Please Explain (actually explain) The Monty Hall Problem

[phyti]

Yes, it does make a difference. There is a difference between an accidental result of rare luck versus the intentional act of avoiding opening one door. Consider the example (again) of the game with 100 doors. In the first case of luck, the only inference that can be made from opening 98 doors without revealing the prize is that a rare, lucky event was just witnessed.

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.
The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

 

[PeroK]

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.
The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

Are you saying that if you stick with your original choice then you win 50% of the time?

 

[FactChecker]

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.

In the Monte Hall game, the player only has one choice that he does before Monte opens any door.

The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.
All goat doors are identical thus they have no special status.
The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

So you don't think that Rule 2 and the fact that Monte did not open that remaining door gives any sort of hint that should be used to adjust the odds but that same implication does not apply to your door? As has been explained dozens of times on this thread alone, you are wrong.

 

[DaveC426913]

The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

Except that you can be objectively shown to be wrong.

The simulation in post 27 proves that switching doors will result in winning twice as often.
If your logic/intuition does not lead to that result, then it is demonstrably faulty.

 

[Dale]

The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin

Only if the two options are equal probability, which they are not. Simply having two options does not mean that the probability is 1/2

https://www.ac-psych.org/en/download-pdf/volume/10/issue/4/id/164

 

[sysprog1]

The host must know the car location to comply with the rules:
1. Host cannot open a player door choice until after player last choice.
2. Host cannot open a car door until after player last choice.
The game rules prevent the host from accidently revealing the car location, ending the game, and denying the player a 2nd choice.

If player chooses a car door, it survives by rule 2.
If player chooses a goat door, it survives by rule 1.
The host opening 98 goat doors does not change the contents of any doors.

if there are 100 doors, then you have only 1% chance of having picked the car door when you are offered the option to switch, and if you haven't picked the car door (99% chance), then switching wins (every time), but if you have picked the car door (1% chance), then switching loses (every time).

All goat doors are identical thus they have no special status.

The door you picked is ineligible for opening prior to the switch offer.

The end result is 1 car door and 1 goat door with the player not knowing which one wins.

This is equivalent to tossing a coin.

There's one door you picked, which still has 1% chance of having been and still being the car door, and the remaining door, which has the other 99% chance of being the car door.

 

[Pikkugnome]

I think the sieve method is very good at explaining Monty Hall problem. On the otherhand I prefer another explanation, when the host opens envelopes randomly, I didnt read the posts well, so I am probably repeating. This version is equivalent to the game where contestants are given an envelope and asked one by one to open it. At each stage the ones left are at equal footing, the winning chance increases as long as the envelope was empty. It is easy to imagine being part of the contest and hoping the opened envelope is empty, so I can still win, but everyone else is looking at the situation the same way.
The sieve version kind of means that all the other envelopes are sieved except yours and in the random version no sieving is happening or it done the same way to all envelopes.

 

[phyti]

I disagree. In @Dale 's 100-door example, our selected door initially had a 1/100 chance of being the correct door. The other 99/100 chances were in the other doors. The host is not allowed to open your door or the winning door. He opens 98 other doors. That gives you a lot of information -- there are 98 doors that you know are losers. Furthermore, you know that his selection of those 98 doors did not miss the prize out of blind luck. So the 99/100 chance is now associated with the one remaining door that you did not select.

If there were 100 doors, each would not contain .01 car. The reality would be1 door containing a car and 99 doors containing goats. As the host opens 98 goat doors, the contents of the remaining doors does not change. That would be an illusionist trick you might see in Las Vegas.
If the player chose the car door, it survives since the host cannot open the car door nor the player choice.
If the player chose a goat door, it survives since the host cannot open the player choice.
The goat doors can be opened in any random order since order is not a factor. Any goat door could be the 1st or last or anywhere in between, thus no special status.
The host has reduced the set of doors to 2. There are only 2 possible results. Win a car or win a goat
If you believe the probability is1/3, what is the 3rd choice?.

 

[Dale]

If you believe the probability is1/3, what is the 3rd choice?.

There are two choices. One has 1/3 chance having the car and the other has 2/3 chance to have the car. Your error is in thinking that the two options are equally likely to have the prize.

Everyone agrees that there are two options. Your mistake is thinking that the probability is equal.

Any goat door could be the 1st or last or anywhere in between, thus no special status.

If the player selects a goat then that goat door does have special status. That special goat door cannot be chosen by the host, according to the rules you yourself described.

The standard Monty Hall problem has been simulated and tested experimentally. The probability of winning is 1/3 for staying with the original choice and 2/3 for switching. That is a measurable experimental fact. Since your reasoning leads to a demonstrably false conclusion, you know that your reasoning is wrong. We have pointed out exactly where it is wrong. Your goal now needs to be to understand why it is wrong, not to continue arguing for an empirically wrong position.

 

[javisot]

If there were 100 doors, each would not contain .01 car. The reality would be1 door containing a car and 99 doors containing goats. As the host opens 98 goat doors, the contents of the remaining doors does not change.

