Does Monty Hall Problem have any effect on an equation with multiple choices? [closed]Monty hall problem extended.Help with monty hall problemWhat's wrong with this equal probability solution for Monty Hall Problem?Bayes' theorem - monty hall problem with five doors - incorrect final probabilitiesMonty Hall Problem with twistMonty Hall problem with 7 doorsMonty Hall problem with biased door selection probabilityMonty Hall problem generalized to $n$ doorsHow does the Monty Hall Problem work?A confusion about the Monty Hall problem
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Does Monty Hall Problem have any effect on an equation with multiple choices? [closed]
Monty hall problem extended.Help with monty hall problemWhat's wrong with this equal probability solution for Monty Hall Problem?Bayes' theorem - monty hall problem with five doors - incorrect final probabilitiesMonty Hall Problem with twistMonty Hall problem with 7 doorsMonty Hall problem with biased door selection probabilityMonty Hall problem generalized to $n$ doorsHow does the Monty Hall Problem work?A confusion about the Monty Hall problem
$begingroup$
I am having a discussion with someone that insists the Monty Hall Problem would improve the probability of guessing the correct door. However, the circumstances are different.
You have four doors. Only one has the right answer. We are not concerned about the probability of guessing the correct answer on the first try, but whether or not using Monty Hall Problem will improve your chances on guessing the correct door earlier. You choose one door. It's wrong. You choose again until you get the correct door. Does using the Monty Hall Problem better your chances of guessing the correct door with less guesses?
My argument is that since there are multiple choices, the Monty Hall Problem probability factor doesn't apply. The equation is reset each time you choose a new door, so the base probability factor is simply evenly split amongst the remaining doors.
The other person's argument is following a 4 door Monty Hall Problem that only has 1 door opened. You choose Door 1, you open Door 2. Find out that Door 2 is wrong. (S)he says there is now a 25% chance it is Door 1, and a 37.5% chance it is Door 3 and a 37.5% chance it is Door 4.
Who is right? I'm not sure if I'm overlooking something. I don't believe I am though.
If I am right, how would I thoroughly explain how the Monty Hall Problem doesn't apply here? I've tried explaining it how I have here to no avail and the other person still believes they're right.
probability monty-hall
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as unclear what you're asking by Matthew Towers, Lord Shark the Unknown, Eevee Trainer, Riccardo.Alestra, Shailesh Mar 29 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
I am having a discussion with someone that insists the Monty Hall Problem would improve the probability of guessing the correct door. However, the circumstances are different.
You have four doors. Only one has the right answer. We are not concerned about the probability of guessing the correct answer on the first try, but whether or not using Monty Hall Problem will improve your chances on guessing the correct door earlier. You choose one door. It's wrong. You choose again until you get the correct door. Does using the Monty Hall Problem better your chances of guessing the correct door with less guesses?
My argument is that since there are multiple choices, the Monty Hall Problem probability factor doesn't apply. The equation is reset each time you choose a new door, so the base probability factor is simply evenly split amongst the remaining doors.
The other person's argument is following a 4 door Monty Hall Problem that only has 1 door opened. You choose Door 1, you open Door 2. Find out that Door 2 is wrong. (S)he says there is now a 25% chance it is Door 1, and a 37.5% chance it is Door 3 and a 37.5% chance it is Door 4.
Who is right? I'm not sure if I'm overlooking something. I don't believe I am though.
If I am right, how would I thoroughly explain how the Monty Hall Problem doesn't apply here? I've tried explaining it how I have here to no avail and the other person still believes they're right.
probability monty-hall
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as unclear what you're asking by Matthew Towers, Lord Shark the Unknown, Eevee Trainer, Riccardo.Alestra, Shailesh Mar 29 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
$endgroup$
– Robert Israel
Mar 28 at 20:22
$begingroup$
I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
$endgroup$
– Tarrant Walter
Mar 28 at 20:36
$begingroup$
"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
$endgroup$
– fleablood
Mar 28 at 20:43
1
$begingroup$
You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
$endgroup$
– fleablood
Mar 28 at 21:16
1
$begingroup$
the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
$endgroup$
– fleablood
Mar 28 at 21:19
|
show 3 more comments
$begingroup$
I am having a discussion with someone that insists the Monty Hall Problem would improve the probability of guessing the correct door. However, the circumstances are different.
