The Monty Hall Problem is not a Probability Puzzle (it's a challenge in mathematical modelling)
TThe Monty Hall Problem is nota Probability Puzzle ∗ (It’s a challenge in mathematical modelling) Richard D. Gill †
12 November, 2010
Abstract
Suppose you’re on a game show, and you’re given the choice of threedoors: Behind one door is a car; behind the others, goats. You picka door, say No. 1, and the host, who knows what’s behind the doors,opens another door, say No. 3, which has a goat. He then says to you,“Do you want to pick door No. 2?” Is it to your advantage to switchyour choice?
The answer is “yes” but the literature offers many reasons whythis is the correct answer. The present paper argues that the mostcommon reasoning found in introductory statistics texts, dependingon making a number of “obvious” or “natural” assumptions and thencomputing a conditional probability, is a classical example of solutiondriven science. The best reason to switch is to be found in von Neu-mann’s minimax theorem from game theory, rather than in Bayes’theorem.
In the above abstract to this paper, I reproduced The Monty Hall Problem,as it was defined by Marilyn vos Savant in her “Ask Marilyn” column in
Parade magazine (p. 16, 9 September 1990). Marilyn’s solution to the prob-lem posed to her by a correspondent Craig Whitaker sparked a controversy ∗ v.4. Revision of v.3 arXiv.org:1002.0651 , to appear in Statistica Neerlandica † Mathematical Institute, University Leiden; during 2010–11 the author is Distin-guished Lorentz Fellow at the Netherlands Institute of Advanced Study, Wassenaar. ∼ gill a r X i v : . [ m a t h . HO ] N ov hich brought the Monty Hall Problem to the attention of the whole world.Though MHP probably originated in a pair of short letters to the editor in The American Statistician by biostatistician Steve Selvin (1975a,b), from1990 on it was public property, and has sparked research and controversyin mathematical economics and game theory, quantum information theory,philosophy, psychology, ethology, and other fields, as well as having becomea fixed point in the teaching of elementary statistics and probability.This has resulted in an enormous literature on MHP. Here I would like todraw attention to the splendid book by Jason Rosenhouse (2009), which hasa huge reference list and which discusses the pre-history and the post-historyof vos Savant’s problem as well as many variants. My other favourite isRosenthal (2008), one of the few papers where a genuine attempt is made toargue to the layman why MHP has to be solved with conditional probability.Aside from these two references, the English language wikipedia page on theMonty Hall Problem, as well as its discussion page, is a rich though every-changing resource. Much that I write here was learnt from the many editorsof those pages, both allies and enemies in the never ending edit-wars whichplague it.The battle among wikipedia editors could be described as a battle betweenintuitionists versus formalists, or to use other words, between simplists versusconditionalists. The main question which is endlessly discussed is whethersimple arguments for switching, which typically show that the unconditional probability that the switching gets the car is 2/3, may be considered rigor-ous and complete solutions of MHP. The opposing view is that vos Savant’squestion is only properly answered by study of “the” conditional probabil-ity that the switching gets the car, given the initial choice of door by theplayer and door opened by the host. This more sophisticated approach re-quires making more assumptions, and that leads to the question whetherthose supplementary conditions are implicitly implied by vos Savant’s words.What particularly interests me, however, is that the conditionalists take ona dogmatic stance: their point of view is put forward as a moral imperative.This leads to an impasse, and the clouds of dust thrown up by what seems areligious war conceal what seem to me to be much more interesting, thoughmore subtle, questions.My personal opinion on the wikipedia-MHP-wars is that they are fightsabout the wrong question. Craig Whitaker, through the voice of Marilynvos Savant, asks for an action, not a probability. I think that game theorygives a more suitable framework in which to represent our ignorance of themechanics of the set-up (where the car is hidden) and of the mechanics ofthe host’s choice, than subjectivist probability.Therefore, though Rosenhouse’s book is a wonderful resource, I strongly2isagree with him, as well as with many other authors, on what the essentialMonty Hall problem is (and that is the main point of this paper). Decidingunilaterally (Rosenhouse, 2009) that a certain formulation is canonical isasking for schism. Calling a specific version original (Rosenthal, 2008) isasking for contradiction. Rosenthal states without any argument at all thatadditional assumptions are implicitly contained in vos Savant’s formulation.Selvin (1975a) did state all those assumptions explicitly but strangely enoughdid not use all of them. His second paper Sevin (1975b) gave a new solutionusing all his original assumptions but the author