Don’t Switch! Why Mathematicians’ Answer to the Monty Hall Problem is Wrong

Don’t Switch! Why Mathematicians’ Answer to the Monty Hall Problem is Wrong


The Monty Hall problem is one of those rare curiosities – a mathematical problem that has made the front pages of national news.  Everyone now knows, or thinks they know, the answer but a realistic look at the problem demonstrates that the standard mathematician’s answer is wrong.  The mathematics is fine, of course, but the assumptions are unrealistic in the context in which they are set.  In fact, it is not clear that this problem can be appropriately addressed using the standard tools of probability theory and this raises questions about what we think probabilities are and the way we teach them.

1. Introduction

The Monty Hall problem hit the headlines in 1990, when Craig F. Whitaker of Columbia, Maryland, asked Marilyn vos Savant: ‘Suppose you’re on a game show, and you’re given the choice of three doors: behind one door is a car; behind the others, goats.  You pick a door, say No. 1, and the host, who knows what’s behind the other 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 take the switch?’^1

Vos Savant wrote a column called ‘Ask Marilyn’ in the popular magazine Parade, in which she responded to readers’ questions.  According to the Guinness Book of Records, at the time she was the woman with the highest IQ in the world.

Vos Savant responded to Whitaker in her column of 9 September 1990: she said you should switch and that you double your chances of winning if you do.  The result was a torrent of criticism and abuse – much of it from mathematicians – such as:

  • ‘May I suggest that you obtain and refer to a standard textbook on probability before you try to answer a question of this type again?’ (Charles Reid, PhD, University of Florida)
  • ‘You blew it, and you blew it big! … There is enough mathematical illiteracy in this country, and we don’t need the world’s highest IQ propagating more. Shame!’ (Scott Smith, PhD, University of Florida)
  • ‘You made a mistake, but look at the positive side. If all those PhD’s were wrong, the country would be in some very serious trouble.’ (Everett Harman, PhD, US Army Research Institute)

She followed it up with another article on 2 December, addressing the problem in more detail, to which there was more criticism and abuse.  Vos Savant said she received 10,000 letters about her articles, including almost 1,000 from PhDs.  ‘Of the letters from the general public, 92% are against my answer, and of the letters from universities, 65% are against my answer.  Overall, nine out of ten readers completely disagree with my reply.’  In a third article on 17 February 1991, she spelt out her reasoning in even more detail, and suggested school classes carry out empirical trials.  These were carried out and supported her view that switching increases your chances of winning, and now most of her correspondents agreed with her.  On Sunday, 21 July 1991, John Tierney wrote a front page article in the New York Times that was very firmly on her side.^2

It looks as if America is in some very serious trouble!

There were some complaints that vos Savant’s first article did not fully spell out all the assumptions underlying her answer and which constituted a subtle extension of the original question.  Nor were her arguments mathematically rigorous.  She accepted these points but argued that her respondents had clearly not been confused, and that mathematical rigour had no place in what was essentially a light-hearted magazine column.  We will come back to the question of those assumptions later.

2. A brief history of the Monty Hall problem

Monty Hall, real name Maurice Halperin (born 1921), was a Canadian TV personality, who hosted the American television game show Let’s Make a Deal in the 1960s and 1970s.  In this programme, Monty offered many different types of challenge to contestants and the Monty Hall problem is supposedly based on one of them, though in fact the game as described above did not appear on the show.  The ideas behind the Monty Hall problem were far from new, though.  Joseph Bertrand’s Box paradox, which he described in 1889 [1], was based on similar ideas and Martin Gardner’s Three Prisoners problem of 1959 [2] was equivalent to it in mathematical terms.

The modern version, but without the goats, was described in a letter to American Statistician by Professor Steve Selvin of the University of California in 1975 [3].  This introduced Monty Hall (but spelt Monte) and Let’s Make a Deal.  Like vos Savant, Selvin received letters claiming his solution was incorrect so he expanded on his original solution in a second letter later the same year.  He noted, in particular, that: ‘Benjamin King pointed out the critical assumptions about Monty Hall’s behavior that are necessary to solve the problem’ [4]. It was in this second letter that the name ‘Monty Hall problem’ first appeared in print. The goats appear to have originated with Whitaker.

3. The mathematics of the Monty Hall problem

There are various ways to solve the Monty Hall problem mathematically.  Here is one using Bayes’ theorem, which tells us how to compute certain types of conditional probabilities.  It is sometimes called the ‘probability of causes’ theorem, because if we have an outcome that can arise in different ways (’causes’), it tells us how likely those causes are in the light of the outcome.  Despite the fact that it can be deduced relatively easily from the basic rules and definitions of probability theory and is absolutely sound mathematically, its use sometimes gives rather unintuitive results.

The general form of Bayes’ theorem is:

    \[ P(B_i|A) = \frac{P(A|B_i)P(B_i)}{\sum_i P(A|B_i)P(B_i)} \]

where A is the result that we have and the B_i are a set of mutually exclusive events that includes all the possible ’causes’.  Bayes’ theorem has the advantage that, not only does it give the correct answer, it forces us to be explicit about our assumptions.

It deals very easily with the Monty Hall problem.  Let C_i be the event ‘the car is behind door i‘ and D_i be the event ‘Monty shows you a goat behind door i‘.  Suppose you choose door 1 and Monty shows you a goat behind door 3.  What we need to know is the conditional probability P(C_1|D_3), the probability that the car is behind the door you have chosen, now that Monty has shown us a goat behind door 3.

By Bayes’ theorem, assuming that Monty will always open a door, never open the door you have chosen and never show us the car (so the D_i are the only possibilities we need to take into account):

    \begin{multline*} P(C_1|D_3) = \\ \frac{P(D_3|C_1) P(C_1)} {P(D_3|C_1) P(C_1) + P(D_3|C_2) P(C_2) + P(D_3|C_3) P(C_3)}. \end{multline*}

If we assume that the car is initially equally likely to be behind any door then the P(C_i) are easy, they are all 1/3 but the C‘s and D‘s are not independent so what are the conditional probabilities?  Let us start with P(D_3|C_1).  Now door 1 conceals the car so both doors 2 and 3 conceal goats, so if Monty always chooses at random P(D_3|C_1) = 1/2.  And what is P(D_3|C_2)?  Now our door (door 1) conceals a goat and door 2 conceals the car so Monty has no choice but to open D_3 as he never shows us the car, so P(D_3|C_2) = 1.  Similarly P(D_3|C_3) = 0.  Plug these numbers in:

    \[ P(C_1|D_3) = \frac{1/2 \times 1/3}{1/2 \times 1/3 + 1 \times 1/3 + 0 \times 1/3} = \frac{1}{3} \]

and we see that the probability our original door conceals the car is still 1/3.  The car is not behind door 3 so the probability that it is behind door 2 must be 2/3.  So switch.