You have four doors. Only one has the right answer. We are not concerned about the probability of guessing the correct answer on the first try, but whether or not using Monty Hall Problem will improve your chances on guessing the correct door earlier. You choose one door. It's wrong. You choose again until you get the correct door. Does using the Monty Hall Problem better your chances of guessing the correct door with less guesses?
My argument is that since there are multiple choices, the Monty Hall Problem probability factor doesn't apply. The equation is reset each time you choose a new door, so the base probability factor is simply evenly split amongst the remaining doors.
The other person's argument is following a 4 door Monty Hall Problem that only has 1 door opened. You choose Door 1, you open Door 2. Find out that Door 2 is wrong. (S)he says there is now a 25% chance it is Door 1, and a 37.5% chance it is Door 3 and a 37.5% chance it is Door 4.
Who is right? I'm not sure if I'm overlooking something. I don't believe I am though.
If I am right, how would I thoroughly explain how the Monty Hall Problem doesn't apply here? I've tried explaining it how I have here to no avail and the other person still believes they're right.
probability monty-hall
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am having a discussion with someone that insists the Monty Hall Problem would improve the probability of guessing the correct door. However, the circumstances are different.
You have four doors. Only one has the right answer. We are not concerned about the probability of guessing the correct answer on the first try, but whether or not using Monty Hall Problem will improve your chances on guessing the correct door earlier. You choose one door. It's wrong. You choose again until you get the correct door. Does using the Monty Hall Problem better your chances of guessing the correct door with less guesses?
My argument is that since there are multiple choices, the Monty Hall Problem probability factor doesn't apply. The equation is reset each time you choose a new door, so the base probability factor is simply evenly split amongst the remaining doors.
The other person's argument is following a 4 door Monty Hall Problem that only has 1 door opened. You choose Door 1, you open Door 2. Find out that Door 2 is wrong. (S)he says there is now a 25% chance it is Door 1, and a 37.5% chance it is Door 3 and a 37.5% chance it is Door 4.
Who is right? I'm not sure if I'm overlooking something. I don't believe I am though.
If I am right, how would I thoroughly explain how the Monty Hall Problem doesn't apply here? I've tried explaining it how I have here to no avail and the other person still believes they're right.
probability monty-hall
probability monty-hall
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
asked Mar 28 at 20:13
Tarrant WalterTarrant Walter
11
11
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tarrant Walter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as unclear what you're asking by Matthew Towers, Lord Shark the Unknown, Eevee Trainer, Riccardo.Alestra, Shailesh Mar 29 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Matthew Towers, Lord Shark the Unknown, Eevee Trainer, Riccardo.Alestra, Shailesh Mar 29 at 11:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
$endgroup$
– Robert Israel
Mar 28 at 20:22
$begingroup$
I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
$endgroup$
– Tarrant Walter
Mar 28 at 20:36
$begingroup$
"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
$endgroup$
– fleablood
Mar 28 at 20:43
1
$begingroup$
You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
$endgroup$
– fleablood
Mar 28 at 21:16
1
$begingroup$
the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
$endgroup$
– fleablood
Mar 28 at 21:19
|
show 3 more comments
2
$begingroup$
The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
$endgroup$
– Robert Israel
Mar 28 at 20:22
$begingroup$
I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
$endgroup$
– Tarrant Walter
Mar 28 at 20:36
$begingroup$
"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
$endgroup$
– fleablood
Mar 28 at 20:43
1
$begingroup$
You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
$endgroup$
– fleablood
Mar 28 at 21:16
1
$begingroup$
the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
$endgroup$
– fleablood
Mar 28 at 21:19
2
2
$begingroup$
The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
$endgroup$
– Robert Israel
Mar 28 at 20:22
$begingroup$
The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
$endgroup$
– Robert Israel
Mar 28 at 20:22
$begingroup$
I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
$endgroup$
– Tarrant Walter
Mar 28 at 20:36
$begingroup$
I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
$endgroup$
– Tarrant Walter
Mar 28 at 20:36
$begingroup$
"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
$endgroup$
– fleablood
Mar 28 at 20:43
$begingroup$
"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
$endgroup$
– fleablood
Mar 28 at 20:43
1
1
$begingroup$
You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
$endgroup$
– fleablood
Mar 28 at 21:16
$begingroup$
You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
$endgroup$
– fleablood
Mar 28 at 21:16
1
1
$begingroup$
the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
$endgroup$
– fleablood
Mar 28 at 21:19
$begingroup$
the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
$endgroup$
– fleablood
Mar 28 at 21:19
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
To do the Monty Hall problem you need Monty Hall. Without him you have this.
