# Monty Hall Problem

The **Monty Hall problem** is a famous, seemingly paradoxical problem in conditional probability and reasoning using Bayes' theorem. Information affects your decision that at first glance seems as though it shouldn't.

In the problem, you are on a game show, being asked to choose between three doors. Behind each door, there is either a car or a goat. You choose a door. The host, Monty Hall, picks one of the other doors, which he knows has a goat behind it, and opens it, showing you the goat. (You know, by the rules of the game, that Monty will always reveal a goat.) Monty then asks whether you would like to switch your choice of door to the other remaining door. Assuming you prefer having a car more than having a goat, do you choose to switch or not to switch?

The solution is that switching will let you win twice as often as sticking with the original choice, a result that seems counterintuitive to many. The Monty Hall problem famously embarrassed a large number of mathematicians with doctorate degrees when they attempted to "correct" Marilyn vos Savant's solution in a column in *Parade Magazine*. [1]

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## Possible Outcomes

One way to see the solution is to explicitly list out all the possible outcomes, and count how often you get the car if you stay versus switch. Without loss of generality, suppose your selection was door \(1\). Then the possible outcomes can be seen in this table:

In two out of three cases, you win the car by changing your selection after one of the doors is revealed. This is because there is a greater probability that you choose a door with a goat behind it in the first go, and then Monty is *guaranteed* to reveal that one of the other doors has a goat behind it. Hence, by changing your option, you double your probability of winning.

Another way of seeing the same set of options is by drawing it out as a decision tree, as in the following image:

## Using Bayes' Theorem

There are two aspects of the Monty Hall problem that many struggle to agree with. First, why aren’t the odds 50-50 after the host opens the door? Why is it that switching doors has a 2 in 3 chance of winning when sticking with the first pick only has a 1 in 3 chance? Secondly, why is it the case that if Monty opened a door truly randomly and happened to show a goat, then the odds of staying vs. switching doors are now 50-50? Bayes' theorem can answer these questions.

Bayes’ theorem is a formula that describes how to update the probability that a hypothesis is correct, given evidence. In this case, it’s the probability that our initially picked door is the one with the car behind it (that staying is right) given that Monty has opened a door showing a goat behind it: that Monty has shown us that one of the choices we didn’t pick was the wrong choice. Let \(H\) be the hypothesis "door 1 has a car behind it," and \(E\) be the evidence that Monty has revealed a door with a goat behind it. Then the problem can be restated as calculating \(P(H \mid E)\), the conditional probability of \(H\) given \(E\).

Since every door either has a car or a goat behind it, the hypothesis "\(\text{not} H\)" is the same as "door 1 has a goat behind it."

In this case, Bayes' theorem states that

\[P(H \mid E) = \frac{P(E \mid H)} {P(E)} P(H) = \frac{ P(E\mid H) \times P(H)} {P(E\mid H) \times P(H) + P(E\mid \text{not} H) \times P(\text{not} H)}. \]

The problem as stated says that Monty Hall **deliberately** shows you a door that has a goat behind it.

Breaking down each of the components of this equation, we have the following:

- \( P(H) \) is the prior probability that door 1 has a car behind it, without knowing about the door that Monty reveals. This is \(\frac{1}{3}\).
- \( P(\text{not} H) \) is the probability we did not pick the car with the door behind it. Since the door either has the car behind it or not, \(P(\text{not} H) = 1 - P(H) = \frac{2}{3}\).
- \( P(E \mid H) \) is the probability that Monty shows a door with a goat behind it, given that there is a car behind door 1. Since Monty always shows a door with a goat, this is equal to \(1\).
- \( P(E\mid \text{not} H) \) is the probability that Monty shows the goat, given that there is a goat behind door 1. Again, since Monty always shows a door with a goat, this is equal to \(1\).

Combining all of this information gives \[ P(H \mid E) = \frac {1 \times \frac13} {1 \times \frac13 + 1\times \frac23} = \frac {\hspace{1mm} \frac13\hspace{1mm} }{1} = \frac{1}{3}. \]

The probability that the car is behind door 1 is completely unchanged by the evidence. However, since the car can only either be behind door 1 or behind the door Monty didn't reveal, the probability it is behind the door that is not revealed is \(\frac{2}{3}\). Therefore, switching is twice as likely to get you the car as staying.

## Variant: The Random Monty Hall Problem

This result depends crucially on the fact that Monty was always guaranteed to open a door with a goat behind it, regardless of what door you picked initially. That is, \(P(E \mid H) = P(E \mid \text{not}H)\). Now consider what would happen if Monty *randomly* opened a door we did not pick and it contained a goat. What is the probability that our first pick is correct, regardless of which specific door we picked?