4. The assumptions

The mathematics is correct, so you do indeed seem to double your chances by switching but only provided certain assumptions hold.  As the words in italics above show, there are actually a number of assumptions:

  1. Monty will always open a door.
  2. Monty never opens the door you have chosen.
  3. Monty never opens the door with the car behind it.
  4. The car is equally likely to be behind any door.
  5. Given a choice of doors, Monty chooses at random.

Where do these assumptions come from and just how plausible are they?

If we had watched Monty play this game many times we might have been able to spot a pattern that would justify them but, as we have seen, Monty Hall never played this game; no comfort from that source.  Where do they come from?  They were not in Craig Whitaker’s original query given above but vos Savant’s added some crucial words in her response: ‘the host, who knows what’s behind the doors and will always avoid the one with the prize‘ (my emphasis).  Vos Savant has done what all academics do when putting this problem to students (and I do when I am putting it to mine), she has turned a real problem that mathematicians cannot answer because they don’t know what the true probabilities are into one they can answer, by making plausible assumptions.  Except that it is not clear in this case how plausible the assumptions are.

Let us start by thinking about Monty’s objectives and motivations.  As a game show host it is reasonable to assume that he has a number of things to worry about:

  1. He has to manage the game; that is, he has to ensure that all the contestants have a reasonable hearing and that it finishes on time.
  2. He has to entertain the audience; this is likely to mean, among other things, that the contestants win cars reasonably often.
  3. He is not likely to want to give away too many cars because of the cost to his employers.

He may have other motivations but these will do for now.

In the light of these, how plausible are the assumptions above? Let us take them one by one:

Assumption 1:  That Monty will always open a door.  This seems entirely reasonable.  It would be a very odd game show where there wasn’t a clear outcome at the end.

Assumption 2:  That Monty never opens the door you have chosen.  It is not at all clear why this should be the case.  Why should Monty, after the usual banter associated with game shows (‘Do you want to switch?  Are you sure you don’t want to switch?’), not simply open the door you have chosen and tell you whether you have won or not?  Indeed, if he is running short of time, if he knows there is a car behind the door and no-one has won for a while or he knows there is a goat behind the door and a number of people have won cars recently, is this not what he is likely to do?

Assumption 3:  That Monty never opens the door with the car behind it.  This assumption is again rather dubious.^3 Why shouldn’t Monty simply open a door and show you the car, particularly if he is running out of time or wants to engineer a particular outcome.  In practice he is likely to open the door you have chosen rather than the door with the car behind it if you have lost and retain an air of mystery over the location of the car, but the effect is the same.

Assumption 4:  That the car is equally likely to be behind any door.  There is no reason to believe that a particular door is likely to be preferred, so this seems reasonable.

Assumption 5:  That, given a choice of doors, Monty chooses at random.  This assumption does not necessarily hold – Monty may be inherently lazy or have a bad leg and so have a tendency to open the available door nearest to him – but it is easy to show that if all the other assumptions hold you cannot lose by switching, though you don’t necessarily gain, so you might as well switch.

The key questionable assumptions, then, are 2 and 3 – that Monty will never open the door you have chosen and will never open the door with the car behind it – and it is these two assumptions that are at odds with our intuitive understanding of the way game shows work.  We all know that it is more fun, for the audience at least, if someone is conned out of a winning choice, so there is always a suspicion that that is what Monty is trying to do.

We also know that if a game show is to be entertaining it has to be varied.  If Monty behaves too simplistically the game will become predictable and so less entertaining.  That leads on to a rather deeper point – it is not at all clear that we can apply probability theory, at least in the traditional sense, to this problem at all.  Traditionally, probability theory applies to certain types of repeatable event where the outcomes are random (in a sense we need not dwell on here) but Monty Hall is not a random event.  He is not a coin to be tossed or a die to be thrown; he is a rational being with free will (or at least the will of his producers) and certain objectives to achieve.  Because a primary objective is to entertain, his choices will not be random in the way that simple probability theory assumes.  His decisions will not be independent of each other and will depend on numerous factors not included in our mathematical model.

I have not carried out any detailed experiments to see exactly how people’s reactions to the Monty Hall problem change if all the assumptions are spelt out at the beginning – if people are told before they answer that Monty will always show them another door, that he will never show them the car, and so on – but I suspect that fewer people would get it wrong.  There is, though, another source that might help.

5. The Three Prisoners problem

The Three Prisoners problem appeared in Martin Gardner’s Mathematical Games column in Scientific American in 1959 [2].  He put it like this:

Three men – A, B and C – were in separate cells under sentence of death when the governor decided to pardon one of them.  He wrote their names on three slips of paper, shook the slips in a hat, drew out one of them and telephoned the warden, requesting that the name of the lucky man be kept secret for several days.  Rumor of this reached prisoner A.  When the warden made his morning rounds, A tried to persuade the warden to tell him who had been pardoned.  The warden refused.

‘Then tell me,’ said A, ‘the name of one of the others who will be executed.  If B is to be pardoned, give me C’s name.  If C is to be pardoned, give me B’s name.  And if I’m to be pardoned, flip a coin to decide whether to name B or C.’

‘But if you see me flip the coin,’ replied the wary warden, ‘you’ll know that you’re the one pardoned.  And if you see that I don’t flip a coin, you’ll know it’s either you or the person I don’t name.’

‘Then don’t tell me now,’ said A.  ‘Tell me tomorrow morning.’

The warden, who knew nothing about probability theory, thought it over that night and decided that if he followed the procedure suggested by A, it would give A no help whatever in estimating his survival chances.  So next morning he told A that B was going to be executed.

