Why doesn't the solution to the Monty Hall problem make sense?

[r0bHadz]

Homework Statement

You have 3 doors, with 2/3 chance of being wrong.

A host opens a door, and there is no prize.

You are now left with two doors. I would like an explanation why the car is still equally likely to be behind any three doors still after the host opens a door.

I have a hard time that Paul Erdos could not logically come to this conclusion as well. Something is off about this problem

Homework Equations

The Attempt at a Solution

 

[PeroK]

You could write a computer program to simulate it.

Or, you could simulate the experiment using playng cards. Probablity is relative frequency, so if you simulate the experiment a large number of times, you can estimate the probability.

 

[r0bHadz]

You could write a computer program to simulate it.

Or, you could simulate the experiment using playng cards. Probablity is relative frequency, so if you simulate the experiment a large number of times, you can estimate the probability.

Is there any logical way of explaining of why the probability of picking the car doesn't change

 

[PeroK]

Is there any logical way of explaining of why the probability of picking the car doesn't change

That's been discussed on here about a hundred times. The key point is that the host knows where the car is and never accidentally reveals it. The problem is different if one show in three on average Monty ruins things by revealing the car by mistake.

Note that, as a student of mathematical probability, there is nothing funny about this problem. It's only controversial because it's in the public domain and debated by people with no knowledge of how to analyse probabilities. The sort of people who think that if a coin comes up heads 3-4 times in a row it is more and more likely to be heads next time.

It's important, therefore, to analyse both problems:

1) The actual scenario in the show.

2) The scenario where Monty opens a door (without knowing where the car is) and by luck gets an empty door.

In terms of scenario 1, I would imagine there are 100 doors. You pick one. Monty then opens 98 doors, all empty. That leaves two doors. Then I think it's more obvious that there is still only a 1% chance that you picked the right door first time and a 99% chance that the car is behind the one door that Monty didn't choose to open! It's fairly obvious to me where the car is likely to be in that scenario.

 

[r0bHadz]

That's been discussed on here about a hundred times. The key point is that the host knows where the car is and never accidentally reveals it.
1) The actual scenario in the show.In terms of scenario 1, I would imagine there are 100 doors. You pick one. Monty then opens 98 doors, all empty. That leaves two doors. Then I think it's more obvious that there is still only a 1% chance that you picked the right door first time and a 99% chance that the car is behind the one door that Monty didn't choose to open! It's fairly obvious to me where the car is likely to be in that scenario.

Yes, I am questioning the situation in the show, where Monty knows where everything is.

Your post didn't really answer my question though: why wouldn't the probability of picking the car not change?

If he opens 98 doors, leaving two doors, you say there is a 1% chance that he chose the right door from the start. How could it possibly be 99% to 1% if he could have chosen that other door that is left, instead of the one he chose, then it would be 99% to 1% the other way.

 

[PeroK]

Yes, I am questioning the situation in the show, where Monty knows where everything is.

Your post didn't really answer my question though: why wouldn't the probability of picking the car not change?

If he opens 98 doors, leaving two doors, you say there is a 1% chance that he chose the right door from the start. How could it possibly be 99% to 1% if he could have chosen that other door that is left, instead of the one he chose, then it would be 99% to 1% the other way.

That makes no sense.

Simulate the experiment if you don't believe me. Or, bet on it! I'd be happy to take your money!

 

[stockzahn]

You have 3 doors, with 2/3 chance of being wrong.

A host opens a door, and there is no prize.

You are now left with two doors. I would like an explanation why the car is still equally likely to be behind any three doors still after the host opens a door.

I have a hard time that Paul Erdos could not logically come to this conclusion as well. Something is off about this problem

In the scenario you described you can start the game with just two doors left, since the host opens one empty door before you even chose one. He could have done that before you enter the room and it wouldn't change anything, so in your described scenario the chance of getting the car should be 50-50. The original problem is based on the assumption that you have to choose one door before the host opens an empty one:

1. Choose a door
2. The host opens an empty door
3. You have to decide if you switch to third door or stick with your initial choice

In that case you can choose one of three doors: the car door $C$ or one of the empty doors $E1$, $E2$ - each of the choices is made with the same probability:

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

In two thirds of your initial choices changing leeds two winning the car, in one third you lose. Therefore you should change.

