Likelihood-based solution to the Monty Hall puzzle and a related 3-prisoner paradox
aa r X i v : . [ s t a t . O T ] O c t L IKELIHOOD - BASED SOLUTION TO THE M ONTY H ALL PUZZLEAND A R ELATED PR ISONER PAR ADOX
Yudi Pawitan
Department of Medical Epidemiology and BiostatisticsKarolinska [email protected] 7, 2020 A BSTRACT
The Monty Hall puzzle has been solved and dissected in many ways, but always using probabilisticarguments, so it is considered a probability puzzle. In this paper the puzzle is set up as an orthodoxstatistical problem involving an unknown parameter, a probability model and an observation. Thismeans we can compute a likelihood function, and the decision to switch corresponds to choosing themaximum likelihood solution. One advantage of the likelihood-based solution is that the reasoningapplies to a single game, unaffected by the future plan of the host. I also describe an earlier versionof the puzzle in terms of three prisoners: two to be executed and one released. Unlike the goats andthe car, these prisoners have consciousness, so they can think about exchanging punishments. Whentwo of them do that, however, we have a paradox, where it is advantageous for both to exchange theirpunishment with each other. Overall, the puzzle and the paradox are useful examples of statisticalthinking, so they are excellent teaching topics.
First, here is the Monty Hall puzzle:You are a contestant in a game show and presented with 3 closed doors. Behind one is a car, andbehind the others only goats. You pick one door (let’s call that Door 1), then the host will openanother door that reveals a goat. With two un-opened doors left, you are offered a switch. Shouldyou switch from your initial choice?It is well known that you should switch: doing so will increase your probability of winning the car from 1/3 to 2/3.The problem is actually how to provide a convincing explanation for those who think the chance of winning for eitherof the un-opened door is 50-50, so there is no reason to switch. The puzzle, first published by Selvin (1975ab), wasinspired by a television game show
Let’s Make a Deal originally hosted by Monty Hall. It later became famous afterappearing in Marilyn vos Savant’s ”Ask Marilyn” column in Parade magazine in 1990. It generated its own literatureand many passionate arguments, even among those who agree that you should switch.Martin Gardner (1959)’s article on the same puzzle in terms of three prisoners (below) was titled ‘Problems involvingquestions of probability and ambiguity.’ He started by referring to the American polymath Charles Peirce’s observationthat ‘no other branch of mathematics is it so easy for experts to blunder as in probability theory.’ The main problem isthat some probability problems may contain subtle ambiguities. What is so clear to me – as the statement of the MontyHall problem above – may contain hidden assumptions I forget to mention or even am not aware of. For example, aswill be clear later, the standard solution assumes that, if you happen to pick the winning door, then the host will opena door randomly with probability 0.5. For me that is obvious so as to make the problem solvable and the solution neat.If we explicitly drop this assumption, then there is no neat solution; but our discussion is not going to that direction.Instead, we head to a more interesting paradox. In Martin Gardner’s version there are three prisoners A, B and C, twoof whom will be executed and one released. Being released is like winning a car. Suppose A asks the guard: since
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7, 2020for sure either B or C will be executed, there is no harm in telling me which one. Suppose the guard says C will beexecuted. Should A asks to switch his punishment with B’s? With the same logic as in the Monty Hall problem, itmust be yes, A should switch with B. But, what if B asks the same question to the guard, which of A or C will beexecuted, and the guard also answers C? Then it seems also advantageous for B to switch with A. Now we have aparadox: how can it be advantageous for both A and B to switch?We can make another version of the paradox: Suppose A just heard the breaking news that C has been executed (noguard is involved). Should A ask to switch his punishment with B? What logic should apply here? If the same asbefore, then he should switch. But then, the same logic applies to B, and we arrive at the same paradox. If not thesame, then having the guard to answer the questions matters. But why does it matter how A finds out that C is (to be)executed?In probability-based reasoning we are considering: what is the probability of winning if you stay with your initialchoice vs if you switch, This is relatively easy – hard enough for some people – with the Monty Hall puzzle, but itgets really challenging if we want to explain the 3-prisoners paradox. So, instead, the puzzle and the paradox will beset up as an orthodox statistical problem involving an unknown parameter, a probability model and an observation.This means we can compute a likelihood function, and the decision to switch corresponds to choosing the maximumlikelihood solution. In the Monty Hall puzzle, one advantage of the likelihood-based solution is that the reasoningapplies to a single game, unaffected by the future plan of the host. In addition, the likelihood construction will showexplicitly all the technical assumptions made in the game.
