5.jpg" /> is odd. Therefore, Bob can win the game with certainty.</p><p>Next, we denote a generalized <img src="4-8601005\3e8ef77c-6101-492c-a927-33977e1a8bc0.jpg" />-coin flipping game.</p><p><img src="4-8601005\05d9f222-322e-4e38-a1dd-43962cd47f42.jpg" />-Coin Flip(G): First, Alice prepares <img src="4-8601005\6b5c19ca-a075-4c78-9cc0-83e31384c3e3.jpg" /> coins and puts them into a box in the state of (H,D<img src="4-8601005\a887d874-3e6a-4c59-8a73-3046b5a07b75.jpg" />,…, D<img src="4-8601005\1dca69b2-d30b-450e-b134-8bb821af3686.jpg" />), where <img src="4-8601005\a86a26cf-3d95-46b5-8c05-4dd350c208d1.jpg" /> (<img src="4-8601005\a2eb5184-d2a7-4c0f-8524-f3e0e2ad7633.jpg" />). Suppose that any player cannot see the inside of the box and that Bob does not know D<img src="4-8601005\9d8e769a-5820-44af-b909-cb445c76b36e.jpg" />. Next, Bob flips some coins(or not), Alice flips <img src="4-8601005\ebc35dcf-75e2-4cec-a526-17cb13aca461.jpg" /> coins(<img src="4-8601005\9695c8a9-0f7a-402d-b4af-2703884588a6.jpg" />) under keeping the value of <img src="4-8601005\7146585b-38eb-44e8-a1cb-1f5555ca61b0.jpg" /> secret from Bob, and Bob flips some coins(or not). Finally, they open the box. Then, Bob wins if all the coins are in the following state: the state of the coins is (H,D<img src="4-8601005\e6a54b6f-710d-4417-9a28-10b0a22eb742.jpg" />,…, D<img src="4-8601005\343f60cb-e8f1-4a33-8d5e-e29894b6dfe4.jpg" />) if <img src="4-8601005\67a438a6-f019-43e8-a890-290434edc99f.jpg" /> is even, or the state of the coins is (T,D<img src="4-8601005\32946030-80a4-4d66-b6f8-d6f5d7f1de11.jpg" />,…, D<img src="4-8601005\c248089c-9695-4ac6-b813-971dc8c6a1ba.jpg" />) if <img src="4-8601005\f3893d02-3c84-4223-92ec-06a448a6801d.jpg" /> is odd. Otherwise Alice wins.</p><p>We show the payoff matrix in <xref ref-type="fig" rid="fig6">Figure 6</xref>. Also in this case, it is obvious that if they use classical strategies, neither Alice nor Bob can win the game with certainty and that Bob cannot win the game with certainty even if Meyer’s strategy is used. However, if Bob can use our quantum strategy, he wins the game with certainty. If Bob wishes to know only whether <img src="4-8601005\5432ed4b-8cce-4e44-b17b-d1811d8d0bab.jpg" /> is even or odd, we can easily construct the following protocol.</p><p><img src="4-8601005\7a373f6f-322a-4ab1-a930-04ed00feec83.jpg" /></p><p>where <img src="4-8601005\3d3a3730-6860-473b-8e72-95695a96bcd6.jpg" /> (<img src="4-8601005\895b9cad-aa30-4945-a8a7-ae41d6f42747.jpg" />).</p><p>Now, we show a quantum strategy for this game with which Bob wins with certainty.</p><p>Theorem 3.4 For <img src="4-8601005\943c3daa-b378-4256-82df-849ec56346bf.jpg" />-Coin Flip(G), there exists a quantum strategy with which Bob wins with certainty.</p><p>Proof. Alice prepares<img src="4-8601005\90dcbf1b-17eb-4ed1-8976-4f77e41d34d5.jpg" />, where <img src="4-8601005\b62e2978-0f0a-415d-9d15-3f2698998aa7.jpg" /> (<img src="4-8601005\e0b38bf0-add7-42e0-9c6e-9baa674e72c8.jpg" />). First, Bob executes the following operation.</p><p><img src="4-8601005\519c5dba-30a8-4e4b-a90f-91fffa9a3f83.jpg" /></p><p>where <img src="4-8601005\219e296e-50bf-4c0a-9b71-e736863a245f.jpg" /> and<img src="4-8601005\3e91e750-7ee3-4830-b1a6-078dc19c1978.jpg" />.