After the warden left, A smiled to himself at the warden’s stupidity.  There were now only two equally probable elements in what mathematicians like to call the ‘sample space’ of the problem.  Either C would be pardoned or himself, so by all the laws of conditional probability, his chances of survival had gone up from 1/3 to 1/2.

Did A reason correctly?

As we have already noted, this is mathematically equivalent to the Monty Hall problem; the three prisoners correspond to the three doors, being pardoned to the car, being executed to the goats and the warden to Monty Hall.  We can, therefore, solve it using Bayes’ theorem exactly as above and if we make the same assumptions we get the same answer.  A’s chances of being pardoned have not changed but C’s chances have doubled.

So why, if it is essentially the same problem, did this problem not appear on the front page of the New York Times?

There are all sorts of explanations, of course.  The late 1950s was a different era from the early 1990s; Scientific American was a different sort of magazine from Parade, with a different sort of readership; the problem appeared as part of a series of similar problems; or it just might not have caught the public’s imagination in the same way.  There are, though, more interesting potential reasons.  The first is that while the problems may be the same mathematically, they are in different settings.  And while the assumptions made in the Monty Hall problem (whether made explicit or not) conflict with our intuitive notions about how game shows work, it seems more reasonable that A has not learnt anything about his fate from the warden (and the focus here is on A’s chances – which don’t change – whereas the Monty Hall problem focusses on the unselected and unopened door, whose chances do change).  Furthermore, Gardner spells out all the assumptions explicitly and makes sure that they are satisfied:

Assumption 1:  The warden will always reveal a name.  This is somewhat implausible – it would seem that the warden is best off staying silent, but then there would be no problem to solve.

Assumption 2:  We are told that the warden is under instruction not to reveal who is to be pardoned.  Strictly speaking, this does not rule out the warden telling A his outcome if he is to be executed – but then his chances of being pardoned are 0.

Assumption 3:  We are told that the warden knows who is to be pardoned and is under instruction not to reveal who it is.

Assumption 4:  We are told that the name of the prisoner to be pardoned is chosen by drawing a name out of a hat.

Assumption 5:  We are told that, by a rather convoluted mechanism, the warden chooses at random when he has a choice.

In other words, whereas Monty is a free agent making decisions according to his own motivations, Gardner so constrains the warden that once he has decided to cooperate with A he is, in effect, a random variable.  Hence, not only are the assumptions themselves inherently more plausible in this scenario, the application of Bayes’ theorem is incontestable.

Interestingly, this problem appeared in an article devoted to the ambiguities that can arise if problems are not unambiguously specified.  Gardner began his discussion of it with:

A wonderfully confusing little problem involving three prisoners and a warden, even more difficult to state unambiguously, is now making the rounds.^4

And stating real problems unambiguously may involve making assumptions that are approximations, implausible or simply guesswork, so you have not necessarily solved the real problem at all.

6. To switch or not to switch

I don’t mean ‘Don’t Switch’, of course, but ‘Don’t Necessarily Switch’ isn’t as catchy a title.  Whether or not you should switch depends on what assumptions you make and the standard ones made are not necessarily reasonable.  Gardner and Selvin had already appreciated the importance of stating problems so that the assumptions about the probabilities are unambiguous but we seem to have forgotten that in the Monty Hall problem.  Having spent much of my career trying to solve real problems using mathematics, I have found that the hardest part is not usually doing the mathematics but finding out what assumptions you should make and in problems involving chance that includes being clear what the probabilities mean as well as what they are.

So if you ever find yourself in a game that looks like the Monty Hall problem should you switch or not?  Mathematics only helps if you know how to estimate the probabilities and that is much harder to do than simply making plausible assumptions.  My best advice is to look Monty in the eye and see if you can work out if he is trying to con you or not, or maybe if he is genuinely trying to give you another chance.  Think about how many cars he has given away so far and assess whether Monty might be trying to encourage more winners or more losers.  How long is there to go before the end of the game, and is Monty trying to spin it out or bring it to a halt?  When you have decided that you can do all the maths.  In practice, you should switch unless you think Monty is trying to con you out of a car, because in most cases you are no worse off by switching and you may gain, but how can you know?  There are ways of using probability theory to tackle problems like this that are very different from the way the subject is usually taught – but that is a topic for another article.

Clive Rix
University of Leicester

About the author

Clive Rix is a part-time teaching fellow at the University of Leicester but has spent most of his career in the commercial world in a range of roles, including operational research, marketing and strategic planning.  Most recently he has been a freelance business advisor.  The views expressed in this article are his own and do not necessarily reflect the views of the University.

Notes

  1. For what follows, see http://www.marilynvossavant.com/articles/gameshow.html (accessed 11 November 2014).
  2. Available at http://tinyurl.com/NYTimesMH (accessed 16 November 2014).
  3. There is a prior assumption here that Monty knows which door the car is behind, which is not necessarily true.  If not, there are more options to consider but Bayes’ theorem shows easily enough that both unopened doors have probability 1/2, so  it makes no difference whether you switch or not.  To keep things simple, therefore, we will assume he does know.
  4. My emphasis.

References

  1. Bertrand J. (1889) Calcul des probabilités, Gauthier-Villars et fils, Paris, pp. 2–3.
  2. Gardner, M. (1959) Problems involving questions of probability and ambiguity, Scientific American, vol. 201, no. 4, pp. 180–182.
  3. Selvin S. (1975a) A Problem in Probability, The American Statistician, vol. 29, no. 1, p. 67.
  4. Selvin S. (1975b) On the Monty Hall Problem, The American Statistician, vol. 29, no. 3, p. 134.

Reproduced from Mathematics Today, August 2015

Download the article, Don’t Switch! Why Mathematicians Answer to the Monty Hall Problem is Wrong (pdf)

Image credit: The Monty Hall Problem by David Simmonds / Flickr / CC BY-SA 2.0
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47 thoughts on “Don’t Switch! Why Mathematicians’ Answer to the Monty Hall Problem is Wrong”

  1. Even though both problems seem to be the same; having three options, showing one wrong answer, the host or warden knowing the answer and the contestant or prisoner who is figuring out his odds; Both are different.