 

[PeroK]

... Monty Hall rides again!

 

[r0bHadz]

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

In two thirds of your initial choices changing leeds two winning the car, in one third you lose. Therefore you should change.

Sorry I meant the original monty hall problem, I didn't feel like it needed to be written out since I thought everyone knows about it.

My question was: Is there any logical way of explaining of why the probability of picking the car doesn't change

and your post doesn't really answer it.

You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ (why doesn't the probability chance to 1/2, 1/2 here, since we only have two doors left?) change = win, stick = lose

 

[PeroK]

My question was: Is there any logical way of explaining of why the probability of picking the car doesn't change

The simplest answer is that the host can always open an empty door. What he does, therefore, does not affect the existing probability that you picked the correct door.

The second event, therefore, carries no information about the door you picked. It carries information only about the doors you did not pick.

 

[stockzahn]

Sorry I meant the original monty hall problem, I didn't feel like it needed to be written out since I thought everyone knows about it.

My question was: Is there any logical way of explaining of why the probability of picking the car doesn't change

and your post doesn't really answer it.

You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ (why doesn't the probability chance to 1/2, 1/2 here, since we only have two doors left?) change = win, stick = lose

It's that: The original problem is based on the assumption that you have to choose one door before the host opens an empty one. But I just can agree with @PeroK. Take three cards and play the game hundred times - you are done within one hour.

EDIT: Following these possible paths seems to me as logical as it can be:

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

 

[r0bHadz]

The simplest answer is that the host can always open an empty door. What he does, therefore, does not affect the existing probability that you picked the correct door.

The second event, therefore, carries no information about the door you picked. It carries information only about the doors you did not pick.

The host will always open an empty door. If it doesn't change the existing probability that I picked the correct door, why doesn't the probability of the remaining door change?

Mathematically it would be because door 1 is 1/3 then the remaining door would have to be 2/3 to satisfy probability laws.

Which is why I feel as though the probability of the first door changing would have to happen, or the entire model would have to change.

 

[r0bHadz]

It's that: The original problem is based on the assumption that you have to choose one door before the host opens an empty one. But I just can agree with @PeroK. Take three cards and play the game hundred times - you are done within one hour.

EDIT: Following these possible paths seems to me as logic as it can be:

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

I mean I guess I will have to do it as I've searched all over math stack exchange and there was a question word for word like mine and I haven't found a single answer that satisfied me. I will try out the experiment but it is hardly mathematical and I wouldn't consider it a proof

 

[stockzahn]

I mean I guess I will have to do it as I've searched all over math stack exchange and there was a question word for word like mine and I haven't found a single answer that satisfied me. I will try out the experiment but it is hardly mathematical and I wouldn't consider it a proof

What's mathematically wrong with decision trees?

 

[PeroK]

I mean I guess I will have to do it as I've searched all over math stack exchange and there was a question word for word like mine and I haven't found a single answer that satisfied me. I will try out the experiment but it is hardly mathematical and I wouldn't consider it a proof

This is what I warned you about: people who say "it's more likely to come up heads next time". Then, when you do an experiment with a real coin and it's shown that it's 50-50 every time, they say "it's mathematically more than 50%, but in an experiment it's 50%".

What you're saying is that "mathematically the probabilty is 50-50". It's just that reality differs from your mathematics. In which case your mathematics is wrong!

 

[r0bHadz]

What's mathematically wrong with decision trees?

I was talking about the act of carrying an experiment out. I could carry out experiments until I die and it would not constitute a proof under "mathematics" am i right?

This is what I warned you about: people who say "it's more likely to come up heads next time". Then, when you do an experiment with a real coin and it's shown that it's 50-50 every time, they say "it's mathematically more than 50%, but in an experiment it's 50%".

What you're saying is that "mathematically the probabilty is 50-50". It's just that reality differs from your mathematics. In which case your mathematics is wrong!

I completely agree, but this is a mathematical problem. It's in a textbook, not the real world. So it only feels like it should be able to be worked out through logic and proven.

I will have to go talk to my prof in a couple of hours here since I don't feel like I convert my doubts and my understanding from reading text is limited I guess but hopefully I can move along soon. I have wasted a lot of time on this already -_-

 

[stockzahn]

I was talking about the act of carrying an experiment out. I could carry out experiments until I die and it would not constitute a proof under "mathematics" am i right?