Monty Hall problem
Let θ be the location of the car, which is completely unknown to you. Your choice is called Door 1; for you the carcould be anywhere, but of course the host knows where it is. Let y be the door opened by the host. Since he has toavoid opening the prized door, y is affected by both θ and your choice. The probabilities of y under each θ are givenin this table: θ = 1 θ = 2 θ = 3 b θy = 1 y cannot be 1, because the host cannot open your door (duh!), so it has zero probabilityunder any θ . If your Door 1 is the winning door ( θ = 1 ), the host is assumed to choose randomly between Doors 2and 3. If θ = 2 , he can only choose Door 3. Finally, if θ = 3 , he can only choose Door 2. So, the ‘data’ in this gameis the choice of the host. Reading the table row-wise gives the likelihood function of θ for each y , so the maximumlikelihood estimate is obvious:If y = 2 , then b θ = 3 (so you should switch from 1 to 3.)If y = 3 , then b θ = 2 (switch from 1 to 2.)The increase in the likelihood of winning by switching is actually only 2 folds, so it is not enormous. But the increasein prize from a goat to a car is enormous, so a switch is warranted.In this likelihood formulation, the problem is a classic statistical problem with data following a model indexed by anunknown parameter. For orthodox non-Bayesians, there is no need to assume that the car is randomly located, so noprobability is involved on θ ; it is enough to assume that you are completely ignorant about it. From the table, it is clearalso that, when θ = 1 , the host does not need to randomize with probability 0.5; he can do that any probability at alland you will not reduce your likelihood of winning by switching. Probability or likelihood?
Previous solutions of the Monty Hall puzzle are typically given in terms of probability. Why bother with the likelihood?Imagine a forgetful host in the Monty Hall game: he forgets which door has the car, so he opens a door at random. Ifit reveals a goat, then the game can go on; if it reveals the car, then he makes excuses, and the game is cancelled andnobody wins. Suppose in your particular game, a goat is revealed. Is it still better to switch? As before, let’s call your2
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7, 2020chosen door as Door 1. Define the data y as the un-opened door (other than yours) if the opened one is a goat (so thegame is on); if the opened one is a car, then set y ≡ (and the game is off). The probability table is now: θ = 1 θ = 2 θ = 3 b θy = 2 { } { } y = 2 or 3, and the game still on, your likelihood of winning the car with your original or the other unopeneddoor is equal, and there is no benefit of switching. Comparing with the original version, this means that the evidenceof an open door revealing a goat is not sufficient to say that switching is beneficial. We must also know whether it wasintentional or accidental. But, how about the future plan? Should it also affect your reasoning? In technical probabilityreasoning it should also matter, because probability is not meant to apply for a specific game.Say in the current game the host intentionally opens a door with a goat, but in the future he plans to open a doorrandomly. What logic applies to the current game? Non-Bayesians certainly cannot apply the orthodox probability-based argument to the current game. But, the likelihood-based reasoning still applies to the current game, unaffectedby the future plan of the host. The likelihood-based reasoning also helps solve the 3-prisoner paradox. Let θ be the identity of the prisoner to bereleased; to avoid confusion, set θ =