    In the tv show the contestant has two choices to make while the prisoner has no choice. The knowledge of who else is going to be executed doesn’t improve his odds because the probabilities are already chosen, while in the show you can change the outcome.

  2. No, the problems are indeed equivalent. The correspondence is not between the contestant in the Monty Hall problem and prisoner A. The three prisoners correspond to the three doors. One will be freed, corresponding to winning the car, and two will be executed, equivalent to a goat.
    The warden corresponds to Monty Hall. He knows who will be freed (where the car is) and will not tell the prisoner what will happen to him (open his door) but will tell him the name of one of the other two who will be executed (show him a goat).
    The contestant corresponds to you, as an outside observer, deciding whether prisoner A’s reasoning is correct.
    Any differences between the two problems – such as that prisoner A asks the warden but the contestant asks Monty Hall – are solely down to the different setting, and do not affect the logic.

    1. They are not equivalent because the first is a free choice between 3 objects. A choice which can then be changed. The whole point is whether or not your odds improve by changing that choice.

      In the prisoner problem you are not making a choice of 3, you ARE one of the three and there is no choice involved that you can then reconsider. Essentially, you are the door and it’s about whether or not a car is behind you. That isn’t the same thing.

      In the Monty Python problem your original choice and its circumstances matter as a point of comparison. Are your odds improved by switching? The prisoner problem is missing this factor. Whether odds are in his favor depends on what you’re measuring them by. I’d say his odds are 1 in 2 because prisoner B has been rendered irrelevant. There is no comparison to made involving him so it’s just a matter of current odds. The Monty Python problem is not about the current odds it’s about a comparison of odds. (I hope I’m wording this right.
      I’ve had little formal education)

      The equivalent prisoner problem would be the prisoner guessing who is to be pardoned (he doesn’t have to guess himself) and if he changes his guess later is he more likely to be correct . In that scenario, Prisoner B is still a factor despite being a living dead man. But in the scenario posed…? It’s
      more like “Prisoner who? Dead man walking? Oh, he doesn’t matter anymore.”

  3. They are not equivalent because in the Monty Hall game you are able to switch your choice which changes everything. (This is with the assumption that Monty will select a door that is not yours and does not reveal the car). You see there is a 33% chance of picking a car and a 66% chance of picking a goat upon your first choice, that is undeniable. Monty will then reveal a door with a goat in it. If you stay with your first choice then obviously you needed to have picked a car to win which is 33%. If you switch however that means your initial pick needed to be a goat which is a 66% chance.You will always have a 66% chance when switching.

    The prisoner scenario is completely different because there is no element of switching. Prisoner A simply knows that his odds go from 1/3 to 1/2.

    If in the Monty Hall problem he revealed a door with a goat before the contestant made any decision, then it would reflect the same logic as the prisoner scenario.

  4. I’m confused since the original question is and has always been “should I switch”?

    Your cases, such as Monty opening the door you chose first, don’t lead to a second choice and are therefore not the Monty Hall problem.

    Likewise, if Monty opens the door with the car behind it, why would he then give you the option to switch?

    The question does not require us to ask what Monty ‘always’ does but rather what did Monty do before giving you a non-trivial choice to switch.

    The brilliant game-theory contribution that you have made is to consider Monty’s interest as well. If Monty is trying to minimize his losses (such as if this game was being played in a Casino), we can imagine that being offered to switch would be associated with having the winning first choice. In this case, the solution is not obvious.

    If, at this casino, it was $5 to play and $10 jackpot. We may even describe the casino strategy that maximizes winnings. I.e. when and how often should the game give the Monty Hall option. Thanks

    1. The answer is that you should switch. Monty knows where the prize is but won’t show it and neither will show your guess.

      You are two-in-three likely, with your first guess, to pick either of the non-prize windows (of three in total).

      Pick either of those and the host will show you the other widow, also, NOT the prize, leaving the prize window, alone, for you to switch to if you switch.

      Of course, if your first guess is the prize and you switch, you’ll lose, but that is only one-in-three likely, because there is only one prize window of three.

  5. The usual explanation of this has always puzzled me, since it seems to focus on only one of the three possible pairs of doors.

    As everyone knows, it proposes that if you pick door A, there is a 1/3 chance that the prize is behind door A, and there must therefore be a 2/3 chance that it is behind “either B or C”.

    Once it is revealed not to be behind B, that 2/3 chance resides entirely in C, so it pays to switch to C.

    So far so good. But it seems to ignore other aspects.

    At the beginning, there was also a 2/3 chance that the prize would be found behind “either A or B” and “either A or C”.

    So, once the door to B is opened, the chances are as follows:

    “Either A or B” – 2/3 chance now resides entirely in A.
    “Either A or C” – 2/3 chance remains equally split between A and C.
    “Either B or C” – 2/3 chance now resides entirely in C.

    This gives A and C equal chances of having the prize.

    I’m sure I’m missing something, but I can’t see why what I tried to explain above wouldn’t be the logical extension of the usual explanation…

    1. What you’re missing is what the host knows but will never do. The host knows where the prize window is, but will never reveal it, meaning that, as a contestant, if you pick either of the non-prize windows (two in three likely) the host will show you the other window, also, not the prize, leaving the prize window, alone, for you to switch to if you switch.

      You will only switch and lose if your first guess IS the prize, one in three likely.

    2. I also have come to the conclusion, negating the hosts choice of the 2 possible goats, negates another 2/3, and when you count for that, the math returns to a 50/50 chance

  6. @Barnaby you are right , in fact the probability of both gates is equal unless you make certain assumptions about how monty chooses which aren’t even mentioned in the question.

  7. This is for those having trouble understanding the equivalence between the Monty Hall and Prisoner problems. I have had success explaining the solution to the Monty Hall when considering a more extreme (and obvious) version of the problem.

    Assuming all the formal assumptions laid out in this article suppose that instead of 3 doors with 1 car you have 100 doors with 1 car. After your choice, Monty chooses 98 of the remaining doors, avoiding the winning door if it is present. The only way the remaining door not doesn’t contain the car is if you had chosen the car to begin with, so switching will win you the car 99% of the time.

    Similarly, consider 100 prisoners where one is chosen to survive at random. Inmate A asks the warden to reveal 98 other inmates that will be executed, which the warden does. The inmate not named has a 99% chance of being pardoned since the only way for that not to be, is if inmate A was pardoned, which only occurs 1% of the time.