</p><p>Next, Alice flips <img src="4-8601005\9b749bf0-afe4-42af-888e-482efe12c536.jpg" /> coins using the operation<img src="4-8601005\df7577ee-c2df-4182-9ce7-5f61324c82fd.jpg" />. Then, the state becomes</p><p><img src="4-8601005\cd3acb0f-0e2f-48b9-9af6-fb0e88a10a1f.jpg" /></p><p>where <img src="4-8601005\cc580b44-5076-409b-85c2-80421e22e25c.jpg" /> (<img src="4-8601005\707f8acb-ba63-46f4-812a-0d0cf432184c.jpg" />) and<img src="4-8601005\9285c1fa-8788-4376-bd7b-230106588626.jpg" />.</p><p>By using Lemma 2.2, Bob obtains</p><p><img src="4-8601005\13b4e7e7-c8fb-4f6e-b985-eb0f3b033833.jpg" /></p><p>where<img src="4-8601005\7123bd00-77f8-478c-a673-266209f6f412.jpg" />. Moreover, he executes the following operation.</p><p><img src="4-8601005\0ef4d23a-b788-4484-82b4-53d111ed3243.jpg" />&#160;</p><p>Then, <img src="4-8601005\235330e2-4a3f-4bb1-bb0c-47d092b4158d.jpg" />if <img src="4-8601005\7e0628ab-bfb2-4106-bab6-a1361c658a2e.jpg" /> is even; otherwise <img src="4-8601005\d8dff22d-36b0-4fb9-8601-64b2c679ea75.jpg" /> if <img src="4-8601005\15fbb011-7692-42b8-9bf6-a5c7e96aca22.jpg" /> is odd. Therefore, Bob can win the game with certainty.</p><p>Finally, we denote a game combining the Even-Odd game and the Flipping game.</p><p>Extended <img src="4-8601005\1be75411-1d6c-48f4-9d3b-3a94c2174589.jpg" />-Coin Flip: First, Alice prepares <img src="4-8601005\37bb028d-1895-46a9-9be7-7bf057994f6a.jpg" /> coins and puts them into a box in the state of all the coins being heads, i.e., (H, H,…, H). Suppose that any player cannot see the inside of the box. Next, Bob flips some coins(or not), Alice flips <img src="4-8601005\d4baa6af-76a2-403c-a64b-3c70e8aa3d40.jpg" /> coins (<img src="4-8601005\ee497c2c-ef61-4187-850c-8e6254bd58d7.jpg" />) under keeping the value of <img src="4-8601005\a9554a6f-f012-4c55-878c-118ae24059cd.jpg" /> secret from Bob, and Bob flips some coins(or not), where for some positive even number<img src="4-8601005\009a5c5c-56f4-4a2f-bcd1-6f7c75f97160.jpg" />,<img src="4-8601005\b95b5302-5aed-4510-b32a-fd129d7c5e81.jpg" /> and <img src="4-8601005\ffa7b5b7-1897-49be-88b7-52717a80c9d2.jpg" /> Finally, they open the box. Then, Bob wins if all the coins are in the following state: the number of H is even if <img src="4-8601005\9c0cd25f-10ac-48c1-bddd-d97d716e795f.jpg" /> is an element in<img src="4-8601005\1426e34e-5eb1-4e0a-b9db-f9dae231d536.jpg" />, or the number of H is odd if <img src="4-8601005\0a68ceb8-5a78-4a1e-bd40-cf36d0321562.jpg" /> is an element in<img src="4-8601005\bd85715d-fad2-4557-bacc-69b1cc2fba6f.jpg" />. Otherwise Alice wins.</p><p>We show the payoff matrix in <xref ref-type="fig" rid="fig7">Figure 7</xref>, and construct a quantum strategy such that Bob wins with certainty.</p><p>Theorem 3.5 For Extended <img src="4-8601005\60f74821-3df8-4e47-a2d7-95b1f2997410.jpg" />-Coin Flip, there exists a quantum strategy with which Bob wins with certainty.</p><p>Proof. Alice prepares<img src="4-8601005\702d3caf-432a-400b-ae26-736e78bcfbf7.jpg" />. First, Bob executes the following operation:</p><p><img src="4-8601005\454e3a96-7d9c-4bf6-bbd2-582c12691a74.jpg" /></p><p>Next, Alice flips <img src="4-8601005\32c68df2-31d0-4e4a-ab8c-834a157ccd22.jpg" /> coins using the operation<img src="4-8601005\058434e6-9edf-43d6-91d4-3ffdb4b196c0.jpg" />. Then, the state becomes</p><p><img src="4-8601005\e625580d-c224-4afc-86aa-cc1885307b14.jpg" /></p><p>where <img src="4-8601005\a694b1b8-c5ed-4fc1-a190-c8a1f064ab4b.jpg" /> (<img src="4-8601005\214a8739-8f75-4c47-92dc-ba53148b6cfc.jpg" />) and<img src="4-8601005\427975c3-58ef-4e22-9ae1-09a186a4abfb.jpg" />.