1, 2 or 3 for the three prisoners A, B and C. Let A be the guard’s answer toprisoner A, also denote the answer by 1, 2 and 3. Then we have indeed the same probability table as for the gameshow: θ = 1 θ = 2 θ = 3 b θA = 1 to prisoner A, and he will randomize whenever possible. So, we get the samemaximum-likelihood estimate as above, leading to switching as a good strategy.When B asks the same question, let A be the guard’s answer. The joint distribution of ( A , A ) is as follows. It canbe derived under the same requirements: the guard cannot tell the questioner that he would be executed, and he mustrandomize whenever possible. A = 1 A = 2 A = 3 θ = 1 , A = 1 A = 2 A = 3 θ = 2 , A = 1 A = 2 A = 3 θ = 3 , A = 1 A = 2 A = 3 θ = 1 θ = 2 θ = 3 b θ Better for A Better for B ( A , A ) =(2,1) 0 0 1 3 Switch with C Switch with C(2,3) 0.5 0 0 1 Don’t switch! Switch with A(3,1) 0 0.5 0 2 Switch with B Don’t switch!(3,3) 0.5 0.5 0 { } None NoneTotal 1 1 1Based on the relevant outcome of the story ( A = 3 , A = 3) , the prisoners A and B actually have equal likelihood ofbeing released. So, a switch confers no advantage to either side, and there is no paradox.3 PREPRINT - O
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7, 2020How can the paradox appear? Let’s consider who has access to what information. If A only knows A = 3 , but doesnot know nor presume the existence of A , then his reasoning is incomplete. Similarly, B only knows A . So, theparadox appears because of incomplete information by each side. An external agent – e.g. the guard – who knowsboth answers A and A can see there is no advantage in switching.Suppose, both A and B know that each has asked the relevant question, but A only knows his own answer A and Bonly knows A . Furthermore, assume that a switch can only happen by mutual consent. Suppose A = 3 , then, forA, a switch is advantageous only when A = 1 , but neutral when A = 3 . But A = 1 means B is told that A willbe executed, so he will not be willing to switch with A. The same logic applies of B. So, when both sides are willing,they know the switch must be neutral, and there is no paradox. How about the news break version? As before,let θ be the identity of the prisoner to be released. Now, consider thedata y is the first prisoner reported to be executed, assuming that execution is in random order. Then we can derive theprobabilities under different θ ’s. θ = 1 θ = 2 θ = 3 b θy = 1 { } { } { } Total 1 1 1So, when y = 3 , A and B have equal likelihood of being released. In other words, when both A and B heard that Chas been executed, there is no advantage for either of them to switch.Why does the explanation look so trivial? Well, we can make it more complicated. The fact that A heard the newsmeans that he was not the first to be executed. Suppose A was in fact told that, if he is to be executed (which he mightnot), then he will not be the first. Does that affect he should react to the news? The probability table becomes: θ = 1 θ = 2 θ = 3 b θy = 1 y = 3 , then b θ = 2 , so B should not switch with A. If both are told – and both know this – thatthey won’t be the first to be executed, then the table becomes: θ = 1 θ = 2 θ = 3 b θy = 1 { } So, when y = 3 , we are back to the situation of neutral switch. Not surprisingly, different decisions by the player in the Monty Hall problem or by the prisoners depend on differentmodel setup/assumptions and available data. The same data, e.g. an open Door 3 revealing a goat, can be interpreteddifferently depending on the setup. The likelihood-based calculation requires explicit assumptions, so it clarifiesthem. We also compare the probability- vs likelihood-based reasoning; the advantage of the latter is its applicabilityto a single game, unaffected by future intention or plans. Gill (2011) provided a clear discussion of the assumptionsassociated with the standard probability-based solution of the puzzle as well as a two-person game that shows switchingas a minimax solution. The model formulation here agrees with Gill that the Monty Hall problem is not a probabilitypuzzle; he considered it a mathematical modelling problem. Here I have phrased it as an orthodox statistical problemso it – and the related paradox – can be a useful example in likelihood-based modelling.4
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Gardner, Martin (1959). ”Mathematical Games: Problems involving questions of probability and ambiguity”.
Scien-tific American , 201 (4): 174–182.Gill, Richard (2011). The Monty Hall problem is not a probability puzzle* (It’s a challenge in mathematical mod-elling).
Statistica Neerlandica , 65 (1), 58–71.Selvin, Steve (1975a). ”A problem in probability”.
American Statistician , 29 (1), 67–71.Selvin, Steve (1975b). ”On the Monty Hall problem”.