    1. An excellent comment, Josh. Conditional probability is a creepy thing, and extending it to these larger numbers can often reveal what makes the Monty Hall problem so frustrating — the psychological / cognitive fallacy of ‘anchoring.’

      However, what if someone were, when asked by Monty if they wanted to stick with door #1 or switch to door #100 (having revealed #2-#99 as goats), to themselves *flip a coin* to decide? Would that disrupt the chain of probability, and essentially reset it as a binary choice between doors A and B? I think it’s the sequential aspect of the game that is confusing.

      1. To understand the coin flip situation I imagine a different game show where we are asked to pick one of two boxes, knowing one has a prize inside. After losing, we are told one of the boxes had a 99% chance of having the prize, while the other box only had a 1% chance. This information is useless to us because we had no way of knowing which box was the 99% chance box at the moment of picking. Our chance of winning was always 50%.

        However in the 100 door Monty Hall problem we do know which box is the 99% chance box at the moment of picking, which makes all the difference.

        The problem is formulated in a misleading way that seems to imply that the switching of the choice or the revealing of the goats affects the outcome in some way, which is not true. In the end you’re just being asked if you want to enter a 100 ballot raffle with 1 ballot or 99 ballots, in a really complicated way 😛

  8. Focus only on the problem as stated. Don’t get distracted by assumptions or scenarios that aren’t part of the problem, such as what might be motivating the behavior of the host. The problem, as presented by Marilyn Vos Savant, contains *all* the information necessary to give an unequivocal answer as to whether to switch or not. In fact, more information than necessary. Is Monty restricted to opening a different door than the one you chose? Who cares? He did. What if Monty opened his door at random? He might reveal the car. But he didn’t. It doesn’t matter a whit if the host knows where the car is or not. You don’t need to be told in order to figure things out.

    Hint: Does the probability the car is behind the two doors you didn’t choose depend upon if those doors are opened simultaneously or one at a time? Answer: NO. Of course not. Agree?

    The point being that if you agree, and the prize is not revealed when the first door is opened, then the probability it will be revealed when the remaining door is opened is still 2/3. Which means that the probability it’s behind the door you picked has remained 1/3. And that the probability the prize is behind that remaining door you didn’t pick has changed from 1/3 to, of course, 2/3. Not to mention that the probability it was behind the door that was opened has changed from 1/3 to 0. Sort of blows the whole Monty Hall “problem” out of the water, no?

    Obviously, Switch.

  9. It was really upsetting to read this article. In this day and age there is no room for academics to be unclear on this problem. The Monty Hall Problem is irrefutably 2/3 probability if you switch. Do you believe in the Law of Large Numbers? Surely proving the difference between a 1/3 and 2/3 probability should not be difficult. Just run the tests, the scientific method shall set you free!

    1. Yes, at this time there is academical solution. And at same time it should be similiary clear to us that real life problems cannot be solved academically. Since you cannot assume subjective decision. And even if you can affect it, that creates room to voluntarily change it just to reverse outcome.

  10. Once the car is placed behind a door, the probability it is behind that door is 1 and the probability it is behind any other door is 0. The probability of winning on a single game is the probability of selecting the door times the probability the selected door has the car behind it. Hence, the probability of winning is 0 if the car is not behind the selected door and 1 if it is.
    The simulations only show that for many, many games 2/3 of the times switching wins. It does not show that on a single game, switching has a 2/3 probability of winning.
    How else do we know? When playing the game several times, on each game you do count it as 0 or 1 if you lost or won when switching. You don’t count it as 1/3 or 2/3.

    1. The probability is the same no matter the number of games.

      The probability of winning on a single game is the probability of selecting the door times the probability the selected door has the car behind it.

      That probability is 1 in 3. Whether it’s one game or 1,000 games, it is 1 in 3. The odds are 2 to 3 that your choice of doors is wrong. When the first door is opened you’re basically being told that IF your choice is wrong then the remaining door has the car. What would you do?

  11. Great. I like thinking like this. Isn’t the game explaining it wrong but sounding like you’re right even if it’s only for a few seconds which is fun cos you are being wrong. Here’s my effort. I was thinking try imagining you are monty. Then that half the time the contestant switches and half the time they stick. So 1 third of the time they choose goat 1. You open a door. Assuming the contestant is halftime-changing-halftime-sticking, half of the time you give them the car. The second third of the time the contestant chooses goat 2. You opens a door. Again in this situation if contestant is halftime-stick-halftime-switch, half of the time you give them the car. The third third of the time he chooses the car. You open a door and again if the contestant is halftime-stick-halftime -switch, half of the time you give them the car. So EVERY TIME if the contestant is half of the time switching or sticking, half of the time you give them the car. Therefore if the contestants choices are 50-50
    Then there’s a 50-50 chance you give away a car. Therefore as the contestant your chances are 50-50 if you stick or twist.

    1. The 50-50 don’t really work. If you pick the car at first you will only win if you don’t switch but odds are 1/3 this happens. If you pick a goat first time you will only win if you switch and odds are 2/3 this happens.

    2. Maybe more i should state it more precisely:
      1/3 of the time you will pick the car and you will only win if you DONT switch.
      2/3 of the time you will pick a goat and you will only win if you DO switch.
      So odds are you picked a goat and should switch and you will win the car 66% of the time and only lose if you picked the car and then switched but that would only be 33%.

  12. Monty’s motivations are irrelevant. It doesn’t matter if Monty sometimes reveals the car. He didn’t this time. It doesn’t matter if Monty sometimes doesn’t reveal anything at all. This time he did. It doesn’t matter whether Monty even knows where the car is. Maybe in one out of three games Monty shoots the contestant instead of opening a door. Even that doesn’t matter because in this case, he didn’t. The only thing that matters is whether or not Monty revealed a goat behind one of the other doors. If he did, you double your winning chance by switching.

    Since the problem states that Monty did indeed reveal a goat behind one of the other doors, the correct answer is that you should switch in this instance. No other assumption is needed.

    1. You don’t double the chances of winning if the host opens a door without knowing where the car is, it’s 50/50 if the door happens to reveal a non-prize.