</p><p>Bob applies <img src="4-8601005\ac05aa04-f508-457d-83db-a16b3de8d098.jpg" /> to the state.</p><p><img src="4-8601005\38820647-9aac-49d8-bca1-7000f291be1d.jpg" /></p><p>Moreover, by using Lemma 2.1, Bob obtains</p><p><img src="4-8601005\c8b7de68-1236-400b-889b-f20ca9d84b3c.jpg" /></p><p>if<img src="4-8601005\f4e9c1b0-28b8-46e4-85c5-021b4f8dbfe7.jpg" />; otherwise he obtains</p><p><img src="4-8601005\be1c70c0-24ed-43c3-91ff-50952afe069a.jpg" /></p><p>if<img src="4-8601005\39790cb4-fcd7-4cbc-8d09-942b7e2de1f4.jpg" />.</p><p>Thus, <img src="4-8601005\3eb36e30-11d9-4ad9-a2bf-4bb24da95e97.jpg" />is an element in <img src="4-8601005\c94e7c6f-48f5-4332-bc4f-2e8552ae6f81.jpg" /> and <img src="4-8601005\58c17fe3-a26f-434d-a61d-c966163132c7.jpg" /> if<img src="4-8601005\7e70d759-2308-4a15-a8f5-00c0705f8088.jpg" />; otherwise <img src="4-8601005\f2b3df3b-5e93-4b89-8698-c8fcad7ef3ab.jpg" /> is an element in <img src="4-8601005\e54b7861-279a-444e-8a61-7b4a79c4de0e.jpg" /> and</p><p><img src="4-8601005\f2a7d941-a43e-4ff7-bc8a-5338fd62ed61.jpg" />if<img src="4-8601005\c6054b30-92e6-4b47-8a13-a8fcab024cd5.jpg" />. Therefore, Bob can win the game with certainty.</p><p>When <img src="4-8601005\74430c0b-e439-4f0b-896a-1a9f78c049de.jpg" /> is odd, Bob can always win without operating to <img src="4-8601005\3af4e51a-c1e5-4b04-8c04-97a0bcd1c054.jpg" /> coins. On the other hand, our strategy succeeds even if <img src="4-8601005\d10989d2-b679-4434-b5ac-2845a17d6b72.jpg" /> is even.</p></sec></sec></sec><sec id="s4"><title>4. Applications to Newcomb’s Paradox</title><p>In this section, we study Newcomb’s paradox (Free Will problem) and show some quantum strategies for this problem. A quantum solution of this problem is shown by using Meyer’s quantum strategy by Piotrowski and Sładkowski [<xref ref-type="bibr" rid="scirp.1438-ref18">18</xref>]. We study this problem by using our results in previous section. A problem is as follows:</p><p>Newcomb’s problem: Let Bob have the ability to predict Alice’s will. Now, Bob prepares two boxes, Box<sub>1</sub> and Box<sub>2</sub>, and Alice can select either Box<sub>2</sub> or both boxes. Box<sub>1</sub> contains $1. Box<sub>2</sub> contains $1,000 only if Alice selects only Box<sub>2</sub>; otherwise Box<sub>2</sub> is empty($0). Which is better for Alice?</p><p>The payoff matrix is shown in <xref ref-type="fig" rid="fig8">Figure 8</xref>. The focus of this problem is whether Bob can really predict Alice’s will, or whether Bob can control Alice’s will. Obviously, Alice’s strategy is selecting both boxes if Bob cannot predict Alice’s will. Now, we modify this problem as simplified problems. In addition, we observe only strategies on Box<sub>2</sub>.</p><p>Newcomb1: First, Alice decides whether she selects either Box<sub>2</sub> or not, but Bob cannot know her selection. Box<sub>2</sub> contains $1,000 only if Alice selects Box<sub>2</sub>; otherwise Box<sub>2</sub> is empty($0). Can Bob let Alice select her first will even if Alice changes her will after the first selection?