How to solve the Monty Hall problem using Bayes Theorem

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I've recently come across the Monty hall problem and while the reasoning behind switching doors makes sense intuitively to me I can't seem to understand the maths behind it.

I've seen many proofs online using Bayes Theorem and I manage to understand the majority of it aside from one thing.

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Then, using Bayes Theorem, we have

$$P(A mid B)=fracP(B mid A)P(A)P(B)$$

Now, $P(B mid A)=frac12$ because if the car is behind door 1 then Monty can choose either the second or third door. $P(A)=frac13$ because there's a one in three chance of the car being behind the first door.

This all makes sense to me - my struggle comes in finding $P(B)$.

I understand that there are 3 separate scenarios:

-$ textbfThe car is behind door 1 $
As above in this case $P(B)=frac12.$

• $textbfThe car is behind door 2 $
  Clearly, $P(B)=0$ as Monty cannot reveal a goat behind door 2.




• $textbfThe car is behind door 3 $
  If the car is behind door 3, then Monty is forced to open door 2 and so in this case $P(B)=1$.

The way I see it, by combining these three scenarios $$P(B)=frac12 + 0 + 1=frac32$$

But in every proof that I have seen they divide this by 3 for some unknown reason. I feel like I may be missing something blindingly obvious but I don't understand why this is true.

Could someone explain the reason that $P(B)=frac12$ as opposed to $frac32$?

probability bayes-theorem monty-hall

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asked Jul 25 at 19:44

Inspector gadget

11

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A probability cannot exceed $1$
  – Peter
  Jul 25 at 19:56
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

I've recently come across the Monty hall problem and while the reasoning behind switching doors makes sense intuitively to me I can't seem to understand the maths behind it.

I've seen many proofs online using Bayes Theorem and I manage to understand the majority of it aside from one thing.

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Then, using Bayes Theorem, we have

$$P(A mid B)=fracP(B mid A)P(A)P(B)$$

Now, $P(B mid A)=frac12$ because if the car is behind door 1 then Monty can choose either the second or third door. $P(A)=frac13$ because there's a one in three chance of the car being behind the first door.

This all makes sense to me - my struggle comes in finding $P(B)$.

I understand that there are 3 separate scenarios:

-$ textbfThe car is behind door 1 $
As above in this case $P(B)=frac12.$

• $textbfThe car is behind door 2 $
  Clearly, $P(B)=0$ as Monty cannot reveal a goat behind door 2.




• $textbfThe car is behind door 3 $
  If the car is behind door 3, then Monty is forced to open door 2 and so in this case $P(B)=1$.

The way I see it, by combining these three scenarios $$P(B)=frac12 + 0 + 1=frac32$$

But in every proof that I have seen they divide this by 3 for some unknown reason. I feel like I may be missing something blindingly obvious but I don't understand why this is true.

Could someone explain the reason that $P(B)=frac12$ as opposed to $frac32$?

probability bayes-theorem monty-hall

share|cite|improve this question

asked Jul 25 at 19:44

Inspector gadget

11

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A probability cannot exceed $1$
  – Peter
  Jul 25 at 19:56
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I've recently come across the Monty hall problem and while the reasoning behind switching doors makes sense intuitively to me I can't seem to understand the maths behind it.

I've seen many proofs online using Bayes Theorem and I manage to understand the majority of it aside from one thing.

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Then, using Bayes Theorem, we have

$$P(A mid B)=fracP(B mid A)P(A)P(B)$$

Now, $P(B mid A)=frac12$ because if the car is behind door 1 then Monty can choose either the second or third door. $P(A)=frac13$ because there's a one in three chance of the car being behind the first door.

This all makes sense to me - my struggle comes in finding $P(B)$.

I understand that there are 3 separate scenarios:

-$ textbfThe car is behind door 1 $
As above in this case $P(B)=frac12.$

• $textbfThe car is behind door 2 $
  Clearly, $P(B)=0$ as Monty cannot reveal a goat behind door 2.