You are right. Then differently: Why don't you accept the decision tree as mathematical method?

 

[r0bHadz]

You are right. Then differently: Why don't you accept the decision tree as mathematical method?

Are you talking about this:

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

If so, the problem to me doesn't seem like "if you choose E1," "if you choose E2" ... and so on (if we look at this problem with the 100 door example)

It's: "You've chosen a door. A door with no prize is eliminated. There are now two doors, you can have the same choice or change it"

I'm looking for something that satisfies the question "why does the probability not change"

We can look at this problem after a trail of 100, even 1 million, but this problem is mathematically a trail of just 1. (Actually, on a game show it is too. Youre probably not going to get invited again.)

 

[PeroK]

I was talking about the act of carrying an experiment out. I could carry out experiments until I die and it would not constitute a proof under "mathematics" am i right?
-

That's not true of applied mathematics and especially probability theory and computer science.

It's not necessarily about proving something rigorously but about informing your logic.

 

[stockzahn]

It's: "You've chosen a door. A door with no prize is eliminated. There are now two doors, you can have the same choice or change it"

It's: You have three choices of equal probability (paths). Two of the three paths lead to the car, if you change. The third path leads to the car, if you don't change. Therefore in two out of three choices, changing is the winning strategy.

I'm looking for something that satisfies the question "why does the probability not change"

Because you made your first door choice before the host opens one of the empty doors. The choice was made at 1/3 probability - that doesn't change even another door is opened afterwards (consider the 100 doors example of @PeroK).

 

[mfb]

If it doesn't change the existing probability that I picked the correct door, why doesn't the probability of the remaining door change?

A door you didn't pick can get opened, a door you picked cannot.

A very simple way to consider the question:
If you keep your door, you win if and only if you pick the right door initially.
If you switch your door, you win if and only if you picked one of the two wrong doors initially.
Which case is more likely?

 

[DrClaude]

If Monty Hall always choses a door where the car is not, offering to change after opening the door is the same as offering you the chance between

• keeping your original choice
• choosing all the other doors together

 

[OmneBonum]

The OP's original question was 'Why doesn't the solution to the Monty Hall problem make sense?'
The answer to that question isn't strictly mathematical. It has to do with the human brain, and how it evolved to do the kinds of things that it needs to do on a regular basis to ensure the survival of the species.
I remember when I first was presented with the problem, on an episode of MythBusters. I actually thought to myself, "If this is true, if switching beats staying pat, then I am ready to believe it is a magical phenomenon!" But then, I had a good think on the throne and I thought to myself, "Oh, wait. Monty knows where the prize is. Information is entering the system!" I eventually worked it out, but I will never forget, there was that very weird moment when my mind just could not fathom it! There's a nice, quick article on this in WIRED which I think the OP might have read or even linked to previously: https://www.wired.com/2014/11/monty-hall-erdos-limited-minds/

 

[StoneTemplePython]

That's been discussed on here about a hundred times. The key point is that the host knows where the car is and never accidentally reveals it. The problem is different if one show in three on average Monty ruins things by revealing the car by mistake.

Note that, as a student of mathematical probability, there is nothing funny about this problem. It's only controversial because it's in the public domain and debated by people with no knowledge of how to analyse probabilities...

The only thing I'd add, is a lot of people who should know better have messed this problem up. I think the bulk of it comes from people who don't understand basic probability. But a fair amount of mathematicians have messed it up (most famously Erdos) because there's something of a linguistic sleight of hand in how the problem is typically posed (e.g. by vos Savant).

If the problem is posed and the question asker addresses underlined part explicitly, then I think people who should know better tend to do much better.

- - - -
as for mathy-er solutions:
my vote goes to a Bayes formulation, or posing it as a renewal rewards problem.

 

[Ray Vickson]

Sorry I meant the original monty hall problem, I didn't feel like it needed to be written out since I thought everyone knows about it.

My question was: Is there any logical way of explaining of why the probability of picking the car doesn't change

and your post doesn't really answer it.

