# Monty Hall problem

Bertrand’s box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his Calcul des probabilités, published in 1889.

There are three boxes:

1. a box containing two gold coins,
2. a box containing two silver coins,
3. a box containing one gold coin and one silver coin.

After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is 12; in fact, the probability is actually 23. Two problems that are very similar are the Monty Hall problem and the Three Prisoners problem.

These simple but slightly counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including theKolmogorov axioms.

The Monty Hall problem is a brain teaser, in the form of a probability puzzle (Gruber, Krauss and others), loosely based on the American television game show Let’s Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader’s letter quoted inMarilyn vos Savant‘s “Ask Marilyn” column in Parade magazine in 1990 (vos Savant 1990a):

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 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 switch your choice?

Vos Savant’s response was that the contestant should switch to the other door. (vos Savant 1990a)

The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975b) and by vos Savant (1991a), that the host always opens a different door from the door chosen by the player and always reveals a goat by this action—because he knows where the car is hidden. Leonard Mlodinow stated: “The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated.” (Mlodinow 2008) It is also assumed that the contestant prefers to win a car, rather than a goat.

Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance. One explanation notices that 2/3 of the time, the initial choice of the player is a door hiding a goat. The host is then forced to open the other goat door, and the remaining one must, therefore, hide the car. “Switching” only fails to give the car when the player picks the “right” door to begin with, which only has a 1/3 chance.

Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).

The Monty Hall problem has attracted academic interest from the surprising result and simple formulation. Variations of the Monty Hall problem are made by changing the implied assumptions and can create drastically different consequences. For one variation, if Monty only offers the contestant a chance to switch when the contestant initially chose the door hiding the car, then the contestant should never switch. For another variation, if Monty opens another door randomly and happens to reveal a goat, then it makes no difference (Rosenthal, 2005a), (Rosenthal, 2005b).

The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand’s box paradox.

The problem continues to attract the attention of cognitive psychologists. The typical behaviour of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: 1) the endowment effect (Kahneman et al., 1991); people tend to overvalue the winning probability of the already chosen – already “owned” – door; 2) the status quo bias (Samuelson and Zeckhauser, 1988); people prefer to stick with the choice of door they have already made; 3) the errors of omission vs. errors of commission effect (Gilovich et al., 1995); all else considered equal, people prefer that any errors that they are responsible for to have occurred through ‘omission’ of taking action rather than through having taken an explicit action that later becomes known to have been erroneous. Experimental evidence confirms that these are plausible explanations which do not depend on probability intuition (Kaivanto et al., 2014Morone and Fiore, 2007).

### Criticism of the simple solutions

As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given the contestant initially picks door 1 and the host opens door 3; various ways to derive and understand this result were given in the previous subsections. Among these sources are several that explicitly criticize the popularly presented “simple” solutions, saying these solutions are “correct but … shaky” (Rosenthal 2005a), or do not “address the problem posed” (Gillman 1992), or are “incomplete” (Lucas et al. 2009), or are “unconvincing and misleading” (Eisenhauer 2001) or are (most bluntly) “false” (Morgan et al. 1991). Some say that these solutions answer a slightly different question – one phrasing is “you have to announce before a door has been opened whether you plan to switch” (Gillman 1992, emphasis in the original).

The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability 2/3, and hence that switching is the winning strategy, if the player has to choose in advance between “always switching”, and “always staying”. However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given the player has picked door 1 and the host has opened door 3. As one source says, “the distinction between [these questions] seems to confound many” (Morgan et al. 1991). This fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation the following two questions have different answers:

1. What is the probability of winning the car by always switching?
2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?