• $textbfThe car is behind door 3 $
  If the car is behind door 3, then Monty is forced to open door 2 and so in this case $P(B)=1$.

The way I see it, by combining these three scenarios $$P(B)=frac12 + 0 + 1=frac32$$

But in every proof that I have seen they divide this by 3 for some unknown reason. I feel like I may be missing something blindingly obvious but I don't understand why this is true.

Could someone explain the reason that $P(B)=frac12$ as opposed to $frac32$?

probability bayes-theorem monty-hall

share|cite|improve this question

asked Jul 25 at 19:44

Inspector gadget

11

I've recently come across the Monty hall problem and while the reasoning behind switching doors makes sense intuitively to me I can't seem to understand the maths behind it.

I've seen many proofs online using Bayes Theorem and I manage to understand the majority of it aside from one thing.

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Then, using Bayes Theorem, we have

$$P(A mid B)=fracP(B mid A)P(A)P(B)$$

Now, $P(B mid A)=frac12$ because if the car is behind door 1 then Monty can choose either the second or third door. $P(A)=frac13$ because there's a one in three chance of the car being behind the first door.

This all makes sense to me - my struggle comes in finding $P(B)$.

I understand that there are 3 separate scenarios:

-$ textbfThe car is behind door 1 $
As above in this case $P(B)=frac12.$

• $textbfThe car is behind door 2 $
  Clearly, $P(B)=0$ as Monty cannot reveal a goat behind door 2.




• $textbfThe car is behind door 3 $
  If the car is behind door 3, then Monty is forced to open door 2 and so in this case $P(B)=1$.

The way I see it, by combining these three scenarios $$P(B)=frac12 + 0 + 1=frac32$$

But in every proof that I have seen they divide this by 3 for some unknown reason. I feel like I may be missing something blindingly obvious but I don't understand why this is true.

Could someone explain the reason that $P(B)=frac12$ as opposed to $frac32$?

probability bayes-theorem monty-hall

share|cite|improve this question

asked Jul 25 at 19:44

Inspector gadget

11

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 19:44

Inspector gadget

11

asked Jul 25 at 19:44

Inspector gadget

11

asked Jul 25 at 19:44

Inspector gadget

11

11

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A probability cannot exceed $1$
  – Peter
  Jul 25 at 19:56
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  A probability cannot exceed $1$
  – Peter
  Jul 25 at 19:56
  

  
  
  
  

2

2

A probability cannot exceed $1$
– Peter
Jul 25 at 19:56

A probability cannot exceed $1$
– Peter
Jul 25 at 19:56

add a comment | 

2 Answers
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0
down vote

Because each of the scenarios has probability $frac13$. You're applying the law of total probability, and each term contains a factor $frac13$.

Let's say it rains tomorrow with probability $frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(Bmid A)$. This is $frac13$, not $frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
  – Inspector gadget
  Jul 28 at 18:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$mathsf P(A_1mid B)~=dfracmathsf P(Bmid A_1)mathsf P(A_1)mathsf P(Bmid A_1)mathsf P(A_1)+mathsf P(Bmid A_2)mathsf P(A_2)+mathsf P(Bmid A_3)mathsf P(A_3)\=dfractfrac 12tfrac 13tfrac 12tfrac 13+0+tfrac 11tfrac 13\=tfrac 13$$

share|cite|improve this answer

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Because each of the scenarios has probability $frac13$. You're applying the law of total probability, and each term contains a factor $frac13$.

Let's say it rains tomorrow with probability $frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(Bmid A)$. This is $frac13$, not $frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
  – Inspector gadget
  Jul 28 at 18:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

Because each of the scenarios has probability $frac13$. You're applying the law of total probability, and each term contains a factor $frac13$.