      The car would simply be randomly placed 1/3 of the time behind the host’s door, 1/3 of the time behind the contestant’s door, and 1/3 of the time behind the door neither of them picked.

      1. The host doesn’t have a door and it doesn’t matter what he knows. He is irrelevant, he’s a door opener and the door he opens has a goat. It’s not about your choice vs. his choice, it’s about your choice and its probability, nothing else.

        You’re choosing one of three and odds are that you are making the wrong choice. It’s 2 to 3 against you.
        When the first door is opened you now know which door has the car IF you were wrong with your initial choice. Which is 2/3 of the time. That’s why it’s better to switch. It’s not a matter of 50/50 because your previous choice and its circumstances are the core of the problem. It’s about whether or not your chances improve by switching, not whether you can correctly guess between two doors.

        1. You’re right that two-in-three of the time, your first guess will be either of the non-prize windows. But when it is, the host’s knowledge of where the prize window is is what also makes sure that if you switch, you win.

          The host, faced with actually NOT a choice between showing you the prize window, or showing you the (only remaining) non-prize window, must show the latter, leaving the prize window for you to switch to.

          If the prize is A

          You can choose B or C

          And switch

          And win.

          If the prize is B

          You can choose A or C

          And switch

          And win.

          If the prize is C

          You can choose A or B

          And switch

          And win.

          Concise version: if your first guess is NOT the prize (two-in-three likely) you will win if you switch.

          The host will show you the other non-prize window.

          Leaving the prize window for you to switch to if you switch.

          You will only switch and lose if your first guess is the prize window (one-in-three likely).

        2. No, despite you start choosing wrong 2/3 of the time, the host actions can decide how much of those 2/3 you will have the opportunity to win by switching. If the host knows the positions and so always reveals a goat and offers the switch, then you win with the same frequency as you start getting it wrong: 2/3. Now, go to the other extreme: imagine the host knows the positions but only reveals a goat and offers the switch if you started picking the car door (if you started picking a goat he does not reveal anything and the game ends). That would mean that if he opened a door with a goat, it is because it is 100% sure that the car is behind your door, which implies that it is impossible to win by switching.

          As you see, the revelation of the goat tells you that you are in a specific subset of the total cases (here only in the subset in which the car is behind your door), and in a subset the proportions don’t need to remain the same.

          Now, when the host randomly reveals a door that is not which the contestant chose, it is a midpoint between the two previous examples. These three things could happen:

          1) In 1/3 the player starts picking the car door and later the host necessarily reveals a goat -> staying wins.
          2) In 1/3 the player starts picking a goat door and later the host reveals the other goat door -> switching wins.
          3) In 1/3 the player starts picking a goat door and later the host reveals the car.

          So, if a goat is revealed, you know you are only under one of the first two cases, where each of them represents the half of that subset, and so each has 1/2 probability now.

          If you like one more example, suppose another scenario in which you are in front of 30 persons, where 20 of them are engineers and 10 of them are doctors, but you don’t know the respective profession of none of those people. You have a list of their names, so you call one by his/her name with a speaker and that person approaches you. We both agree that it is 2/3 likely that the person is an engineer, right?

          Now, suppose that you know that all of the doctors are using blue jacket, but only half of the engineers are using blue jacket (the other 10 are using white jacket). So, in total we have:

          10 doctors use blue jacket –> 1/3 of the total people.
          10 engineers use blue jacket –> 1/3 of the total people.
          10 engineers use white jacket –> 1/3 of the total people.

          So, if you see that the person that approaches you is using blue jacket, would you still say that he/she is 2/3 likely to be an engineer?

          Note that in total there are only 20 people using blue jacket, from which 10 are doctors and 10 are engineers (50% chance for each). So once you see the blue jacket, what you have to do is to restrict your sample space to the only 20 people that use blue jacket, not count as if all the 30 people used blue jacket.

          Same with the revelation of the goat: just because you saw a goat revealed that does not mean that you should still count all the 2/3 cases in which you start failing as if that revelation would occur in all of them. Instead, you must restrict to the subset in which it happens.

      2. You’re right that the host’s knowledge of where the prize is is vital to understanding how the Monty Hall principle works.

        If the host has no clue where the prize window is, he is likely as not to wreck the game by revealing the prize window when he opens one.

  13. Monty CAN’T open randomly. If Monty is managing the game he plausibly knows before he starts if it’s going to be the “switch doors” game or “pick a door and we’ll see if you won” game. If it’s the “switch doors” game he CAN’T show the car because that obviates the need to switch doors and it degenerates into the “pick a door” game.

    1. Absolutely irrelevant. All you need to know is he opened a door and it has a goat. Why he chose that door or whether he already knew it had a goat is as important as knowing if the doors are red or blue. You’re not competing against the host in any way. You’re choosing one of three, host opens a door, say hi to the goat. Why wasn’t it a car? Because the question says it wasn’t.

  14. In order to test the probability of each scenario, the Monty Hall model needs to be traversed 36 times with the CAR behind door #1 twelve times, door #2 twelve times and door #3 twelve times.

    If the car was behind door #1, and I as the contestant had twelve tries, based on probability, I should select Door #1 four times, Door #2 four times and Door #3 four times (see table below).

    For the four times I selected Door #1, Monty should show me the goat behind Door #2 two times (once I would switch and get the goat, the other I would not switch and win the CAR). Monty should also show me the goat behind Door #3 two times (once I would switch and get the goat, the other I would not switch and win the CAR).

    For the four times I selected Door #2, Monty should show me the goat behind Door #3 all four times since he can’t show me the CAR behind door #1. Twice I would switch and win the CAR, twice I would not switch and get the goat.

    For the four times I selected Door #3, Monty should show me the goat behind Door #2 all four times (again, he can’t show the CAR). Twice I would switch and win the CAR, twice I would not switch and get the goat.

    The table below lists all options and the only options for the CAR behind Door #1. 4 of the options are repeated to balance the distribution of the options and equal the probability of each.