So you have set A with 1 box and set B with 99 boxes.

You'll agree that the car is more likely to be in set B, It's more likely that the car is in one of the 99 boxes you didn't choose.

When you remove boxes from set B, it's still more likely that the car is in set B. Finally, when only 2 boxes remain, the car is more likely to be in the remaining box from set B than in the box from set A that you initially chose.

 

[Ibix]

If you believe the probability is1/3, what is the 3rd choice?.

The possibilities are:

1. Contestant chose the car, host opened the left hand goat door $p=1/6$.
2. Contestant chose the car, host opened the right hand goat door, $p=1/6$.
3. Contestant chose the left hand goat, host opened the other goat door, $p=1/3$.
4. Contestant chose the right hand goat, host opened the other goat door, $p=1/3$.

Only in cases 1 and 2 does the contestant win by sticking, with a total probability of $1/3$. As @Dale pointed out, just because there are equal numbers of possible outcomes it does not mean that the probabilities of those outcomes are equal.

 

[jack action]

The host has reduced the set of doors to 2. There are only 2 possible results. Win a car or win a goat
If you believe the probability is1/3, what is the 3rd choice?.

It is true that what you get at the end is a 50/50 odd. But the important information comes from the first round, about the odds you picked a goat:

Look at it the other way around; what are the probabilities it is a losing door instead of a winning door? Facts:

• On the first round, he has knowledge of his first chosen door, that it's 2/3 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.

2/3 > 1/2, therefore his first chosen door has a bigger probability that it is a losing door; he better select the other one.

Repeating the process with 100 doors and Monty opening 98 doors makes it even clearer:

• On the first round, he has knowledge of his first chosen door, that it's 99/100 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.

Why would you stay with a door that you are 99% sure it hides a goat when you could choose one that only has 50% chance of hiding a goat?

 

[FactChecker]

The real difference is between luck versus the sure thing of a rule.
If Monte blindly opens a goat door, that is luck that changes the odds of both your door and the remaining unopened door equally.
In the other hand, if Monte opens a door following the rule that he can not open the prize door, than there is no luck involved. It was guaranteed. The odds of your door having the prize are unchanged at 1/3 and the odds of the other door having the prize are changed to 2/3.

 

[sysprog1]

If there were 100 doors, each would not contain .01 car. The reality would be1 door containing a car and 99 doors containing goats. As the host opens 98 goat doors, the contents of the remaining doors does not change.

What's behind each door does not change, and the collective chance of the non-selected doors (that's (n-1)/n, in the case of 100 doors, that's 99/100) does not change, but opening the non-selected non-car doors causes that 99/100 chance to be aggregated into and distributed over fewer doors; opened doors are no longer eligible, but any unopened door could still have the car behind it. When 98 doors have been opened, the number of non-selected over which the 99% chance is distributed has been reduced to 1 door.

If the player chose the car door, it survives since the host cannot open the car door nor the player choice.
If the player chose a goat door, it survives since the host cannot open the player choice.

Correct.

The goat doors can be opened in any random order since order is not a factor. Any goat door could be the 1st or last or anywhere in between, thus no special status.

In the 3-door game, the concept of the host making a 50-50 random selection arises for when he encounters 2 doors eligible to be opened, which happens only when the contestant has picked the car door.

The host has reduced the set of doors to 2. There are only 2 possible results. Win a car or win a goat
If you believe the probability is1/3, what is the 3rd choice?.

It's not 1/3 for 100 doors; it's 1/3 car and 2/3 non-car for 3 doors; for 100 doors, it's 1/100 car and 99/100 non-car $-$ those are the chances for the contestant when he picks a door.

When Monty opens a door, the chance for that door reverts to the remaining unopened doors (not including the one the contestant picked $-$ that chance doesn't change; it remains 1/n $-$ one out of the number of doors). The contestant can keep his 1/100 chance by sticking, or he can take his 99/100 chance of having not picked the car door, and switch to the other door that now has all of that 99/100 chance of being the car door. That accounts for all 100/100 chance.

So, with 100 doors, you have 99/100 chance of picking a non-car door, and a 1/100 chance of picking the only car door. Your chance if you stick stays the same, so if you picked the car, switching loses, but that will happen only 1/100 of the time; the other 99/100 of the time, you didn't pick the car door, so switching wins.

 

[sysprog1]

It is true that what you get at the end is a 50/50 odd. But the important information comes from the first round, about the odds you picked a goat:

Why would you stay with a door that you are 99% sure it hides a goat when you could choose one that only has 50% chance of hiding a goat?

It's not 50%.
Among car-door pickers (1/3 of contestants),
all stickers win, all switchers lose;
Among goat-door pickers (2/3 of contestants),
all switchers win, all stickers lose.

You can't know in advance which kind of picker you are, whether you're a car-door picker or a goat-door picker, but you can decide in advance whether to be a switcher or a sticker, and if you're a switcher, you win the 2/3 of the time that you didn't pick the car door, but if you're a sticker, you have to make do with the 1/3 chance you have of picking the car door at the outset.