You pick Door 1 and open Door 2.
1 out of 4 times the car will be in door 1 and Door 2 is a goat.
1 out of 4 times the car will be in door 2 and you see the car.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.
The car was behind door 1
The car was behind door 3
The car was behind door 4
You are right.
But if you had a monty hall then what would happen is
The door you choose is called A and the door monty shows you is B. The other two are C and D.
1 in 4 the car is behind A.
3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.
Your friend is right.
But this is only if you have a host.
If you don't:
1 in 4 it is behind A.
1 in 4 it is actually behind B and you SEE it (that never happens with the host)
1 in 4 is is behind C.
and 1 in 4 it is behind D.
==== if you actually do the Monty Hall problem====
1 out of 4 cases you choose the right door the first time.
Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.
There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.
There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.
1 out of 4 times this will happen.
3 out of 4 times you will choose a door with a goat.
There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.
A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.
A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.
This happens 3 out of 8 times.
B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.
That happes 3 out of 8 times.
So the Monty Hall method works $5$ out of $8$ times.
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
$endgroup$
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
add a comment |
$begingroup$
You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
$endgroup$
add a comment |
$begingroup$
Using Bayes theorem:
Your scenario WITHOUT a host.
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 14frac 34=frac 13$.
So probability one of the other two doors is correct. $1 - frac 13 = frac 23$. And each is equally likely so probability a specific other door is correct is $frac 13$
But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(textshown door is wrong) = 1$
So
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 141=frac 14$.
So probability one of the other two doors is correct. $1 - frac 14 = frac 34$. And each is equally likely so probability a specific other door is correct is $frac 38$
.... or ... WITHOUT a host these happen.
Car is behind door 1. 1-4 times.
1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)
Car is behind door 2. 1-4 times.
$colorredtext1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)$
1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)
1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)
Car is behind door 3. 1-4 times.
1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)$
1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)
Car is behind door 4. 1-4 times.
1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)
1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)$
If we know the red ones didn't happen each of the others are equally likely.
And WITH a host:
Car is behind door 1. 1-4 times.
1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)
Car is behind door 2. 1-4 times.
1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)
1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)
Car is behind door 3. 1-4 times.
1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)
1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)
Car is behind door 4. 1-4 times.
1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)
1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)
WITH a host we know that a car will never be shown and that changes everything.
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To do the Monty Hall problem you need Monty Hall. Without him you have this.
You pick Door 1 and open Door 2.
1 out of 4 times the car will be in door 1 and Door 2 is a goat.
1 out of 4 times the car will be in door 2 and you see the car.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.
The car was behind door 1
The car was behind door 3
The car was behind door 4
You are right.
But if you had a monty hall then what would happen is
The door you choose is called A and the door monty shows you is B. The other two are C and D.
1 in 4 the car is behind A.
3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.
Your friend is right.
But this is only if you have a host.
If you don't:
1 in 4 it is behind A.
1 in 4 it is actually behind B and you SEE it (that never happens with the host)
1 in 4 is is behind C.
and 1 in 4 it is behind D.