The answer to the first question is 2/3, as is correctly shown by the “simple” solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty’s preference for rightmost doors means he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3). For this variation, the two questions yield different answers. However as long as the initial probability the car is behind each door is 1/3, it is never to the contestant’s disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2. (Morgan et al. 1991)

Players who STAY have won 49040 cars out of 145987 games yielding a winning percentage of 34%
players who SWITCH have won 68356 cars out of 103063 games yielding a winning percentage of 66%

## Video Transcript:

You’re on a game show and there are three doors in front of you. The host, Monty Hall, says, “Behind one door is a brand new car. Behind the other two doors are goats. Pick a door!” You think, “Well, it doesn’t matter which door I choose, every door has a 1/3 chance of having the car behind it.” So, you choose door number 1. Now it gets interesting. Monty, the host, who knows where the car is, opens door number 2 and reveals a goat. The host always opens a door to reveal a goat. The host says, “If you want, you can switch to door number 3.” What should you do? Stay with your original choice or switch to the other door? All right, so what are you going to do? Stay or switch? Well, it’s a fifty-fifty chance of winning the car in either door. Right? [Wrong!] You actually double your chances of winning the car by switching doors. And that is why the Monty Hall Problem is so evasive!
Choose an explanation to the Monty Hall Problem:

1/3 vs 2/3 – Solution #1 to the Monty Hall Problem
There is a 1/3 chance of the car being behind door number 1 and a 2/3 chance that the car isn’t behind door number 1. After Monty Hall opens door number 2 to reveal a goat, there’s still a 1/3 chance that the car is behind door number 1 and a 2/3 chance that the car isn’t behind door number 1. A 2/3 chance that the car isn’t behind door number 1 is a 2/3 chance that the car is behind door number 3.
100 Doors! – Solution #2 to the Monty Hall Problem
Imagine that instead of 3 doors, there are 100. All of them have goats except one, which has the car. You choose a door, say, door number 23. At this point, Monty Hall opens all of the other doors except one and gives you the offer to switch to the other door. Would you switch? Now you may arrogantly think, “Well, maybe I actually picked the correct door on my first guess.” But what’s the probability that that happened? 1/100. There’s a 99% chance that the car isn’t behind the door that you picked. And if it’s not behind the door that you picked, it must be behind the last door that Monty left for you. In other words, Monty has helped you by leaving one door for you to switch to, that has a 99% chance of having the car behind it. So in this case, if you were to switch, you would have a 99% chance of winning the car.
Pick a Goat – Solution #3 to the Monty Hall Problem
To win using the stay strategy, you need to choose the car on your first pick because you’re planning to stay with your initial choice. The chance of picking the car on your first pick is clearly one out of three. But, in order to win using the switch strategy, you only need to pick a goat on your first pick because the host will reveal the other goat and you’ll end up switching to the car. So you want to use the strategy that lets you win if you choose a goat initially because you’re twice as likely to start by picking a goat.
Scenarios – Solution #4 to the Monty Hall Problem
To understand why it’s better to switch doors, let’s play out a few scenarios. Let’s see what will happen if you were to always stay with your original choice. We’ll play out three scenarios, one for each door that the car could be behind (door number 1, door number 2, or door number 3). And it doesn’t matter which door you start out with, so, to keep it simple, we’ll always start by choosing door number 1.
Stay strategy, scenario 1: the car is behind door number 1. You choose door number 1, then the host reveals a goat behind door number 2 and because you always stay, you stay with door number 1. You win the car! Stay strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host reveals a goat behind door number 3, and you’re using the stay strategy so you stay with door number 1. You get a goat and don’t win the car. Stay strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you stay with door number 1, and you get a goat. So, using the stay strategy, you won the car one out of three times. That means that in any one instance of playing the game, your chance of winning the car if you choose to stay is 1/3 or about 33%.

Now let’s try switching doors. Again, we’ll always start by picking door number 1. Switch strategy, scenario 1: the car is behind door number 1. You choose door number 1, the host opens door number 2 to reveal a goat, you are using the switch strategy so you switch to door number 3. You get a goat. Switch strategy, scenario 2: the car is behind door number 2. You start by picking door number 1, the host opens door number 3 to reveal a goat, you switch to door number 2 and win the car! Switch strategy, scenario 3: the car is behind door number 3. You pick door number 1, the host opens door number 2 to reveal a goat, you switch to door number 3 and win the car again! So, with the switch strategy you won the car 2 out of 3 times. That means, that in any one instance of the game, your chance of winning the car if you choose to switch doors is 2/3 or about 67%.

Therefore, if you play the game three times and stay, on average you’ll win the car once. But if you play the game three times and switch each time, on average you’ll win the car twice. That’s twice as many cars!