Let's say it rains tomorrow with probability $frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(Bmid A)$. This is $frac13$, not $frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
  – Inspector gadget
  Jul 28 at 18:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Because each of the scenarios has probability $frac13$. You're applying the law of total probability, and each term contains a factor $frac13$.

Let's say it rains tomorrow with probability $frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(Bmid A)$. This is $frac13$, not $frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

Because each of the scenarios has probability $frac13$. You're applying the law of total probability, and each term contains a factor $frac13$.

Let's say it rains tomorrow with probability $frac12$. If it rains, I brush my teeth. If it doesn't rain, I also brush my teeth. Is the probability that I brush my teeth tomorrow $1$ or $2$?

You also have an error in your calculation of $P(Bmid A)$. This is $frac13$, not $frac12$. This follows by symmetry – conditioning on an event that doesn't distinguish one of the three doors from the others can't make the probabilities for the doors being opened non-uniform.

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:13

joriki

164k10180328

answered Jul 25 at 20:13

joriki

164k10180328

answered Jul 25 at 20:13

joriki

164k10180328

164k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
  – Inspector gadget
  Jul 28 at 18:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
  – Inspector gadget
  Jul 28 at 18:13
  

  
  
  
  

I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
– Inspector gadget
Jul 28 at 18:13

I thought $P(Bmid A)$ would be a half, simply for the reason that Monty knows what is behind each door and so if the car is behind the first door, he then has a choice of two doors to pick.
– Inspector gadget
Jul 28 at 18:13

add a comment | 

up vote
0
down vote

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$mathsf P(A_1mid B)~=dfracmathsf P(Bmid A_1)mathsf P(A_1)mathsf P(Bmid A_1)mathsf P(A_1)+mathsf P(Bmid A_2)mathsf P(A_2)+mathsf P(Bmid A_3)mathsf P(A_3)\=dfractfrac 12tfrac 13tfrac 12tfrac 13+0+tfrac 11tfrac 13\=tfrac 13$$

share|cite|improve this answer

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$mathsf P(A_1mid B)~=dfracmathsf P(Bmid A_1)mathsf P(A_1)mathsf P(Bmid A_1)mathsf P(A_1)+mathsf P(Bmid A_2)mathsf P(A_2)+mathsf P(Bmid A_3)mathsf P(A_3)\=dfractfrac 12tfrac 13tfrac 12tfrac 13+0+tfrac 11tfrac 13\=tfrac 13$$

share|cite|improve this answer

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

up vote
0
down vote

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$mathsf P(A_1mid B)~=dfracmathsf P(Bmid A_1)mathsf P(A_1)mathsf P(Bmid A_1)mathsf P(A_1)+mathsf P(Bmid A_2)mathsf P(A_2)+mathsf P(Bmid A_3)mathsf P(A_3)\=dfractfrac 12tfrac 13tfrac 12tfrac 13+0+tfrac 11tfrac 13\=tfrac 13$$

share|cite|improve this answer

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

In the classic setting of the Monty Hall problem let $A$ be the event that the car is behind the first door that I chose. Let $B$ the event that Monty reveals a goat behind door 2.

Ah! You want $B$ to be the event that Monty chooses door 2 (rather than door 3) since you always select door 1.

Let $A_n$ be the event that the car is behind door $n$.   $A_1$ is the event that it is behind the door you choose.

$$mathsf P(A_1mid B)~=dfracmathsf P(Bmid A_1)mathsf P(A_1)mathsf P(Bmid A_1)mathsf P(A_1)+mathsf P(Bmid A_2)mathsf P(A_2)+mathsf P(Bmid A_3)mathsf P(A_3)\=dfractfrac 12tfrac 13tfrac 12tfrac 13+0+tfrac 11tfrac 13\=tfrac 13$$

share|cite|improve this answer

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 23:30

edited Jul 25 at 23:30

edited Jul 25 at 23:30

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

answered Jul 25 at 21:50

Graham Kemp

80.1k43275

80.1k43275

add a comment | 

add a comment | 

 

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