==== if you actually do the Monty Hall problem====
1 out of 4 cases you choose the right door the first time.
Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.
There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.
There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.
1 out of 4 times this will happen.
3 out of 4 times you will choose a door with a goat.
There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.
A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.
A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.
This happens 3 out of 8 times.
B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.
That happes 3 out of 8 times.
So the Monty Hall method works $5$ out of $8$ times.
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
$endgroup$
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
add a comment |
$begingroup$
To do the Monty Hall problem you need Monty Hall. Without him you have this.
You pick Door 1 and open Door 2.
1 out of 4 times the car will be in door 1 and Door 2 is a goat.
1 out of 4 times the car will be in door 2 and you see the car.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.
The car was behind door 1
The car was behind door 3
The car was behind door 4
You are right.
But if you had a monty hall then what would happen is
The door you choose is called A and the door monty shows you is B. The other two are C and D.
1 in 4 the car is behind A.
3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.
Your friend is right.
But this is only if you have a host.
If you don't:
1 in 4 it is behind A.
1 in 4 it is actually behind B and you SEE it (that never happens with the host)
1 in 4 is is behind C.
and 1 in 4 it is behind D.
==== if you actually do the Monty Hall problem====
1 out of 4 cases you choose the right door the first time.
Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.
There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.
There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.
1 out of 4 times this will happen.
3 out of 4 times you will choose a door with a goat.
There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.
A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.
A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.
This happens 3 out of 8 times.
B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.
That happes 3 out of 8 times.
So the Monty Hall method works $5$ out of $8$ times.
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
$endgroup$
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
add a comment |
$begingroup$
To do the Monty Hall problem you need Monty Hall. Without him you have this.
You pick Door 1 and open Door 2.
1 out of 4 times the car will be in door 1 and Door 2 is a goat.
1 out of 4 times the car will be in door 2 and you see the car.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.
The car was behind door 1
The car was behind door 3
The car was behind door 4
You are right.
But if you had a monty hall then what would happen is
The door you choose is called A and the door monty shows you is B. The other two are C and D.
1 in 4 the car is behind A.
3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.
Your friend is right.
But this is only if you have a host.
If you don't:
1 in 4 it is behind A.
1 in 4 it is actually behind B and you SEE it (that never happens with the host)
1 in 4 is is behind C.
and 1 in 4 it is behind D.
==== if you actually do the Monty Hall problem====
1 out of 4 cases you choose the right door the first time.
Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.
There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.
There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.
1 out of 4 times this will happen.
3 out of 4 times you will choose a door with a goat.
There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.
A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.
A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.
This happens 3 out of 8 times.
B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.
That happes 3 out of 8 times.
So the Monty Hall method works $5$ out of $8$ times.
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
$endgroup$
To do the Monty Hall problem you need Monty Hall. Without him you have this.
You pick Door 1 and open Door 2.
1 out of 4 times the car will be in door 1 and Door 2 is a goat.
1 out of 4 times the car will be in door 2 and you see the car.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.
You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.
The car was behind door 1
The car was behind door 3
The car was behind door 4
You are right.
But if you had a monty hall then what would happen is
The door you choose is called A and the door monty shows you is B. The other two are C and D.
1 in 4 the car is behind A.
3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.
Your friend is right.
But this is only if you have a host.
If you don't:
1 in 4 it is behind A.
1 in 4 it is actually behind B and you SEE it (that never happens with the host)
1 in 4 is is behind C.
and 1 in 4 it is behind D.
==== if you actually do the Monty Hall problem====
1 out of 4 cases you choose the right door the first time.
Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.
There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.
There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.
1 out of 4 times this will happen.
3 out of 4 times you will choose a door with a goat.
There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.
A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.
A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.
This happens 3 out of 8 times.
B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.
That happes 3 out of 8 times.
So the Monty Hall method works $5$ out of $8$ times.