does not seem to notice thedifference. At the same time, he quotes with approval a simplist solution ofMonty Hall himself, who sees randomness in the choice of the player ratherthan in the actions of the team who prepare the show in advance, and thequiz-master himself. Vos Savant did not use the full set of assumptionswhich others find implicit in her question. Her relatively simple explanationof why one should switch seems to satisfy everyone except for the writersof elementary texts in statistics and probability. I have the impression thatwords like original, canonical, standard, complete are all used to hide thepaucity of argument of the writer who needs to make that extra assumptionin order to be able to apply the tool which they are particularly fond of,conditional probability.One of the most widely cited but possibly least well read papers in MHPstudies is Morgan et al. (1991a), published together with a discussion by Sey-mann (1991) and a rejoinder Morgan et al. (1991b). Morgan et al. (1991a)firmly rebuke vos Savant for not solving Whitaker’s problem as they con-sider should be done, namely by conditional probability. They use only theassumption that all doors are initially equally likely to hide the car; this as-sumption is hidden within their calculations. The paper was written duringthe peak of public interest and heated emotions about MHP which arosefrom vos Savant’s column. It actually contains an unfortunate error whichwas only noticed 20 years later by wikipedia editors Hogbin and Nijdam(2010): if the player puts a non-informative and hence symmetric Bayesianprior on the host’s bias in opening a door when he has a choice, it will beequally likely (for the player) that the host will open either door when he hasthe opportunity. Morgan et al. (2010) acknowledge the error and also repro-duce part of Craig Whitaker’s original letter to Marilyn vos Savant whosewording is even more ambiguous than vos Savant’s.Rosenhouse (2009), Rosenthal (2005, 2008), Morgan et al. (1991a,b, 2010),and Selvin (1975b) (but not Selvin, 1975a) solve MHP using elementary con-ditional probability. In order to do so they are obliged to add mathematicalassumptions to vos Savant’s words, without which the conditional probabil-ity they are after is not determined. Actually, and I think tellingly, almost3o author gives any argument at all why we must solve vos Savant’s questionby computing a conditional probability that the other door hides the car,conditional on which door was first chosen by the player and which openedby the host.For whatever reasons, it has become conventional in the elementary statis-tics literature, where MHP features as an exercise in the chapter on Bayes’theorem in discrete probability, to take it for granted that the car is initiallyhidden “at random”, and the host’s choice, if he is forced to make one, is“at random” too. Morgan et al. (1991a) are notable in only making the firstassumption. Many writers also have the player’s initial choice “at random”too. “At random” is a code phrase for what I would prefer to call completely at random. The student is apparently supposed to make these assumptionsby default, though sometimes they are listed without motivation as if theyare always the right assumptions to make.In my opinion, this approach to MHP is an example of solution drivenscience , and hence an example of bad practise in mathematical modelling.Taking for granted that unspecified probability distributions must be uniformor normal, depending on context, is the cause of such disasters as the mis-carriage of justice concerning the Dutch nurse Lucia de Berk, or the dopingcase of the German skater Claudia Pechstein. Of course, MHP does indeedprovide a nice exercise in conditional probability, provided one is willing tofill in gaps without which conditional probability does not help you answerthe question whether you should stay or switch. Morgan et al. (1991a)’s orig-inal contribution is to notice the minimal condition under which conditionalprobability does give an unequivocal solution.Precisely because of all these issues, MHP presents a beautiful playgroundfor learning the art of mathematical modelling. For me, MHP is the problemof how to build a bridge from vos Savant’s words to a mathematical problem,solve that problem, and translate the solution back to the real world. If weuse probability as a tool in this enterprise, we are going to have to motivateprobabilistic assumptions. We must also interpret probabilistic conclusions .Like it or not, the interpretation of probability plays a role, going bothdirections.Real world problems are often brought to a statistician because the personwith the question, for some reason or other, thinks the statistician must beable to help them. The client has often already left out some complicatingfactors, or made some simplifications, which he thinks that the statisticiandoesn’t need to know. The first job of the consulting statistician is to find outwhat the real question is with which the client is struggling, which may oftenbe very different from the imaginary statistical problem that the client thinkshe has. The first job of the