In this case, \( P(E \mid H), \) the probability that Monty will show us a door with a goat behind it given that our first pick was correct and does have a car behind it, is still \(1\). If you picked the door with the car, the two other doors are guaranteed to have a goat.

But \( P(E \mid \text{not} H) \) changes. This represents the probability that Monty picks a door with a goat behind it given that your original choice of door had a goat behind it. In this case, Monty chooses randomly between one door that has a goat behind it and one door that doesn't have a goat behind it. Therefore, the probability that he picks a door with a goat is \(P(E \mid \text{not} H) = \frac{1}{2}\).

Laying this out, we get

\[ P(H \mid E) = \frac{ P(E\mid H) \times P(H)} {P(E \mid H) \times P(H) + P(E \mid \text{not} H) \times P(\text{not} H)} = \frac{1\times \frac13} {1\times \frac13 + \frac12 \times \frac23} = \frac{\hspace{1mm} \frac13\hspace{1mm} } {\hspace{1mm}\frac23\hspace{1mm}} = \frac12. \]

There is a \(\frac12,\) or 50%, chance of our first pick being correct when Monty **randomly** opens a door and it happens to have a goat behind it. The naive reasoning that fooled so many mathematicians works in this case.

## Example Problems

Suppose you're a contestant on a game show and the host shows you 5 curtains. The host informs you that behind 3 of the curtains lie lumps of coal, but behind the other 2 curtains lie separate halves of the same $10,000 bill. He then asks you to choose 2 curtains; if these 2 curtains are the ones with the 2 bill halves behind them then you win the $10,000, otherwise you go home with nothing.

After you choose your 2 curtains, the host opens one of the remaining 3 curtains that he knows has just a lump of coal behind it. He then gives you the following options:

(0) stick with the curtains you initially chose,

(1) swap either one of your curtains for one of the remaining curtains, or

(2) swap both of your curtains for the remaining 2 curtains.

Let \(p(k), k = 0, 1, 2\), be the respective probabilities of winning the money in scenarios \((k)\) as outlined above. Then \[p(2) - p(0) - p(1) = \dfrac{a}{b},\] where \(a\) and \(b\) are coprime positive integers. Find \(a + b\).

You find yourself on a game show and there are 10 doors. Behind nine of them are assorted, random farm animals including a variety of pigs, goats, and geese, but behind one is a brand-new, shiny, gold car made of real gold!

The game proceeds as follows:

- You pick a door.
- The game show host then opens a door you didn't choose that he knows has only farm animals behind it.
- You are then given the option to switch to a different, unopened door, after which he opens another door with farm animals behind it.

This process continues until there are only two doors left, one being your current choice. (Every time the host opens a door, you are given another option to switch or stick with the one you currently have selected.)

If you play optimally, your chances of winning the car are of the form \( \frac ab\), where \(a\) and \(b\) are coprime positive integers.

Find \(a+b\).

\(\)

**Details and Assumptions:**

- You know ahead of time that this process will continue until you are down to only two doors.
- Assume that you prefer the car over any of the farm animals!

**Photo credit:** http://iloboyou.com/

More probability questions:

Check out my other problems.

You find yourself on a game show, and the host presents you with four doors. Behind three doors are an assortment of gummy bears, and the remaining door has a pure gold car behind it!

You pick a door, and then the host reveals a door behind which he knows is only a couple of red gummy bears. Then you are given the choice to stick with your first choice or switch to one of the other two unopened doors.

If the probability that you will win the car if you switch is \( \frac ab\), where \(a\) and \(b\) are coprime positive integers, what is \(a+b\)?

More probability questions:

**Photo credit:** http://www.huffingtonpost.com/

On a game show, there is a popular game that is always carried out in the same way. A contestant is given the choice of three doors: behind one door is a car and behind the other two doors are goats. The contestant picks a door, say #1, and the host, who knows what's behind each door, always opens another door, say #3, to reveal a goat. Then, the contestant is given the option to switch to the other unopened door, door #2 in this case.

A long-term fan of the game show has noticed a hint in the staging of the game by the game show host. Thus, this fan can correctly guess the door with the car behind 50% of the time before any door selection is made.

Now, this fan has been selected as a contestant for the game. Using the best possible strategy, what is the probability (as a percentage) that this fan will end up with the car prize?

**Question**: Can you devise your own variations of Monty Hall problem?

## Citations

[1] Crockett, Zachary. The Time Everyone 'Corrected' the World's Smartest Woman. *Priceonomics*. 19 Feb 2015. Retrieved from http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/ on 29 Feb 2016.

**Cite as:**Monty Hall Problem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/monty-hall-problem/