</p><p>Theorem 4.1 For Newcomb1, there exists a quantum strategy that can be positively solved.</p><p>Proof. Alice selects a state <img src="4-8601005\7c460804-5fb1-48ef-8daa-261f8f3efde7.jpg" /> if she selects Box<sub>2</sub>; otherwise she selects a state<img src="4-8601005\1ecd6abc-db39-4b18-963b-a40b98bde311.jpg" />. Note that Bob does not know the state. To this state, Bob applies H, Alice applies X if she changes her will, and Bob applies H. Fi</p><p>nally, the state is <img src="4-8601005\99608b84-233f-42ca-8061-8ab0376cdf0e.jpg" /> if her first selection is Box<sub>2</sub>; otherwise it is<img src="4-8601005\a4bec82a-fe5d-4fb8-b041-6cefde5f0ab1.jpg" />. Then, Bob can let Alice select her first will.</p><p>This strategy uses Meyer’s strategy. Moreover, we can also show other proofs by using Theorem 3.3 or Theorem 3.4. For Newcomb1, Bob does not permit Alice’s change. Next, by using <img src="4-8601005\65dd7acc-031d-4405-9b97-114fab61fa12.jpg" />-Coin Flip(G), we modify this problem to problems such that Bob permits Alice’s change.</p><p>Newcomb2: Alice prepares <img src="4-8601005\6f673169-c5f8-4f10-8e09-6b15eea8bd2e.jpg" /> coins and puts them into a box in the state of (H,D<img src="4-8601005\ce1458cc-c2b6-48f6-9909-b59f866fe233.jpg" />,…, D<img src="4-8601005\0d39cdd1-4b2f-4207-b7df-c612c3f520c9.jpg" />), where <img src="4-8601005\6f573912-df48-4d77-a788-3b3a4db5069f.jpg" /> (<img src="4-8601005\d16fe32c-9d9b-4652-8428-7b5ccd1c7534.jpg" />). Next, Bob flips some coins(or not), Alice flips <img src="4-8601005\2346825a-2c42-406e-be89-843c273ba0e0.jpg" /> coins(<img src="4-8601005\7b587648-5b05-42ea-8ebc-ace9429952eb.jpg" />) under keeping the value of <img src="4-8601005\bc457140-5148-4136-afde-8e8bef1fc3f8.jpg" /> secret from Bob, and Bob flips some coins(or not), where let Alice do not change her will if <img src="4-8601005\7f3a4837-c944-4467-a1f6-d5c134d2ac80.jpg" /> is even; otherwise let she change her will. Can Bob know whether <img src="4-8601005\e997992f-9281-42c6-a812-46311eead201.jpg" /> is even or odd?</p><p>Theorem 4.2 For Newcomb2, there exist a quantum strategy that can be positively solved.</p><p>Proof. This result is immediately obtained by Theorem 3.4.</p><p>We can regard this problem as Newcomb’s problem when the value of <img src="4-8601005\4d785099-dbb6-4d18-ad58-cdb724b748c6.jpg" /> is her will, i.e., she selects Box<sub>2</sub> when <img src="4-8601005\a5bfe76a-f636-4487-84e8-d0b6e5407895.jpg" /> is even; otherwise she selects both boxes. Thus, Bob can predict Alice’s will. We can also modify Newcomb’s problem by using <img src="4-8601005\3a0f9599-f5d9-45e9-bdd3-03e538501ee4.jpg" />-Coin Flip and Extended <img src="4-8601005\bc8c98ec-a7f8-4ae6-94dc-0ef7f8fb4852.jpg" />-Coin Flip.</p><p>Next, we denote that we can also modify Alice’s will in Newcomb2 to <img src="4-8601005\7cfc435e-b65c-4ad8-b02d-d72a69579f16.jpg" />-player’s will.