statistical consultant is to undo the pre-processing4f the question which his client has done for him.In mathematical model building we must be careful to distinguish theparts of the problem statement which are truly determined by the backgroundreal world problem, and parts which represent hidden assumptions of theclient who thinks he needs to enlist the statistician’s aid and therefore hasalready pre-processed the original question so as to fit in the client’s pictureof what a statistician can do. The result of a statistical consultation mightoften be that the original question posed by the client is reformulated, andthe client goes home, happier, with a valuable answer to a more meaningfulquestion than the one he brought to the statistician. Maybe this is the realmessage which the Monty Hall Problem should be telling us? What if vosSavant’s opening words had been “ Suppose you’re going to be on a gameshow tonight. If you make it to the last round, you’ll be given the choice ofthree doors... ”? In this section, I present some elementary mathematical facts, firstly fromprobability theory, secondly from game theory. The results are formulatedwithin a mathematical framework which does not make any assumptionsrestricting the scope of the present discussion. Modelling all the various doorchoices as random variables does not exclude the case that they are fixed. Italso leaves the question completely open how we think of probability: in afrequentist or in a Bayesian sense. I impose only the “structural” conditionson the sequence of choices, or moves, which are universally agreed to beimplied by vos Savant’s story.
I distinguish four consecutive actions:1. Host: hiding the car before the show behind one of three doors,
Car
2. Player: choosing a door, P1
3. Host: revealing a goat by opening a different door,
Goat
4. Player: switching or staying, final choice door P2 The doors are conventionally labelled “1”, “2” and “3”, and we can rep-resent the door numbers specified by the four actions with random variables
Car, P1, Goat, P2 . Since two doors hide goats and one hides a car and the5ost knows the location of the car, he can and will open a door different tothat chosen by the player and revealing a goat. I allow both the locationof the car
Car and the initial choice of the player P1 to be random, and as-sume them to be statistically independent. From different modelling pointsof view, we might want to take either of these two variables to be fixed; theindependence assumption is then of course harmless. Given the location ofthe car and the door chosen by the player, the host opens a different door Goat revealing a goat, according to a probability distribution over the twodoors available to him when he has a choice (which includes the case thathe follows some fixed selection rule). Then the player makes his choice P2 ,deterministically or according to a probability distribution if he likes, butin either case only depending on his initial choice and the door opened bythe host. Finally we can see whether he goes home with a car or a goat byopening his door and revealing his Prize .The probabilistic structure of the four actions together with the finalresult
Prize (whether the player goes home with a car or a goat) can berepresented in the graphical model or Bayes net shown in Figure 1. Thisdiagram (drawn using the gRain package in the statistical package R) was in-spired by Burns and Wieth (2004) who performed psychological experimentsto test their hypothesis that people fail MHP because of their inability tointernalise the collider principle : conditional on a consequence, formerly in-dependent causes become correlated. In this case, the initially statisticallyindependent initial moves
Car and P1 of host and player are conditionally dependent of one another given the door Goat opened by the host.I now write down three simple propositions, each making in turn a strongermathematical assumption, and each in turn giving a better reason for switch-ing.
Proposition 1.
If the player’s initial choice has probability / to hit the car,then always switching gives the player the car with (unconditional) probability / . Proposition 2.
If initially all doors are equally likely to hide the car, then,given the door initially chosen and the door opened, switching gives the playerthe car with conditional probability at least / .Consequently, not only does “always switching” give the player the carwith unconditional probability / , but no other strategy gives a higher successchance. Proposition 3.
If initially all doors are equally likely to hide the car and ifthe host is equally likely to open either door when he has a choice, then, giventhe door initially chosen and the door opened, switching gives the player the ar P1GoatP2Prize Figure 1: A Graphical Model (Bayes Net) for MHP car with conditional probability / , whatever door was initially chosen andwhich door was opened (Morgan et al., 1991a,b). Proof.
Prop. 1: This implication is trivial once we observe that a “switcher” wins ifand only if a “stayer” loses.Prop. 2: We use Bayes’ theorem in the form posterior odds equals prior odds times likelihood ratio.
