To switch or not to switch
To switch or not to switch
The Monty Hall problem is named for its similarity to the Let's Make a Deal television game show hosted by Monty Hall. The problem is stated as follows. Assume that a room is equipped with three doors. Behind two are goats, and behind the third is a shiny new car. You are asked to pick a door, and will win whatever is behind it. Let's say you pick door 1. Before the door is opened, however, someone who knows what's behind the doors (Monty Hall) opens one of the other two doors, revealing a goat, and asks you if you wish to change your selection to the third door (i.e., the door which neither you picked nor he opened). The Monty Hall problem is deciding whether you do.
The correct answer is that you do want to switch. If you do not switch, you have the expected 1/3 chance of winning the car, since no matter whether you initially picked the correct door, Monty will show you a door with a goat. But after Monty has eliminated one of the doors for you, you obviously do not improve your chances of winning to better than 1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3 chance you will win the car (counterintuitive though it seems).
https://mathworld.wolfram.com/MontyHallProblem.html
The Monty Hall dilemma became famous in 1990 when it was presented in the “Ask Marilyn” column in Parade, a magazine inserted in the Sunday edition of hundreds of American newspapers. The columnist was Marilyn vos Savant, known at the time as “the world’s smartest woman” because of her entry in the Guinness Book of World Records for the highest score on an intelligence test. Vos Savant wrote that you should switch: the odds of the car being behind Door 2 are two in three, compared with one in three for Door 1. The column drew ten thousand letters, a thousand of them from PhDs, mainly in mathematics and statistics, most of whom said she was wrong. Here are some examples:
You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I’ll explain. After the host reveals a goat, you now have a one‑in‑two chance of being correct. Whether you change your selection or not, the odds are the same. 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
I am sure you will receive many letters on this topic from high school and college students. Perhaps you should keep a few addresses for help with future columns.
—W. Robert Smith, PhD, Georgia State University
Maybe women look at math problems differently than men.
—Don Edwards, Sunriver, Oregon
Among the objectors was Paul Erdös (1913–1996), the renowned mathematician who was so prolific that many academics boast of their “Erdös number,” the length of the shortest chain of coauthorships linking them to the great theoretician.
Analyzing our Tree Diagram, we notice that exactly half of the outcomes are marked, meaning that the player wins by switching in half of all outcomes. You might be tempted to conclude that a player who switches wins with probability 1/2.
This is wrong. The reason is that these outcomes are not equally likely. As you can see, outcome ABC and outcome BBA have different probabilities why?
The reason why it is not ½ is because of the condition that every time after your pick host is revealing a door that has a goat behind it and this particular condition impacts the likelihood of final outcomes in our sample space.
Explanation with Conditional Probability The conditional probability is the probability of any event A given that another event B has already occurred. The idea here is that the probabilities of an event “maybe” affected by whether or not other events have occurred. The term “conditional” refers to the fact that we will have additional conditions, restrictions, or other information when we are asked to calculate this type of probability.
It is denoted in the following manner – “representing the Probability of A given B has occurred”
Conditional Probability can be calculated as Probability of A intersection B, divided by the probability of event B
P(A | B) = P(A ∩ B) / P(B)
https://www.analyticsvidhya.com/blog/2020/08/probability-conditional-probability-monty-hall-problem/
‘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 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.
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.
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:
- Monty will always open a door.
- Monty never opens the door you have chosen.
- Monty never opens the door with the car behind it.
- The car is equally likely to be behind any door.
- 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:
- 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.
- He has to entertain the audience; this is likely to mean, among other things, that the contestants win cars reasonably often.
- 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. 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.
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.
https://ima.org.uk/4552/dont-switch-mathematicians-answer-monty-hall-problem-wrong/