</p><p><img src="4-8601005\36fb88e8-03cd-476c-8c3c-ed06053c41fe.jpg" />-Newcomb2: First, <img src="4-8601005\6e6c4a54-c7d0-4adb-844f-ca0d69aeca54.jpg" />players, <img src="4-8601005\9765be85-d360-49a9-89f8-6d1c9862ef43.jpg" />, decide whether they selects <img src="4-8601005\31f08565-dc1d-43ac-bc10-c5af2c0a8e77.jpg" /> or not. They prepare <img src="4-8601005\bca316b7-d3bd-4df1-8d5c-af791f62b757.jpg" /> coins and puts them into a box in the state of (H,D<img src="4-8601005\194d7137-ede3-4bba-a80b-6382c6f8a7a3.jpg" />,…, D<img src="4-8601005\d296e9cc-0152-46a2-aed8-bc2f43188eb4.jpg" />), where <img src="4-8601005\a9943c88-3582-41eb-afa2-4488f8a6d27e.jpg" /> (<img src="4-8601005\3ee0a80a-7c1d-465a-93c5-f1341026f42a.jpg" />), and let player <img src="4-8601005\1710561d-a35a-4627-8505-b14452985d6b.jpg" /> (<img src="4-8601005\f462dab5-19ff-49c7-ab54-12d14c9a002e.jpg" />) deal with the <img src="4-8601005\21b49458-8276-4541-8ed4-0d368cd09576.jpg" />-th coin. Now, Bob flips some coins(or not). Next, each <img src="4-8601005\41b439eb-1110-4492-bc40-99d1e5d29d6d.jpg" /> (<img src="4-8601005\b14473fe-ebff-4550-922e-3a497de3149e.jpg" />) flips his/her coin(or not), where let the number of <img src="4-8601005\0f77adce-d15a-4390-91d3-9e65089e6c0b.jpg" />-player’s flips be <img src="4-8601005\e1467e48-06f7-4902-8b13-9ecc15cc2902.jpg" /> (<img src="4-8601005\c707aebd-4132-4242-b532-48f8d47542bc.jpg" />), and let they do not change their will if <img src="4-8601005\5ae6ea85-becf-43b5-937e-345806b64af9.jpg" /> is even; otherwise let they change their will. Finally, Bob flips some coins(or not). Can Bob know whether <img src="4-8601005\b2499d30-a245-4fa4-aeb5-76eca65eb37b.jpg" /> is even or odd?</p><p>The quantum strategy for Bob is same as Newcomb2.</p><p>Finally, we study the Chicken game. The payoff matrix is shown in <xref ref-type="fig" rid="fig9">Figure 9</xref>. In the Chicken game, the Nash equilibrium points are (Swerve, Drive straight) and (Drive straight, Swerve). By classical strategies, the probability that each player selects either Swerve or Drive straight is 1/2. This means that either (Swerve, Swerve) or (Drive straight, Drive straight) may be selected. On the other hand, by our quantum strategies, either (Swerve, Drive straight) and (Drive straight, Swerve) can be selected with certainty because Bob can predict Alice’s will.</p></sec><sec id="s5"><title>5. Conclusions</title><p>In this paper, we proposed quantum strategies with entanglement for <img src="4-8601005\05b6fc48-ac74-45ad-89ea-ee7e805dff4e.jpg" />-coin flipping games. These games are multi-qubit variations of the quantum strategy by Meyer [<xref ref-type="bibr" rid="scirp.1438-ref4">4</xref>]. One player is constrained to play classically but the other player can use a quantum strategy. We then showed that by using a technique in quantum communication complexity theory, our quantum strategies have an advantage over classical ones. For a player using the quantum strategy, the entanglement is used in order to obtain the information of the enemy and the player can always win the games. Moreover, by rewriting Newcomb’s paradox including multi-player’s will, we also showed that we can use our results as its quantum strategies and that a player can have an ability to predict another player’s will.</p><p>Can Bob always win our games even if both players use quantum strategies? This answer is “No”. Meyer also showed that Q cannot always win if Picard can also use a quantum strategy [<xref ref-type="bibr" rid="scirp.1438-ref4">4</xref>]. In the same reason, Bob cannot always win our games if Alice can also use a quantum strategy. 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