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
edited Mar 28 at 21:02
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
answered Mar 28 at 20:39
fleabloodfleablood
73.8k22891
73.8k22891
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
add a comment |
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
I believe you misunderstand the question (or perhaps I did not elaborate enough). In this scenario, nobody reveals anything to you, but rather you open the door yourself to know that it is wrong. The door that is 'revealed' to you is not always wrong, it could be the correct answer. For instance, you 'choose' Door 1 but open Door 2 to ATTEMPT to eliminate it as a possibility. Door 2 is correct, in this case you got it in the first try. The 'swap' tactic has no validity here (regarding how I see it) because you never actually swap anything.
$endgroup$
– Tarrant Walter
Mar 28 at 20:48
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
That is NOT the Monty Hall Problem. That is just guessing. This is called conditional probability. The probability of you being correct is 1 in 4. But GIVEN that door 2 wrong (a probability of 3 out of 4). Then the probability of you being correct given door 2 is wrong is $frac frac 14frac 34=frac 13$ (because we are only considering 3/4 of the cases..
$endgroup$
– fleablood
Mar 28 at 21:09
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
$begingroup$
Bayes theorem is P(A|B) = $frac A)P(A)P(B)$. So probability (YOU ARE RIGHT given OTHER DOOR IS WRONG = $frac textP(OTHER DOOR IS WRONG given YOU ARE RIGHT)P(YOU ARE RIGHT)textP(OTHER DOOR IS WRONG) = frac 1*frac 14frac 34 = frac 13$. Which is just common sense. Each unknown door is equally likely. One is known to be wrong. The others are equally likely.
$endgroup$
– fleablood
Mar 28 at 21:13
add a comment |
$begingroup$
You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
$endgroup$
add a comment |
$begingroup$
You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
$endgroup$
add a comment |
$begingroup$
You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
$endgroup$
You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
answered Mar 28 at 20:22
Especially LimeEspecially Lime
22.7k23059
22.7k23059
add a comment |
add a comment |
$begingroup$
Using Bayes theorem:
Your scenario WITHOUT a host.
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 14frac 34=frac 13$.
So probability one of the other two doors is correct. $1 - frac 13 = frac 23$. And each is equally likely so probability a specific other door is correct is $frac 13$
But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(textshown door is wrong) = 1$
So
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 141=frac 14$.
So probability one of the other two doors is correct. $1 - frac 14 = frac 34$. And each is equally likely so probability a specific other door is correct is $frac 38$
.... or ... WITHOUT a host these happen.
Car is behind door 1. 1-4 times.
1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)
Car is behind door 2. 1-4 times.
$colorredtext1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)$
1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)
1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)
Car is behind door 3. 1-4 times.
1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)$
1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)
Car is behind door 4. 1-4 times.
1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)
1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)$
If we know the red ones didn't happen each of the others are equally likely.
And WITH a host:
Car is behind door 1. 1-4 times.
1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)
Car is behind door 2. 1-4 times.
1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)
1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)
Car is behind door 3. 1-4 times.
1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)
1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)
Car is behind door 4. 1-4 times.
1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)
1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)
WITH a host we know that a car will never be shown and that changes everything.
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Using Bayes theorem:
Your scenario WITHOUT a host.
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 14frac 34=frac 13$.
So probability one of the other two doors is correct. $1 - frac 13 = frac 23$. And each is equally likely so probability a specific other door is correct is $frac 13$
But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(textshown door is wrong) = 1$
So
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 141=frac 14$.
So probability one of the other two doors is correct. $1 - frac 14 = frac 34$. And each is equally likely so probability a specific other door is correct is $frac 38$
.... or ... WITHOUT a host these happen.
Car is behind door 1. 1-4 times.
1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)
Car is behind door 2. 1-4 times.
$colorredtext1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)$
1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)
1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)
Car is behind door 3. 1-4 times.
1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)$
1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)
Car is behind door 4. 1-4 times.
1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)
1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)$
If we know the red ones didn't happen each of the others are equally likely.
And WITH a host:
Car is behind door 1. 1-4 times.
1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)
Car is behind door 2. 1-4 times.
1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)
1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)
Car is behind door 3. 1-4 times.
1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)
1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)
Car is behind door 4. 1-4 times.
