The role of probability in solving the Monty Hall problem

[Chas]

Exactly the reason why I said you people have the right mathematical answer not the right answer.

The mathematics gives the correct answer regardless of your feelings about it.

[ErGingerbreadMandude]

Exactly the reason why I said you people have the right mathematical answer not the right answer.

The mathematics gives the correct answer regardless of your feelings about it.

I don't "feel" your answer is wrong. Logically you are Wrong.

[Jenny A]

The mathematics gives the correct answer regardless of your feelings about it.

I don't "feel" your answer is wrong. Logically you are Wrong.

Still waiting for your logical rebuttal. The correct answer is explained logically here http://atheistforums.org/thread-42000-page-4.html in post 32. It's really quite simple.

[Whateverist]

You can find me any proof you like, it still doesn't make sense that probability helps you in that scenario. I will say it's all luck since you can't explain it.

As I already said, the fact that the moderator takes out all the bad variants but for one doesn't tell you anything about whether you made the right decision or not. Just the fact that he takes out all the other doors, doesn't mean that he made the unchosen left variant any more plausible than the one you already chose. And that's because no matter if you were wrong or right at first he was going to take out the same number of doors after you made your choice.


In some cases people refer to the 2/3 logic. That's bad logic. It's not 2/3 anymore because there's no three left. There are only two options left and so it's 50/50, not 25/75.

You are admitting, even bragging about, being ignorant. 

I suggest you study probability because you really do not understand it.

What is ironic is that when this first appeared in some Sunday newspaper insert a number of math professors wrote in to respond to the columnist precisely as you did EP.  However they scolded her because they agreed with EP -incorrectly of course.  They suggested she purchase a basic textbook on probability and study it so that she would come to understand that switching would not help, since the probability was 'obviously' 1/2 that either remaining door had the big prize.  

The intuition that the odds are 50 to 50 that your original choice was correct (once a wrong choice is revealed) is very powerful.  It arises because of a bad analysis resulting from failing to take proper account of the significance of that wrong choice.  The probability when the wrong choice is revealed does not alter the probability you had at the outset.  I also was flummoxed when I first heard of the problem but also intrigued of course.

A similar problem which probably won't (but might) help is the three domino problem.  One has an X on both sides, one has an O on both sides and one has one X and one O on opposite sides.  You draw one domino from the bag and hold it facing you and see an X.  What is the probability that there is also an X on the other side?

The wrong analysis goes:  

There are two dominos I could be holding, the X-X or the X-O.
So the probability that there is an X on the other side is 1/2.

But the correct analysis is:

I could be looking at one of three X's.
There is another X on the back of two of those X's and an O on the back of the third one.
So the probability is 2/3 that there is an X on the other side.

[Whateverist]

From wikipedia  https://en.wikipedia.org/wiki/Monty_Hall_problem:

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

[Jenny A]

You are admitting, even bragging about, being ignorant. 

I suggest you study probability because you really do not understand it.

What is ironic is that when this first appeared in some Sunday newspaper insert a number of math professors wrote in to respond to the columnist precisely as you did EP.  However they scolded her because they agreed with EP -incorrectly of course.  They suggested she purchase a basic textbook on probability and study it so that she would come to understand that switching would not help, since the probability was 'obviously' 1/2 that either remaining door had the big prize.  

The intuition that the odds are 50 to 50 that your original choice was correct (once a wrong choice is revealed) is very powerful.  It arises because of a bad analysis resulting from failing to take proper account of the significance of that wrong choice.  The probability when the wrong choice is revealed does not alter the probability you had at the outset.  I also was flummoxed when I first heard of the problem but also intrigued of course.

A similar problem which probably won't (but might) help is the three domino problem.  One has an X on both sides, one has an O on both sides and one has one X and one O on opposite sides.  You draw one domino from the bag and hold it facing you and see an X.  What is the probability that there is also an X on the other side?

The wrong analysis goes:  

There are two dominos I could be holding, the X-X or the X-O.
So the probability that there is an X on the other side is 1/2.