You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ (why doesn't the probability chance to 1/2, 1/2 here, since we only have two doors left?) change = win, stick = lose

Look at the "do not switch" strategy. Initially (before any doors are opened) you have P(win) = 1/3, P(lose) = 2/3. If the car is behind your chosen door, Monty opens one of the other doors, so you still win. However, if the car is not behind your door, Monty opens the other door (because you have chosen one door already, and Monty avoids opening the car-door). So, in any case you lose, and the chance of that is 2/3. So, basically, by not switching it does not matter whether Monty opens a door or not; you still have P(win) = 1/3, P(lose) = 2/3. In this case, your statement that the probability of picking the car does not change is correct.

Now look at the "always switch" strategy. In 1/3 of the cases the car is behind your chosen door, so switching will cause you to lose. However, in 2/3 of the cases the car is not behind either your original door or the door opened by Monty, so switching will cause you to win---every single time. So, switching leads to P(win) = 2/3 and P(lose) = 1/3. In this case, by switching, you do, in fact, change the probabilities of winning the car

 

[r0bHadz]

A door you didn't pick can get opened, a door you picked cannot.

A very simple way to consider the question:
If you keep your door, you win if and only if you pick the right door initially.
If you switch your door, you win if and only if you picked one of the two wrong doors initially.
Which case is more likely?

They're both equally likely, because before you switch your door, a door that has nothing behind it is revealed.

"A door you didn't pick can get opened, a door you picked cannot."

Right, and the other door I didn't pick can't get opened either.

I will look into what StoneTemplePython is saying because I need this answered mathematically: "why does the probability not change."

 

[r0bHadz]

Look at the "do not switch" strategy. Initially (before any doors are opened) you have P(win) = 1/3, P(lose) = 2/3. If the car is behind your chosen door, Monty opens one of the other doors, so you still win. However, if the car is not behind your door, Monty opens the other door (because you have chosen one door already, and Monty avoids opening the car-door). So, in any case you lose, and the chance of that is 2/3. So, basically, by not switching it does not matter whether Monty opens a door or not; you still have P(win) = 1/3, P(lose) = 2/3. In this case, your statement that the probability of picking the car does not change is correct.

Now look at the "always switch" strategy. In 1/3 of the cases the car is behind your chosen door, so switching will cause you to lose. However, in 2/3 of the cases the car is not behind either your original door or the door opened by Monty, so switching will cause you to win---every single time. So, switching leads to P(win) = 2/3 and P(lose) = 1/3. In this case, by switching, you do, in fact, change the probabilities of winning the car

Great explanation, and I do seem to understand it. My original question of: why does the probability not change, remains to be unanswered though. I would like an explanation on why after a door is exposed do we have a probability of 2/3d on one door.

 

[mfb]

They're both equally likely

"You initially pick the right door" and "you initially pick a wrong door" are equally likely? Do you really think this?

Ignore what happens afterwards for a moment. Just focus on the initial selection. Do you really think these two options are equally likely?

What about a lottery? "You initially pick all the right numbers" vs. "you initially don't pick all the right numbers"?

 

[r0bHadz]

"You initially pick the right door" and "you initially pick a wrong door" are equally likely? Do you really think this?

Ignore what happens afterwards for a moment. Just focus on the initial selection. Do you really think these two options are equally likely?

What about a lottery? "You initially pick all the right numbers" vs. "you initially don't pick all the right numbers"?

No of course not. Before a door is revealed you have a 1/3 chance picking any of the three doors.

 

[mfb]

Good. Then what is unclear? If you decide to not switch (you can make this decision long in advance, doesn't matter) you are less likely to be successful (1/3). If you decide to switch then you are more likely to be successful (2/3).

 

[StoneTemplePython]

I will look into what StoneTemplePython is saying because I need this answered mathematically: "why does the probability not change."

No of course not. Before a door is revealed you have a 1/3 chance picking any of the three doors.

ok, so if you want to see the symbol manipulation spelled out, here's the Bayes Formulation:

Define $A$ as the event that your initial selection has the prize and $B$ as the event that Monty opens a door, after you've made your selection, and that door does not have a prize behind it.

Now think a carefully about the event B in this problem: Monty opens a door that you haven't selected and that door doesn't have a prize behind it. As I underlined in my earlier post (and a few others have said), this is the key insight to the puzzle. He does this no matter what-- when you have selected the correct door he does this with probability 1 and when you have not selected the correct door he does this with probability 1. That is the event $B$ has a raw probability of 1 -- i.e. $P\big(B\big) =1$, and if we condition on your selection (or anything really) it still has probability of 1.

