# Monty Hall Problem Calculator

Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD and Steven Wooding
Last updated: Jan 18, 2024

Our Monty Hall Problem calculator answers thoroughly the question, "Why is the Monty Hall problem not 50/50?", and much more. Keep reading to discover:

• What the Monty Hall problem is: explaining the game and rules.
• The solution of the Monty Hall problem by Marilyn vos Savan.
• Two approaches to find the answer to the Monty Hall problem: conditional probability and Bayes theorem.

And much more! So, are you switching or not?

## What is the Monty Hall problem?

Before even attempting to calculate the Monty Hall problem, we need a thorough understanding of it and why it's not as intuitive as it may look!

The original formulation of the Monty Hall problem takes place in a televised game. The player has three doors in front of them:

• One of the doors hides a goat;
• Another door hides a car; and
• The third door hides another goat.

We will assume that the player does not enjoy goats and aspires to win the car. The host asks them to choose a door. Once they pick the door, the host opens a door that hides a goat.

The crucial point of the game comes now: the host asks the player if they want to switch doors for the one that was not opened or if they prefer to stick with their initial choice. What is the best strategy for the player?

A first answer, out of instinct, would be "it doesn't matter", as behind the unopened door there are only a goat and the car, the chances are 50/50. Right?

Wrong!!

## The solution to the Monty Hall problem: answer to the switching question

The first printed solution to the Monty Hall problem is by Marilyn vos Savant. In 1990, she — at the time the person with the highest IQ on Earth according to the Guinness World Record — gave an explanation for the Monty Hall problem in her column in the magazine Parade. Albeit correct and accompanied by a satisfying argumentation, the solution sparked an intense debate that didn't cut on the perceived mathematical inferiority of women.

What vos Savant proved is that the player should always change doors. This increases two-fold the probability of winning the car, from $1/3$ to $2/3$. In the next section, we will give you an explanation of the Monty Hall problem with conditional probability and the Bayes Theorem formalism. Still, for now, it's enough to know that the structure of the problem causes the information possessed by the player to change after the door is reveal by the host.

If you think about it, when the player makes the first choice, they have a probability of $1/3$ of choosing the car's door. Together, the other doors have $2/3$ of the probability of hiding the car. When the host opens the door, the last closed door of the two holds all that probability; thus, by switching, the player would change their odds from $1/3$ to $2/3$.

Are you not convinced? Let's slap some numbers on these doors!

## The Monty Hall problem explained

#### Explaining the Monty Hall problem: conditional probability

Conditional probability measures the likelihood that one event occurs given the happened occurrence of another (related) event. A generic conditional probability has form $P(A|B)$, and we read it as the probability of $A$ occurring given $B$. If you are not familiar with this concept, it may be challenging to explain the Monty Hall problem: quickly jump to our conditional probability calculator for an effective introduction to this measure.

We give an answer to the Monty Hall problem using conditional probability in the case the player chooses door number 1. What are the possible scenarios in this case?

• The car is behind door number 1. This happens with probability $1/3$. The host can then open either door number 2 or door number 3 with equal probability $1/2$.

• The car is behind door number 2, with probability $1/3$. As the player chose door number 1, the host can only open door number 3 with probability $1$.

• The car is behind door number 3, with probability $1/3$. The host can only open door number 2 with probability $1$.

What now? Let's calculate the conditional probability of the car being behind door number 1 in case the host opens door number 2. We are interested in the quantity $P(\text{car}_1|\text{host}_2)$, which is the probability of winning by not switching. We'll use the formula:

$\footnotesize P(\text{car}_1|\, \text{host}_2) = \frac{P(\text{car}_1\ \text{and}\ \text{host}_2)}{P(\text{host}_2)}$

To use conditional probability for the Monty Hall problem's solution, we first find the numerator of the fraction above.

We use the product of the probabilities. The probability of the car being behind door number 1 is $1/3$, while the probability of the host opening door number 2, in this case, is $1/2$ (as the host can open either door number 2 or number 3). Thus, $P(\text{car}_1\ \text{and}\ \text{host}_2) = 1/6$. To find this result, we used the and probability formula.

For the denominator, we can use a formula of conditional probability:

$\footnotesize \begin{split} P(\text{host}_2) &=P(\text{car}_1\ \text{and}\ \text{host}_2)\\ &\quad +P(\text{car}_2\ \text{and}\ \text{host}_2)\\ &\quad +P(\text{car}_3\ \text{and}\ \text{host}_2) \end{split}$

We identify three contributions:

• If the car is behind door number 1, the result is the same as we found above, $1/6$.

• If the car is behind door number 2, the result is $0$, as the host can't open door number 2, and the probability is $0$.

• If the car is behind door number 3, according to the rules you learned while discovering what the Monty Hall problem is, the host can only open door number 2 and does this with probability $1$: the contribution is $1/3$.