The initial odds that the car is behind doors 1, 2 and 3 are 1:1:1. Theposterior odds are therefore proportional to the probabilities that the hostopens Door 3 given the player chose Door 1 and the car is behind Door 1, 2and 3 respectively. These probabilities are q , 1 and 0 respectively, where q = Prob( Host opens Door 3 | Player chose Door 1, car is behind same ) . The posterior odds for switching versus staying are therefore 1 : q , so thatstaying does not have an advantage over switching, whatever q might be.Since all doors are initially equally likely to hide the car, the door chosenby the player hides the car with probability 1/3. The unconditional prob-ability that switching gives the car is therefore 2/3. By the law of total7robability, this can be expressed as the sum over all six conditions (doorchosen by player, door opened by host), of the probability of that condi-tion times the conditional probability that switching gives the car, under thecondition. Each of these conditional probabilities is at least 1/2. The strat-egy of always switching can’t be beaten, since the success probability of anyother strategy is obtained from the success probability of always switchingby replacing one or more of the conditional probabilities of getting the carby switching by probabilities which are never larger.Prop. 3: If all doors are equally likely to hide the car then by independenceof the initial choice of the player and the location of the car, the probabilitythat the initial choice is correct is 1 /
3. Hence the unconditional probabilitythat switching gives the car is 2 /
3. If the player’s initial choice is uniformand the two probability distributions involved in the host’s choices are uni-form, the problem is symmetric with respect to the numbering of the threedoors. Hence the conditional probabilities we are after in Proposition 3 areall the same, hence by the law of total probability equal to the unconditionalprobability that switching gives the car, 2/3.Proposition 3 also follows from the inspection of the posterior odds com-puted in the proof of Proposition 2. On taking q = 1 /
2, the posterior oddsin favour of switching are 2:1 (Morgan et al., 1991a).In the literature, Proposition 3 is usually proven by explicit computationor tabulation, i.e., by going back to first principles to compute the conditionalprobability in question. For instance, Morgan et al. (1991a) also give thisdirect computation, attributing it to Mosteller’s (1965) solution of the Pris-oner’s dilemma paradox. It is often offered as an example of Bayes’ theorem,but really is just an illustration of conditional probability via its definition.On the other hand, Bayes’ theorem in its odds form (which I used to proveMorgan et al.’s Proposition 2) is a genuine theorem , and offers to my minda much more satisying route for those who like to see a computation andat the same time learn an important concept and a powerful tool. To mymathematical mind the most elegant proof of Proposition 3 is the argumentby symmetry starting from Proposition 1: the conditional probability is thesame as the unconditional since all the conditional probabilities must be thesame. I learnt this proof from Boris Tsirelson on wikipedia discussion pages,but it is also to be found in Morgan et al. (1991b).This proof also supplies one reason why the literature is so confused asto what constitutes a solution to MHP. One could apply symmetry at theoutset, to argue that we only want an unconditional probability. There is no8oint in conditioning on anything which we can see in advance is irrelevantto the question at hand.The pages of wikipedia, as well as a number of papers in the literature, arethe scene of a furious controversy mainly as to whether Proposition 1 and aproof thereof, or Proposition 3 and a proof thereof, is a “complete and correctsolution to MHP”. These two solutions can be called the simple or popularor unconditional solution, and the full or complete or conditional solutionrespectively. The situation is further complicated by the fact that manysupporters of the popular solution do accept all the symmetry (uniformity)conditions of Proposition 3, for a variety of reasons. I will come back to thisin the next main section, but first consider a rather different kind of resultwhich can be obtained within exactly the same general framework as before.
Let us think of the four actions of the previous subsection as two pairs ofmoves in a two stage game between the host and the player in which theplayer wants to get the car, the host wants to keep it. Von Neumann’sminimax theorem tells us that there exist a pair of minimax strategies forplayer and host, and a value of the game, say p , having the following definingcharacteristics. The minimax strategy of the player (minimizes his maximumchance of losing) guarantees him at least probability p of winning, whateverthe strategy of the host; the minimax stategy of the host (minimizes hismaximum probability of losing) guarantees him at most probability p oflosing. If both player and host play their minimax strategy then indeed theplayer will win with probability p . Proposition 4.