Plugging this into Bayes' Rule, you have

$P\big(A \vert B\big) = \frac{P\big(A\big)\cdot P\big(B\vert A\big) }{P\big(B\big)} = \frac{\frac{1}{3} \cdot 1 }{1}=\frac{1}{3}$ hence the probability of winning given that you stick with your initial selection is equivalent to the probability that your initial selection is right given event $B$ and that is $\frac{1}{3}$.

 

[r0bHadz]

Good. Then what is unclear? If you decide to not switch (you can make this decision long in advance, doesn't matter) you are less likely to be successful (1/3). If you decide to switch then you are more likely to be successful (2/3).

Why would you have a 2/3 success at switching if one of the doors were already revealed? 1/3 of those 2/3 are gone now because it was revealed and the prize is not there.

 

[r0bHadz]

StoneTemplePython, I appreciate the post but that doesn't really say anything about my question: why, after monty opens a door, does one of the two remaining doors have a probability of 2/3 that the prize is behind it, while the one you chose has only 1/3?

 

[StoneTemplePython]

StoneTemplePython, I appreciate the post but that doesn't really say anything about my question: why, after monty opens a door, does one of the two remaining doors have a probability of 2/3 that the prize is behind it, while the one you chose has only 1/3?

Actually it does answer exactly this.

I need this answered mathematically: "why does the probability not change."

And again, it answers exactly this in as direct a mathematical way as I could think of.
- - - -
Evidently, the problem is that you don't understand Bayes Rule and/or events.

 

[r0bHadz]

Actually it does answer exactly this. And again, it answers exactly this in as direct a mathematical way as I could think of.
- - - -
Evidently, the problem is that you don't understand Bayes Rule and/or events.

I understand bayes rules and I understand why you have a 1/3 chance if you remain with your door. And I understand that there has to be another 2/3 somewhere to satisfy 1/3 + x = 1. What I don't understand is how does monty revealing one of the doors give the remaining door that you have the option to switch to, give it a probability of 2/3, since it started off with a probability of 1/3 before he revealed anything.

 

[StoneTemplePython]

I understand bayes rules and I understand why you have a 1/3 chance if you remain with your door. And I understand that there has to be another 2/3 somewhere to satisfy 1/3 + x = 1. What I don't understand is how does monty revealing one of the doors give the remaining door that you have the option to switch to, give it a probability of 2/3, since it started off with a probability of 1/3 before he revealed anything.

do you want symbol manipulation or a heuristic / intuitive argument? My feeling is that page 1 has more than enough for intuition. Note: Erdos was finally convinced by a simulation -- there's no shame in getting your intuition from simulations.

As far as symbol manipulation goes, I've given the most accessible one I can think of...
- - - -
perhaps for intuition:
(i) either just accept it is the fact of how complementary events work (not totally satisfying).
(ii) alternatively, and very roughly, you could think of it as 1/3 of the time you are right to begin with so Monty's maneuvers are irrelevant... but 2/3 of the time you have chose wrongly and Monty is 'merging' all of the other options into one door for you. I.e. 2/3 of the time Monty is doing you a huge favor by narrowing the options down to one door that is correct, and 1/3 of the time what Monty is doing is irrelevant. (Ok this is really just a rehash of (i), I suppose).

From a symbol manipulation standpoint, you could try applying Bayes Rule to some door you didn't select, but it'll be a bit messy. There's something of a reference frame problem underneath this -- the frame of reference is the door you selected and its hard to disentangle from that.

I suspect this is what is inhibiting intuition here but I won't dwell on this. My view is sometimes you need to find (at least) 2 approaches to a problem -- one to get the proof and another complementary one (ha) to reinforce your intuition.

 

[PeroK]

StoneTemplePython, I appreciate the post but that doesn't really say anything about my question: why, after monty opens a door, does one of the two remaining doors have a probability of 2/3 that the prize is behind it, while the one you chose has only 1/3?

Your question has been answered at least a dozen times. Why you can't accept these answers is, frankly, beyond me.

 

[mfb]

1/3 of those 2/3 are gone now because it was revealed and the prize is not there.

The show host will never reveal a prize. The host will actively avoid revealing a prize and will always open an empty door. There is nothing that goes away - apart from the risk to switch from one wrong door to another, something you can't do.