1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)
1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)
WITH a host we know that a car will never be shown and that changes everything.
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Using Bayes theorem:
Your scenario WITHOUT a host.
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 14frac 34=frac 13$.
So probability one of the other two doors is correct. $1 - frac 13 = frac 23$. And each is equally likely so probability a specific other door is correct is $frac 13$
But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(textshown door is wrong) = 1$
So
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 141=frac 14$.
So probability one of the other two doors is correct. $1 - frac 14 = frac 34$. And each is equally likely so probability a specific other door is correct is $frac 38$
.... or ... WITHOUT a host these happen.
Car is behind door 1. 1-4 times.
1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)
Car is behind door 2. 1-4 times.
$colorredtext1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)$
1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)
1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)
Car is behind door 3. 1-4 times.
1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)$
1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)
Car is behind door 4. 1-4 times.
1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)
1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)$
If we know the red ones didn't happen each of the others are equally likely.
And WITH a host:
Car is behind door 1. 1-4 times.
1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)
Car is behind door 2. 1-4 times.
1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)
1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)
Car is behind door 3. 1-4 times.
1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)
1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)
Car is behind door 4. 1-4 times.
1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)
1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)
WITH a host we know that a car will never be shown and that changes everything.
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
$endgroup$
Using Bayes theorem:
Your scenario WITHOUT a host.
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 14frac 34=frac 13$.
So probability one of the other two doors is correct. $1 - frac 13 = frac 23$. And each is equally likely so probability a specific other door is correct is $frac 13$
But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(textshown door is wrong) = 1$
So
$P(textYou pick right door|textshown door is wrong) = frac textYou pick right door)times P(textyou pick right door)P(textshown door is wrong)=$
$frac 1times frac 141=frac 14$.
So probability one of the other two doors is correct. $1 - frac 14 = frac 34$. And each is equally likely so probability a specific other door is correct is $frac 38$
.... or ... WITHOUT a host these happen.
Car is behind door 1. 1-4 times.
1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)
1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)
Car is behind door 2. 1-4 times.
$colorredtext1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)$
1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)
1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)
Car is behind door 3. 1-4 times.
1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)$
1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)
Car is behind door 4. 1-4 times.
1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)
1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)
$colorredtext1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)$
If we know the red ones didn't happen each of the others are equally likely.
And WITH a host:
Car is behind door 1. 1-4 times.
1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)
1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)
Car is behind door 2. 1-4 times.
1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)
1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)
Car is behind door 3. 1-4 times.
1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)
1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)
Car is behind door 4. 1-4 times.
1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)
1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)
WITH a host we know that a car will never be shown and that changes everything.
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
answered Mar 28 at 21:55
fleabloodfleablood
73.8k22891
73.8k22891
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add a comment |
2
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The Monty Hall problem doesn't "do" anything. The probability is whatever it is... But what do you mean by "the equation is reset"? Please explain exactly what are the assumptions in this problem.
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– Robert Israel
Mar 28 at 20:22
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I simply mean that after you choose for the first time, if it's wrong, you choose again. The logic you would use regarding the Monty Hall Problem would start from the beginning again before it gets to the end to provide any useful statistics.
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– Tarrant Walter
Mar 28 at 20:36
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"I simply mean that after you choose for the first time, if it's wrong, you choose again. " That's not in any way resembling the monty hall problem. That's just guessing untill you guess it. Your chances are 100% of guessing the car eventually.
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– fleablood
Mar 28 at 20:43
1
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You can't use the Monty Hall Problem unless you have Monty Hall. Someone has to know what is going on and assure that you only see wrong choices.
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– fleablood
Mar 28 at 21:16
1
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the whole POINT of Monty Hall is that Monty Hall is there thinking. "There will always be a wrong choice; it is my job to always show it". Without that person you will not always see a wrong choice you will on a regular probability see a right choice randomly. In having a host conciously show a wrong choice he is skewing the results. Without a host the choices are not being skewed.
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– fleablood
Mar 28 at 21:19