The player’s strategy “initial choice uniformly at random,thereafter switch” and the host’s strategy “hide the car uniformly at random,open a door uniformly at random when there is a choice” form the minimaxsolution of the finite two-person zero-sum game in which the player tries tomaximize his probability of getting the car, the host tries to minimize it. Thevalue of the game is / .Proof. We must verify two claims. The first is that whatever strategy is usedby the host, the player’s minimax strategy guarantees the player a successchance of at least 2/3. The second is that whatever strategy is used bythe player, the host’s minimax strategy prevents the player from achieving asuccess chance greater than 2/3.For the first claim notice that if the player chooses a door uniformly atrandom and thereafter switches, he’ll get the car with probability exactly2/3; that follows from Proposition 1. 9or the second, suppose the host hides the car uniformly at random andthereafter opens a door uniformly at random when he has a choice. Makingthe initial choice of door in any way, and thereafter switching, gives the playersuccess chance 2/3, and by Proposition 2 (or 3, if you prefer) there is no wayto improve this.Note that I did not use von Neumann’s theorem in any way to get thisresult: I simply made use of the probabilistic results of the previous subsec-tion.MHP was brought to the attention of the mathematical economics andgame theory community by a paper of Nalebuff (1987), which contains anumber of game-theoretic or economics choice puzzles. He considered MHPas an amusing problem with which to while away a few minutes during aboring seminar. After describing the problem he very briefly reproduced theshort solution corresponding to Proposition 1. He enigmatically drops thenames of Neumann-Morgenstern and Bayes as he ponders why most peoplein real life took the wrong decision, but he does not waste any more time onMHP.Variants of the MHP in which the host does not always open a door, orwhere he might be trying to help you, or might be trying to cheat you, lendthemselves very well to game theoretic study, see wikipedia or Rosenhouse(2009) for references.For present purposes, the important point which I think is brought outby a game theoretic approach is that the player does have two decision mo-ments. The player has control over his initial choice. Vos Savant describesthe situation at the moment the player must make up his mind whether toswitch or stay, and most, but not all, people will instinctively feel that this isthe only important decision moment. But the player earlier had a chance tochoose any door he liked. Perhaps he would have been wise to think aboutwhat he would do if he did make it to the last round of the show, beforesetting off to the TV studio. There is no way he can ask the advice of afriendly mathematician as he stands on the stage under the dazzling studiolights while the audience is shouting out conflicting advice.Van Damme (1995; in Dutch) goes a little deeper into the question of whyreal human players did not behave rationally on the Monty Hall show; thisis one of the main questions studied in the psychology, philosophy, artificialintelligence and animal behaviour literature on MHP. Since “rational expec-tations” play a fundamental role in modern economic theory, the actual factsof the real world MHP, where players almost never switched doors, bodes illfor the application of economic theory to real world economics. The usualrationale for human rational expectations in economics is that humans learn10rom mistakes. However, the same person did not get to play several timestimes in the final round of the Monty Hall show, and apparently no-one kepta tally of what had happened to previous contestants, so learning simply didnot take place. Nobody thought there would be a point in learning! Instead,the players used their brains, came to the conclusion that there was no ad-vantage in switching, and mostly stuck to their original choice. At this pointthey do make a rational choice: there would be a much larger emotional lossto their ego on switching and losing, than on staying and losing. Sticking toyour door demonstrates moral fortitude. Switching is feckless and deservespunishment.Interestingly, pigeons (specifically, the Rock Pigeon,
Columba Livia , thepigeon which tourists feed in city squares all over the world) are very goodat learning how to win the Monty Hall game, see Herbranson and Schroeder(2010). They do not burden their little brains thinking about what to do butjust go ahead and choose. There is a lot of variation in their initial decisionswhether to switch or stay, and observing the results gives them a chance tolearn from the past without thinking at all. Only a very small brain is neededto learn the optimal strategy. And these birds are evolutionarily speakingvery succesful indeed.