You would be right if the show host wouldn't know where the prize is. In that case you can lose the option to switch. But this is not the scenario considered.

 

[bhobba]

Its counter-intuitive - simple as that. When you see the answer you go - dah - is it that easy. Yes it is. The answer is opening the door does not change the chances of winning - its an independent action. The chance is 1/3 you are correct. Opening the door doesn't change that. But you now only have two doors so the chance of it being behind the other door is 2/3. A lot of problems in probability are like that, which is why solving such requires well developed intuition - it's the reason the actuarial probability exam is generally considered so hard. Some people have that intuition (or developed it), others do not. I am in that do not group. When I saw the solution to the Montey Hall problem - I went - dah.

Thanks
Bill

 

[FactChecker]

The show host will never reveal a prize. The host will actively avoid revealing a prize and will always open an empty door.

That is the critical point. Monte Hall has intentionally eliminated a door without the prize from the set of doors that you have not chosen. That improves the odds of the other door you have not chosen but does not improve the odds of your door.

This insight can help one to intuitively understand the answer. But a methodical application of Baye's rule should always give the correct answer to deceptive problems like this -- intuitive or not.

 

[Nugatory]

What I don't understand is how does monty revealing one of the doors give the remaining door that you have the option to switch to, give it a probability of 2/3, since it started off with a probability of 1/3 before he revealed anything.

Here's a more extreme example: I toss a fair coin. You cannot see how the coin landed, but you can safely say that when you do look the probability that you will see it heads-up is 50%. Then I look at the coin and tell you truthfully that it landed heads-up - and your 50% probability changes to 100%.

So you shouldn't be surprised that the probabilities change when new information is revealed. The tricky part is correctly calculating how they change, and that's what the other posts in thread are explaining.

 

[FactChecker]

A more extreme case is intuitively obvious. Suppose there were a thousand doors and you picked one. You know that you have almost no chance of having the prize door. Now suppose Monte one-by-one opens the other doors till there is only one other door remaining that you did not pick. Obviously, he has avoided that door because it is the prize door. If you do not switch to that door, you are making a big mistake.

 

[PeroK]

A more extreme case is intuitively obvious. Suppose there were a thousand dors and you picked one. You know that you have almost no chance of having the prize door. Now suppose Monte one-by-one opens the other doors till there is only one other door remaining that you did not pick. Obviously, he has avoided that door because it is the prize door. If you do not switch to that door, you are making a big mistake.

It's even better if Monty goes through the doors in order. Let's say you picked door 1.

He opens doors 2-175, then coughing and shuffling he says "let's miss out door 176", then he opens doors 177-1000.

I wonder why he missed out door 176?

 

[FactChecker]

It's even better if Monty goes through the doors in order. Let's say you picked door 1.

He opens doors 2-175, then coughing and shuffling he says "let's miss out door 176", then he opens doors 177-1000.

I wonder why he missed out door 176?

I think we have a winner! This is the example I will use in the future.

One aspect this brings up is elementary game theory. If he is allowed to use game strategy and knows that you have picked the prize door, he might be tricking you into trading it away.

 

[JeffJo]

Is there any logical way of explaining of why the probability of picking the car can't change

Pardon me for making one small change to what you asked. I did it because "can't" is usually used in explanations. And there isn't a reason why it can't change, because it can. (Don't hit the "reply" button yet.)

Under the assumptions that you should make about how the host chooses a door to open, it won't change. But how this comes about is seldom explained. And I believe that the reason the problem keeps coming back, is because intelligent puzzle-solvers, who are inexperienced in probability, can sense that the assertion "it can't change" hasn't been supported.

Go back to the start of the thread, where you said that your original choice had a 1/3 chance of having the prize. This is a step everybody takes for granted, but you may not know why:

• The reason is that you have no information that can make your assessment of door #1's chances different than door #2, or door #3.
• In such cases, you can only treat them as if they are all equally likely.
• This is called the Principle of Indifference. When there are N options that you cannot distinguish from each other, except by name, then each has a 1/N probability of being the actual case. That way they add up to 1.
  
• This remains true even if there is information, unknown to you, that can make them different. If I ask you to call a coin flip, you have a 50% chance to get it right even if I know that it favor one side. Say the bias is B. Since you don't know which result is favored, there is a B/2 chance that you will pick the favored side and win, and a (1-B)/2 chance that you will pick the unfavored side and win. Since B/2+(1-B)/2=1/2, your chances of winning are 50%.