    Prize Door # Selected Door # Shown Door # Switch? Result
    1 1 2 Y goat
    1 1 2 N CAR
    1 1 3 Y goat
    1 1 3 N CAR
    1 2 3 Y CAR
    1 2 3 Y CAR (repeat)
    1 2 3 N goat
    1 2 3 N goat (repeat)
    1 3 2 Y CAR
    1 3 2 Y CAR (repeat)
    1 3 2 N goat
    1 3 2 N goat (repeat)

    Each of the above 12 options has the same probability of occurrence. 6 of the options result in winning the CAR, 6 get the goat. My claim is that at the time the decision is made to switch or not switch, the probability of winning the CAR is 50% regardless of the decision. The only way to calculate a different probability of winning the CAR as a result of deciding to switch doors would be to show why/how any of the above 12 options has a higher or lower probability of occurrence than any other option in the list. I don’t think it can be done.

    1. You are correct. There are six winning games if you do not switch and six winning games if you do switch. The error most people make is they say: Game A. Prize Door 1, Player picks Door 1 , Monty opens Door 2. is the same game as Game B Prize Door 1, Player picks Door 1, Monty opens Door 3.’ That is incorrect. They are two separate games, two separate ways of winning by not switching. Your table is correct.

  15. The probability is the same no matter the number of games.

    The probability of winning on a single game is the probability of selecting the door times the probability the selected door has the car behind it.

    That probability is 1 in 3. Whether it’s one game or 1,000 games, it is 1 in 3. The odds are 2 to 3 that your choice of doors is wrong. When the first door is opened you’re basically being told that IF your choice is wrong then the remaining door has the car. What would you do?

  16. I’m glad to find this article, as I’ve been thinking for a while that failing to recognise and spell out the behavioural assumptions with the Monty Hall problem can undermine what is otherwise sound mathematically, and it is interesting that this is not universally acknowledged.

    I’ve no doubt that the maths is sound (although I admit that I had to go through every possible combination of events to be convinced), but only under the stringent behavioural assumptions outlined in the article.

    If the game show host is particularly miserly and wants to minimise the chances of you winning the car (and this is perfectly plausible if the show has a low budget) then he / she will only offer to switch doors if the car is behind the door that you’ve already selected. In that case, switching would result in 100% probability of losing, and not switching would result in 100% probability of winning. The probability that the host will behave in that way cannot be estimated unless you’ve got data from previous trials.

  17. Suppose an Observer stands next to a Contestant who picks door 1. After Monty reveals a goat behind door 2, the Contestant accepts Monty’s invitation to change his choice to door 3. Monty then invites the Observer to pick a door and receive a duplicate of whatever is behind it (no prize splitting). Should the Observer now also pick door 3, or is door 1 an equal probability for the Observer?

    1. If the observer watched the whole thing then yes of course his odds are the same, his knowledge is the same as the contestants. If, however, someone were to be shown the two remaining doors not knowing that a door had ever been removed and not knowing that the door removed was definitely a wrong choice then his odds would be 50/50.

  18. the following shows exactly exactly what the probabilities involved in the Monty Hall problem are:

    What are problems like the Monty Hall Problem?

    Here’s one: Originally urn 1 contains 1 black ball and 9 white balls. A ball is picked at random from urn 1 and placed in urn 2. Balls are then drawn from urn 1 (now having one less ball) without replacement: This illustrates/demonstrates the probabilities of Monty revealing the prize,and/or you getting it depending on your choice, as doors are shown not to have the prize behind them.
    Probabilities of what color the ball is on each particular pick are shown below depending on what happened when the first ball was taken from urn 1 and on the previous pick(s). Note N is the number of the “pick” after you’ve chosen which ball to put in urn 2 and the probabilities in a particular row are the apriori probabilities of getting a particular color ball on the pick you are making. The last column represents(9/10 * the probabilities in the columns for when urn 2 contains a white ball) Note that as you go down the yellow shaded column those probabilities are essentially conditional probabilities, in that they reflect having the event(s) in the cells immediately above not occur on the previous picks,i.e. not getting a black ball=not revealing the prize and, correspondingly, the event(s) to the immediate left occur. urn 2 having the black ball corresponds to you having made the correct choice for the prize. we are interested in the probability of getting the prize(black ball) as Monty opens doors equivalent now to picking from urn1. Monty is not allowed to touch urn 2. Of course if he were, the probability he would reveal the prize would be 1/10.

    This is exactly what happens in the Monty Hall problem. Say instead of 3 doors, there were 10 doors, with the prize behind only one. A door is first chosen, but then eliminated from further participation in the selection process. The last row in the table above, highlighted in green, represents the probabilities associated with you choosing the last ball from urn 1 or, equivalently, choosing the final remaining unopened door in the Monty Hall problem, instead of your original choice.
    Just for fun let’s make the same table for urn 1 originally containing 3 balls, 2 White, 1 Black. We have:

    So, the black ball (the prize) has a 2/3 probability of being behind the remaining unopened door, and if there were 10 doors to begin with, a 9/10 chance of the same. And not a word has been spoken about whether Monty knows where the prize is or not. Only tabulating the probabilities for getting the prize (or not) as doors besides the one you originally chose are opened…

  19. Your solution of the three prisoners problem is incorrect. Revealing that prisoner B will die doesn’t increase prisoner A’s chances of survival to 50%, it only reveals that prisoner C has a 67% chance of being the one pardoned. You see, the warden would only reveal B as one of the condemned in 2 possible scenarios: (i) C was to be pardoned, which happens 1/3 of the time, or (ii) A was to be pardoned, and the warden’s coin flip selected B, which happens 1/6 [1/3*1/2] of the time. Therefore, the warden Revealing B as to be executed merely revealed to A that C has a better chance of having been pardoned, nothing about his own chances.

  20. I just heard of this problem for the first time this evening and I am having a hard time understanding why this is so complicated.

    So there are three doors, A, B, and C, and I have picked one of the three, Door A, thinking the car is behind it, not emptiness. Now, one of four possible scenarios will play out.

    1. The car is behind Door C, so the Host must and does open Door B. If I switch, I win.
    2. The car is behind Door B, so the Host must and does open Door C. If I switch, I win.
    3. The car is behind Door A; so the Host can open either, and opens Door B. If I switch, I lose.
    4. The car is behind Door A, so the Host can open either, and opens Door C. If I switch, I lose.

    The key issue I see people going wrong around is treating two distinct cases, Case 3 and Case 4, as if they are one case. They are not.