But the correct analysis is:

I could be looking at one of three X's.
There is another X on the back of two of those X's and an O on the back of the third one.
So the probability is 2/3 that there is an X on the other side.

It took me three reads to get this because you aren't describing dominos.  Dominos have  uniformly patterned backs like cards. Their faces have two different numbers or sets of dots on either side.  But you see the whole face. There are no xes.

But if you have slips of paper with x's or o's on either side it would work.

[Fidel_Castronaut]

The mathematics gives the correct answer regardless of your feelings about it.

I don't "feel" your answer is wrong. Logically you are Wrong.

No, he isn't. 

This isn't even a debate. I can't honestly believe this thread has gone on for so long. It is the definition of wrong to believe this is 50/50. It would be like giving the answer as 5 to what is 2+2.

[Aractus]

Here's the easiest way I've found to convince people why switching is better:

Suppose instead of three doors, you have 100. There's a car behind one and goats behind 99 of them. You get to choose a door, and then 98 other incorrect doors will be eliminated.

So, you choose door 34. The host opens eliminates 1, 2, 3 ... 33, 35, 36... 85, and 87 through 100.

So either the prize is behind door 34, which you picked right off the bat, or 86, which is the only one the guy didn't open.

Intuitively, it's clear that they don't have the same probabilities of being correct, right? Essentially, you've always had a 1% chance of it being in 34, and the other 99% collapses into 86.

But see here's the whole problem. If you don't know that Monty Hall intentionally picks a Goat and his selection is truly random then it actually does not matter if you switch or not.

Here are your choices:

Door A | Door B | Door C
(Goat) | (Goat) | (PRIZE)

In two scenarios you initially pick the wrong door, but in the third you pick the Prize. Therefore if Monty intentionally removes a Goat it is better to switch.

BUT

What happens when Monty picks the doors randomly?

Door A | Door B | Door C
(Goat) | (Goat) |(PRIZE)

Non-Switching Scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 - You pick Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 4 - You pick Door B, Monty Picks Door C, you Lose
Scenario 5 - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 6 - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Now we can see what's going on more clearly - Monty in fact has a 1 in 3 chance that the first door he opens is the Prize, and a 2 in 3 chance that it is a Goat. Assuming no manipulation this is what happens if you switch:

Door A | Door B | Door C
(Goat) | (Goat) |(PRIZE)

Switching Scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 - You pick Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 4 - You pick Door B, Monty Picks Door C, you Lose
Scenario 5 - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 6 - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

It makes no difference. And the reason is that you only have a 4/6 chance (2/3) of having the option to switch, and since your original odds of winning were 1 in 3 they remain 1 in 3 if you switch. You don't do your chances any harm, but there's no benefit. You lose 2/3rds of the time in either scenario. But that's not how Monty plays the game. He never picks the door with the prize first. And that changes the probability like so:

Non-switching scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Or

Switching scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 3 - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

Although there are only four scenarios, each of the first two is twice as likely to happen as the last two, that's because Monty's former choice of "Door C" has been removed, forcing him to chose the Goat even if he has a choice to open the Prize. So while all the outcomes we were talking about had a probability of 1 in 6, now the top two have a probability each of 2 in 6 (1 in 3), while the last two scenarios (formally scenarios 5 and 6) each have a probability of 1 in 6. Therefore:

Non-switching scenarios:
Scenario 1 (p. = 2/6)- You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Or

Switching scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

So I'm not disagreeing about the probabilities, but I'm pointing out that it's only valid to switch if the host is intentionally choosing the goat. If the Host randomly chooses a door it makes no difference.

There is yet one more way for Monty to play the game. This would incredibly cruel, but let's examine it none the less. This time Monty will pick the prize intentionally:

Door A | Door B | Door C
(Goat) | (Goat) | (PRIZE)

Non-Switching Scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Switching Scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

In this scenario switching reduces your chances of winning to zero!

[ErGingerbreadMandude]

I thought and thought about it.
And I just realized I was wrong. Thanks guys.

[robvalue]

No prob

I'm glad we got there in the end!

It is a mindfuck. Probvalue haha