The reason people think that the two remaining doors should each have a 50% probability, is because they apply the Principle of Indifference to them. But the set of information you have about each is different:

Set 1: There is one door that the host could have opened, and did.
Set 2: There is one door that the host could have opened, but did not.

Case A: If it has the prize, he could not have opened it.
Case B: If it does not, that means your door has it. But in this case, the host had to choose between the two doors you didn't pick.​

Set 3: There is one door that the host not could have opened; the door you choose.

Only the door that Set 3 applies to is unaffected by what you learned when the host opened a door. The door that Set 1 applies to is eliminated. The door that Set 2 applies to is different, and that difference is why the probabilities for Cases A and B are different. Specifically, the probability for Case B needs to be reduced to account for the chance that the host would have opened a different door. Since you can only assume (remember the Principle of Indifference?) that the host would have chosen the other door half of the time, Case B is not half as likely as Case A. To make them add up to 100%, you make the probability for Case A 2/3, and the probability for case B 1/3.

+++++

This kind of problem fools many people, including many who should know better. Paul Erdos was one, but he recognized and admitted his error. Here is another, that I will probably get arguments about:

Q1: I have two children. What is the probability that both have the same gender?

There are four possible gender combinations, which I ordered by alphabetizing their first names: BB, BG, GB, and GG. Each is equally likely (remember the Principle of Indifference?), so the probability of (BB or GG) is 1/4+1/4=1/2.

Q2: At least one of them is a boy. Now what is the probability that both have the same gender?

Most teachers will remove the GG possibility, and apply the Principle of Indifference to BB, BG, and GB to get an answer of 1/3. This is the exact same solution that makes people say the answer in the Monty Hall Problem is 1/2.

The PoI does not apply anymore, for the same reason it doesn't in the Monty Hall Problem. There is one combination that I can't have (GG) since I would have had to tell you about a girl, two where I could have told you about a boy or a girl (BG and GB), and one where I could only tell you about a boy (BB). The same reasoning as above applies, but now there are two cases that are reduced by half. The probability I have BB is 1/2, that I have BG is 1/4, and that I have GB is 1/4.

And if you don't believe me, consider how you would have answered:

Q2.1: At least one of them is a girl. Now what is the probability that both have the same gender?
Q2.2: I wrote the gender of at least one on a notepad in front of me. Now what is the probability that both have the same gender?

Q2.1 and Q2.2 must have the same answer as Q2. But it would be a paradox if the information in Q2.2 makes its answer different is than the answer to Q1, since you have no information that is different. This paradox actually has a name: Bertrand's Box Paradox (no, Bertrand didn't refer to the problem as a "paradox,", he used this paradox to explain why the answer to all of the Q2's must be the same as Q1).

 

[sysprog]

This short book on the subject is entertaining and many would find it clarifying: http://faculty.winthrop.edu/abernathyk/Monty Hall Problem.pdf

 

[JeffJo]

This short book on the subject is entertaining and many would find it clarifying: http://faculty.winthrop.edu/abernathyk/Monty Hall Problem.pdf

Read it. In fact, what I said about Bertrand's Box Paradox comes from it, starting ion page 24. You should even be able to see how I constructed my explanation of the Two Child Problem to be parallel to Jason Rosenhouse's explanation of Bertrand's problem.

In particular, note:

Bertrand intended this as a cautionary tale of what happens when you are too cavalier in assigning equal probabilities to events. Lest you find this point too trivial to bother with, I assure you that some very competent mathematicians throughout history have managed to bungle it.

 

[PeroK]

Q1: I have two children. What is the probability that both have the same gender?

There are four possible gender combinations, which I ordered by alphabetizing their first names: BB, BG, GB, and GG. Each is equally likely (remember the Principle of Indifference?), so the probability of (BB or GG) is 1/4+1/4=1/2.

Q2: At least one of them is a boy. Now what is the probability that both have the same gender?

The issue with these questions, especially Q2, is there is no logical way to decide why someone has decided to ask you these questions. These problems all work better if information is given in response to direct questions:

How many children do you have?
Do you have at least one boy?