Propositions 1, 2 and 3 tell us in different ways that switching is a good thing.Notice that the mathematical conditions made are successively stronger andthe conclusion drawn is successively stronger too. As the conditions getstronger, the scope of application of the result gets narrower: there are moreassumptions to be justified. From a mathematical point of view none of theseresults are stronger than any of the others: they are all strictly different .The literature on MHP focusses on variants of Proposition 1, and ofProposition 3. These correspond to what are called the popular or simpleor unconditional solution, and the full or conditional solution to MHP. Thefull solution is mainly to be found in introductory probability or statisticstexts, whereas the simple solution is popular just about everywhere else. Theintermediate “Proposition 2” is only occasionally mentioned in the literature.The full list of conditions in Proposition 3 is often called, at least in the kindof texts just mentioned, the standard or canonical or original MHP. I willjust refer to them as the conventional supplementary assumptions .Regarding the word “original”, it is a historical fact that Selvin (1975a)gave MHP its name, did this in a statistics journal, and wrote down theconventional full list of uniformity assumptions. He proceeded to compute11he unconditional probability that switching gives the car by enumeration of nine equally likely cases, for which he took both the player’s initial choiceand the location of the car as uniform random, and of course independentof one another. In his second note, Selvin (1975b), he computed the condi-tional probability using now his full list of assumptions concerning the host’sbehaviour, and fixing the initial choice of the player, but without noting anyconceptual, let alone technical, difference at all with his earlier solution. Ofcourse, the number “2/3” is the same. In the same note he quotes withapproval from a letter from Monty Hall himself who gave the argument ofProposition 1: switching gives the car with probability 2/3 because the initialchoice is right with probability 1/3. We know Monty will open a door re-vealing a goat. Conditioning on an event of probability one does not changethe probability that the initial choice was right.Thus Selvin set the seeds for subsequent confusion. Let me call his ap-proach the practical-minded approach :The right answer to MHP is “2/3”. There are many ways to getto this answer, and I am not too much concerned how you getthere. As long as you get the right answer 2/3, we’re happy. Afterall, the whole point of MHP is that the initial instinct of everyonehearing the problem is to say “since the host could anyway opena door showing a goat, his action doesn’t tell me anything. Thereare still two doors closed so they still are equally likely to hidethe car. So the probability that switching would give the car is“1/2”, so I am not going to switch, thank you.Selvin’s two papers together gave MHP a firm and more or less standardposition in the elementary statistics literature. There is a conventional com-plete specification of the problem. This enables us to write down a finitesample space and allocate a probability to every single outcome. Usually theplayer’s initial choice is taken, in the light of the other assumptions withoutloss of generality, as fixed. All randomness is in the actions of the host, or in our lack of any knowledge about them. This corresponds to whether thewriter has a frequentist or a subjectivist slant, often not explicitly stated, butimplicit in verbal hints. The question is not primarily “should you switch orstay?”, but “what is the probability, or your probability, that switching willgive the car?” Typically, as in Selvin’s second, conditional, approach, theplayer’s initial choice is already fixed in the problem statement, the host’stwo actions are already seen as completely random, whether because we aretold they are, objectively, or because we are completely ignorant of how theyare made, subjectively. The problem typically features in the chapter which12ntroduces conditional probability and Bayes’ theorem in the discrete setting.Thus the problem is posed by a maths teacher who wants the student to learnconditional probability. The problem is further reformulated as “what is the conditional probability that switching will give the car”.In such a context not much attention is being paid to the meaning ofprobability. After all, right now we are just busy getting accustomed to itscalculus. Most of the examples figure playing cards, dice and balls in urns,and so the probability spaces are usually completely specified (all outcomescan be enumerated) and mostly they are symmetric, all elementary proba-bilities are equal. The student is either supposed to “know”, or he is toldexplicitly, that the car is initially equally likely to be behind any of the threedoors. The host is supposed to choose at random (shorthand for uniformly,or completely, at random) when he has a choice. Since these facts are givenor supposed to be guessed, the initial choice of the player is irrelevant, andwe are indeed always told that the player has already picked Door 1.Well, if MHP is merely an exercise in conditional probability where themathematical model is specified in advance by the teacher, then it is clearhow we are to