    Of the four scenarios that may occur, there are two in which the Host opens Door B (one where Door A has the car, and one where it does not) and two in which the Host opens Door C (one where Door A has the car, and one where it does not).

    Importantly, even though there are two of the four scenarios associated with the car being behind Door A, the car is only behind Door A one third of the time, so the four cases are not on an equal odds footing, that is, the four cases are not each 25 percent likely:

    1/3 of the time car is behind Door C and Host will open B
    1/3 of the time car is behind Door B and Host will open C
    1/6 of the time car is behind Door A and Host will open B
    1/6 of the time car is behind Door A and Host will open C

    The number of times the Host opens Door B is: 1/3 + 1/6 = 3/6 = 1/2
    The number of times the Host opens Door C is: 1/3 + 1/6 = 3/6 = 1/2

    Having picked Door A, and confronted with Door B being opened, I know that this happens 1/2 the time.
    Having picked Door A, and confronted with Door C being opened, I know that this happens 1/2 the time.

    Yet I also additionally know that:

    In cases where Door B is open, 2/3 of those times the car is behind C (based on the 1/3 + 1/6 portions).
    In cases where Door C is open, 2/3 of those times the car is behind B (based on the 1/3 + 1/6 portions).

    This indicates that I should always switch, period.

    Since 1/2 the time it will be Door B open, and 1/2 the time Door C open, if I switch every time, and do so with a 2/3 chance of getting the car by switching, I get the car this many times overall:

    (1/2 * 2/3) + (1/2 * 2/3) = 2/3 of the time

    It seems counter-intuitive, but probabilities are reflect what we know. Being given a choice of three doors, choosing one of them, and then having one of the remaining two doors eliminated as a possibility by the Host, and being offered to choose again, is just not at all the same as being given a choice of three doors, end of story.

    Plain English can help make this less confounding. I know that 2/3 of the times I choose Door A, the car will turn out instead to be behind Door B or Door C. If the Host tips us off that the car is not behind Door C (for example), then (2/3 of the time!) the car is going to be behind Door B.

    1. The most plain English of all is this:

      Whenever you pick a window not the prize and switch, you MUST win.

      The host shows you the other non-prize window, leaving the prize window for you to switch to, if you switch.

      You are two-in-three likely to pick a non-prize window; therefore two-in-three likely to win if you switch.

      One time in three, your first guess will be the prize.

      Then you will switch and lose.

  21. I don’t even know if this is monitored at all but I find this to be a classics problem of over analysis and over thinking. It is true that switching increases the probability but we have to look at each separate choice event. The initial choice is 1/3. If the contestant stays with that selection and does not change then their probability of selecting the prize is still 1/3. But if the contestant switches then they have a 1/2 probability since they are now selecting between two choices, once goat one prize. That’s it. Since the host removes one of the goats it simply renders it a 50/50 of selecting the prize. But this probability only changes once the contestant selects again. Initial selection probability 1/3. If no swap the the probability still remains at 1/3. But if the contestant swaps their choice they are now choosing between 2 choices only thus a 1/2 chance.

    1. I’m with Bill here. The first choice has absolutely nothing to do with the second. The car is behind 1 of 3 doors. Once a door has been shown to not have a car, the second choice is simply a car behind one of two doors. Two completely unrelated choices.

  22. There is nothing weird or out of place with mathematicians setting a problem and outlining the assumptions. What motivates a player or why Monty Hall picks a certain door is irrelevant. It is easy to list all the winning games. There are twelve winning games: six wins from not switching, six wins from switching. Scroll up to the comment by Stefano Edward 2 June 2021. He lists the six wins from not switching.

    1. What are you talking about?? In Stefano’s table there are 6 rows with ‘Y’ for the Always Switch (AS) strategy and 6 rows with ‘N’ for the Never Switch (NS) strategy. Also shown are the outcomes for each row (CAR or goat).
      Among the 6 ‘Y’ rows for the AS strategy, there are 4 wins (CAR), so the probability of winning the CAR for AS is 4/6 = 2/3.
      Among the 6 ‘N’ rows for the NS strategy, there are only 2 wins (CAR), so the probability of winning the CAR is 2/6 = 1/3.
      This reflects the simple, inescapable fact that your first guess has a 1/3 chance of being correct. And since, if your first guess were wrong (which is 2/3 of the time), then the CAR must be behind the door that MONTY does not open and that you did not pick initially.
      So, obviously you should switch!
      Sure, in this game, 1/3 of the time you will LOSE by switching, BUT 2/3 of the time you will WIN by switching. The Always Switch strategy doubles your chances of winning!
      On average, switching is twice as likely to get you a car. AND MONTY cannot change that fact.

  23. There is a flaw in the reasoning behind the Monty Hall problem. Initially there are 3 doors the probability of choosing the car is 33%. Monty opens a door you haven’t chosen revealing a goat. The reasoning behind switching is that the probability of the car being behind the remaining door you haven’t chosen is 66% as the opened door has zero. This is false, there are now 2 unopened doors, the chance of the car being behind one or the other is now 50% irrespective of your initial choice. Your choice to stay or switch is a simple 50:50, your initial choice has no bearing on the outcome.

    1. Assuming we always switch doors, our initial choice determines whether we ultimately win or lose. The only time we lose when switching doors is when we choose the winning door initially. Since we know that there is a 1/3 chance of choosing the winning door from the outset, that gives us a 2/3 chance of winning when we switch.

      We can test every possibility. Let’s say the car is behind door 3.

      We choose door 1, monty opens door 2, we switch to door 3 and win
      We choose door 2, monty opens door 1, we switch to door 3 and win
      We choose door 3, monty opens 1 or 2, we switch to door 1 or 2 and lose

  24. the most important question :
    does he ALWAYS propose to switch ?
    if yes, we have to switch
    if not, it depends

  25. Statistics are based on many* choices/outcomes. If you have one shot, never switch. Yes, in the long run, of course you would switch. In practice, NEVER switch. In horse racing, try betting a non-favorite, where favorites only have a 30% chance of winning and you will feel the pain because you will lose a lot of money. Always stick with your first choice IN THE SHORT RUN. *many means thousands of plays. In the MONTY game, once I pick, if he can’t agree with my choice, he is irrelevant.

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