Sum these results to find the denominator: $1/6+1/3 = 3/6 = 1/2$.

Substitute them now in the formula for the conditional probability of winning without switching:

$\footnotesize \begin{split} P(\text{car}_1|\text{host}_2) &= \frac{P(\text{car}_1\ \text{and}\ \text{host}_2)}{P(\text{host}_2)}\\[1em] &=\frac{1/6}{1/2} = \frac{1}{3} \end{split}$

Uh oh... as you have only two choices, it means that the negation of this probability, that is, the probability of winning by switching, is $1-1/3 = 2/3$.

You can apply the same reasoning for all other combinations of doors, and you'll find the same result: this is the most straightforward explanation of the Monty Hall problem solution. If you want to discover another approach to calculating the Monty Hall problem solution, keep reading!

#### Explaining the Monty Hall problem: Bayes theorem

The same answer to the Monty Hall problem can be obtained using the Bayes theorem. The Monty Hall problem is a great introduction to this powerful tool as it uses a quantity related to our degree of belief in the occurrence of the event, an approach rooted in the Bayesian interpretation of probability.

When calculating the Monty Hall problem's solution using Bayes theorem, we first use the classic form of the theorem itself. Notice that we find the same conditional probability as before:

$\footnotesize P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$

where:

• $P(A|B)$Posterior probability;
• $P(B|A)$Likelihood;
• $P(A)$Prior probability; and
• $P(B)$Normalization constant.

You can find a clear explanation of this formula in our Bayes theorem calculator!

When we try to solve the Monty Hall problem, Bayes theorem takes the following form. If we want to calculate the probability of winning by not switching after choosing door number 1, and when the host opened door number 2, we use the formula:

$\!\footnotesize P(\text{car}_1|\text{host}_2)\! =\! \frac{P(\text{host}_2|\text{car}_1)\cdot P(\text{car}_1)}{P(\text{host}_2)}$
• The likelihood $P(\text{host}_2|\text{car}_1)$ has value $1/2$ as the host can open either door number 2 or door number 3. This quantity reflects our change in information compared to the initial situation.
• The prior probability $P(\text{car}_1)$ has value $1/3$.
• The normalization constant has the same value as before: $P(\text{host}_2)=1/2$.

Put together these values to find the probability of winning by not switching in this situation:

$\!\footnotesize \begin{split} P(\text{car}_1|\text{host}_2)\! &=\! \frac{P(\text{host}_2|\text{car}_1)\cdot P(\text{car}_1)}{P(\text{host}_2)}\\[1em] &\ =\frac{1/2 \cdot 1/3}{1/2} = 1/3 \end{split}$

As before, it turns out that the player's best option is to switch, as it results in victory $2/3$ of the time.

You can also use game theory to find the answer to the Monty Hall problem. It's a game, after all! Visit our prisoner's dilemma calculator to learn more about this fascinating branch of mathematics!

## How to use our Monty Hall problem calculator to play or simulate the Monty Hall problem

Our Monty Hall problem calculator offers you the chance to try to play the Monty Hall problem directly, thus testing the solution we gave in a single round or to simulate the Monty Hall problem for a given number of turns.

If you choose to play a single game of the Monty Hall problem, follow the instructions on the screen, and try to be lucky — or smart!

If you choose to simulate the Monty Hall problem, insert the number of iterations you are interested in, and select one of the strategies: always stay or always switch. Which one gives the best results?

## FAQ

### Why is the Monty Hall problem not 50/50?

After the host opens one of the goat doors, it may look obvious that the other two doors each have a probability of 1/2 of hiding the car. However, the probability of the door you chose first is unchanged at 1/3, while the other door has all the remaining probability of the two doors you didn't pick at the beginning: 2/3.

### Why should I switch doors in the Monty Hall problem?

You should switch doors because the probability of finding a car behind the door you chose at the beginning is 1/3. The remaining doors have a combined probability 2/3 of hiding the car. When the host opens one of the two doors, those 2/3 are entirely assigned to the other closed door. Hence, in all cases, switching gives you a two-fold advantage.

### Who is Marilyn vos Savant for the Monty Hall problem?

Marilyn vos Savant was the first person to thoroughly explain the Monty Hall problem in her column dedicated to mathematical problems in the magazine Parade. She used a simple explanation that highlighted the changes in information in possession of the player, yet even scholars refused to accept it, even for a long time.

### What is the probability of winning a car by switching doors in the Monty Hall problem?

The total probability of winning a car by switching doors is 2/3. We identify three possible cases, assuming you chose door 1:

1. The car is behind door 1. The host reveals a goat. By switching, you'd choose the other (and lose).
2. The car is behind door 2. The host opens door 3. Switching to door 2 gives you a probability of 1 of winning.
3. The car is behind door 3. In this case, too, you have a probability of 1 of winning.

By combining the results, you can find the probability 2/3 of winning the car.

Davide Borchia
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