proceed. But I prefer to take a step back and to imagineyou are on a game show . How could we “know” these probabilities? Thisis especially important when one has the task of “explaining” MHP to non-mathematicians and to non-statisticians.This is where philosophy, or if you prefer, metaphysics, raises its head.How can one “know” a probability; what does it mean to “know” a proba-bility?I am not going to answer those questions. But I am going to compare aconventional frequentist view – probability is out there in the real world – to aconventional Bayesian view – probability is in the information which we pos-sess. I hope to do this neutrally, without taking a dogmatic stance myself. Itis a fact that many amateur users of probability are instinctive subjectivists,not so many are instinctive frequentists. Let us try to work out where eitherinstinct would take us. An important thing to realise is: Bayesian probabil-ities in, Bayesian probabilities out; frequentist in, frequentist out. I will alsoemphasize the difference which comes from seeing randomness in the host’smoves or in the player’s moves , and the difference which comes from seeingthe question as asking for an action , or more passively for a probability .For a subjectivist (Bayesian) the MHP is very simple indeed. We onlyknow what we have been told by vos Savant (Whitaker). The wording “ say ,Door 1” and “ say , Door 3” (my italics) emphasize that we know nothingabout the behaviour of the host, whether in hiding cars or in opening doors.Knowing nothing, the situation for us is indistinguishable from the situationin which we had been told in advance that car hiding and door opening was13ctually done using perfect fair randomizers (unbiased dice throws or cointosses). Probability is a representation of our uncertain knowledge about thesingle instance under consideration. Probability calculus is the unique inter-nally consistent way to manipulate uncertain knowledge. To start off with,since we know nothing, we may as well choose our door initially accordingto our personal lucky number, so we picked Door 1. Having seen the hostopen Door 3, we would now be prepared to bet at odds 2:1 that the car isbehind Door 2. The new situation is indistinguishable for us from a bettingsituation with fair odds 2:1 based on a perfect fair randomizer (by which Isimply refer to the kind of situation in which subjectivists and objectiviststend to agree on the probabilities, even if they think they mean somethingquite different).Does the Bayesian (a subjectivist) need Bayes’ theorem in order to cometo his conclusion? I think the answer is no . For a subjectivist the door num-bers are irrelevant. The problem is unchanged by renumbering of the doors.His beliefs about whether the car would be behind the other door in any ofthe six situations (door chosen, door opened) would be the same. He hasno need to actively compute the conditional probability in order to confirmwhat he already knows. He could use Proposition 3 but is only interestedin Proposition 1. The symmetry argument of my proof of Proposition 3 isthe mathematical expression of his prior knowledge that he may ignore thedoor numbers and just compute an unconditional probability. Do you noticethe symmetry in advance and take advantage of it, or just compute awayand notice it afterwards? It doesn’t matter. The answer is 2 / what is the conditional probability. It isnever against switching.There is just one solution which does not require any prior knowledge atall; instead it requires prior action. Taking our cue from the game theoreticsolution, we realize that the player has two opportunites to act, not one. Weallow ourselves the latitude to reformulate vos Savant’s words as “You are going to be on a game show...”. We advise vos Savant, or her correspondentCraig Whitaker, to take fate into his or her own hands. Before the show,pick your lucky number (1, 2 or 3) by a toss of a fair die. When you make itto the final round, choose that door and thereafter switch. By Proposition1 you’ll come home with the car with probability 2/3, and by Proposition 4that’s the best you can hope for.Both frequentists and subjectivists will agree that you win the car withprobability 2/3 in this way. They will likely disagree about whether theconditional probability that you win, given door chosen and door opened, isalso 2/3. I think the frequentist will say that he does not know it since hedoesn’t know anything about the two host actions, while the subjectivist willsay that he does know the conditional probability (and it’s 2/3) for the verysame reason. So what? The Monty Hall Problem offers much more to the student than a mindlessexercise in conditional probability. It also offers a challenging exercise inmathematical modelling. I notice three important lessons. (1) The more youassume, the more you can conclude, but the more limited are your conclu-sions. The honest answer is not one mathematical solution but a range ofsolutions. (2) Whether you are a subjectivist or a frequentist affects the easewith which you might make probabilistic assumptions but simultaneously af-fects the meaning of the conclusions. (3). Think out of the box. Vos Savantasks for an action , not for a probability . The player has two decision momentsduring the show, not one. 15 eferences
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