But, if someone simply produces this question out of thin air, what sort of logical reasoning can you apply?

Perhaps a good example might be if someone has a five-card hand of cards. If you ask them: do you have the Ace of Hearts, then you can work with the answer. And you might calculate the probability they also have the Ace of Spades. But, if they unilaterally provide you with this information: "I have the Ace of Hearts", then it's not clear on what basis you can start to calculate probabilities.

You could of course assume that they simply picked a card uniformly at random and told you what it was. But, you have no concrete basis for that assumption: unless you told them to do that; and, even then, the fact they picked a "top" card suggests the process may have been far from a uniform selection process.

In particular, it's difficult to deal with the case where someone has both the Ace of Hearts and the Ace of Spades and draw any conclusion about how likely it is that they chose to tell you about the Ace of Hearts.

If we go back to Q2. If someone accosts you in the street and says:

I have two children. At least one of them is a boy. What is the probability that both have the same gender?

Then, what are your assumptions? I might argue that someone who has only one boy would say so. Perhaps not 100% of the time, but most of the time. In fact, most people would tell you both genders: I have two boys or a boy and a girl. So, I might lean towards an answer that they probably have two boys; or, I might lean towards them having only one boy. But, there is no way to know without a statistical and/or psychological study!

In summary, I believe these problems are ill-posed as problems in probability theory. They require an element of statistical analysis to support any assumptions on why someone produces such a question.

On the other hand, information gained from direct questions can be input into straight probability calculations without any such concerns.

 

[PeroK]

Q1: I have two children. What is the probability that both have the same gender?

Q2: At least one of them is a boy. Now what is the probability that both have the same gender?

In fact, I would like to propose an answer to Q2.

Assumption 1: no one in the history of mankind has ever (or will ever) pose that particular question unless they have heard about this riddle and are asking it for that purpose.

Assumption 2: the riddle is almost always given in terms of "boy". Now, it's not impossible that someone might change it to "girl", but I'll assume that the default is to ask the question in terms of boys. Every time I've heard it that has been the case.

Assumption 3: The only (likely) circumstances in which the question gets changed to "at least one girl" is in the case where the questioner has two girls.

Given that this is a riddle, I'm not convinced that the questioner definitely has two children. They may have any number of children and are asking it as a pure riddle. So, I'll have a final assumption:

Assumption 4: the questioner definitely has two children, and is providing genuine information.

With these assumptions, then I can conclude that the questioner has two boys with probability 1/3.

Corollary:

If the questioner asks the question in terms of "at least one girl", then the probability they have two girls is close to 1.

 

[JeffJo]

The issue with these questions, especially Q2, is there is no logical way to decide why someone has decided to ask you these questions.

Yet, when the differences you talk about are not demonstrated as I did, many people will answer them as if they had asked me "Do you have any boys?" See https://en.wikipedia.org/wiki/Boy_or_Girl_paradox.

But I disagree that it is a problem:

• Q1 IS answerable. The answer is 1/2.
• Q2 has the same information content as Q1, and so IS answerable. The answer is 1/2.
  
• Q2 and Q2.1 are just as answerable as Q2.2
• Q2 and Q2.1 can't have a different answer than Q2.2

Probability is not an absolute measure. It measures your uncertainty about a situation. So the uncertain parameters you describe can be handled. Example: I have in front of me a biased coin. It lands on one side 75% of the time, and on the other 25%. If I flip it, what probability should you assign to the proposition that it will land "Heads"? Answer: 1/2. I can assign a different probability only because I know which side is favored.

But, if someone simply produces this question out of thin air, what sort of logical reasoning can you apply?

The genders of a person's children can't affect the form of the information he states, but it can affect the values he states in that form. So, a parent of BB is just as likely to state "I have two, and at least one is a <insert gender here>" as is a parent of BG, GB, or GG. The parents of BB and GG have only one gender they can insert, but the others have two and are equally likely to insert either.

Result: the answer to each question I asked is 1/2.

Perhaps a good example might be if someone has a five-card hand of cards. If you ask them: ...

But I was trying to illustrate how to solve a problem when no question is asked of the informant.

In particular, it's difficult to deal with the case where someone has both the Ace of Hearts and the Ace of Spades and draw any conclusion about how likely it is that they chose to tell you about the Ace of Hearts.

See Principle of Indifference.