Monty Hall Problem

[Robin]

Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?

If the odds were 50/50 and you played the game 1000 times then you would win about 500 cars whichever strategy you took right?

But when you actually play the game 1000 times and never switch you win about 333 cars.

If you play the game 1000 times and always switch you win about 666 cars.

If you doubt this then try it. So yes, you are missing something.

Tell me if the following statement is true or false about the problem:

"Someone who always switches will always get the car if their initial guess was wrong"


If the above statement is true then the only other question is "what is the probability of the initial guess being wrong"

(edited to change the number tries)

 

[epepke]

It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here.

But I'm convinced from empirical evidence that it's not possible to explain this to people. Fortunately, it is possible and a lot more fun to win free drinks off of them.

 

[Robin]

Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here

It has nothing to do with what Monty feels like doing as long as he behaves as stated in the problem. I am convinced that all disagreement in this topic is based around misunderstanding the problem.

Here is how I understand it:

1. There are three doors

2. Behind one door is a car

3. Behind the two other doors are goats

4. You make an initial guess but the door you guessed is not opened yet

5. The host opens one of the doors you didn't pick revealing a goat.

6. You are given the chance to change your guess.

The question posed is do you improve your odds of winning the car by changing your guess and if so by how much?

If you disagree then please state the problem as you understand it.

Step 5 appears to be the sticking point. Some, like gnome, are suggesting that he can arbitrarily leave out this step. But if he can do this then it becomes another problem entirely.

If the problem is as stated then your chances of winning the car are 2/3 if you change, 1/3 if you don't for reasons that have been explained already.

 

[rppa]

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome.

Huh? There are two possibilities on the first pick: You picked a car, or you picked a goat. In 1/3 of the games you are holding a car, in 2/3 a goat.

Do you agree with that much? Because that's the crux of it.

Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat.

No, you had a first choice, and it was completely random. And in 2/3 of the games, that random choice will leave you with a goat.

At 2nd choice time, in 2/3 of the games you chose a goat first and Monty is showing you a door with a car. In 1/3 of the games you chose a car the first time and Monty is showing you a door with a goat.

Am I missing something?

Yes: your assertion that the first choice is irrelevant, since the first choice is all. If your first choice was a goat, then there is a 100% chance that Monty's door is a car.

 

[apoger]

Let me see if I can add some clarity to all this...

Proposition #1
We will flip an unbiased coin three times. What are the odds that heads will come up three times?

Answer:
There are eight patterns that may emerge
hhh
hht
hth
htt
ttt
tth
tht
thh

Thus the probability of three heads in a row is one in eight.

Experiment #1

We flip a coin ----> Heads


Experiemnt #2

We flip a coin ----> Heads


Proposition #2
We will flip an unbiased coin one time. What are the odds that heads will come up this time?

Answer:
There are only two possibilities. Heads or tails. The odds of either are one in two.

Proposition #1 does not affect proposition #2.
The prior coin flips do not affect the odds of proposition #2.

This is a common example used in teaching statistics, and as such will be familiar to most people here.

As far as I can tell the problem at hand is almost the same situation.


Proposition #3
Monty offers you the choice of three doors. One has a car behind it, the other two have goats.

You can think of this as:

Car Goat Goat

Experiment #3
You pick one.
It does not matter what you pick. If you pick "car" then mentally eliminate the left "goat". If you pick a "goat", then mentally eliminate the other goat.

You are left with two doors. One has a car. One has a goat. One was your choice, the other not.


Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4. They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

This is exactly the same as a coin flip.
The odds are 1 in 2, or 50%.

Many people here seem intent on adding bigger numbers, as if that changes anything. It does not. If there are a BILLION doors and I pick one, and only one is a winner, then my odds are one in a billion. However if God comes by and eliminates all but two choices, one of which is the winner, and I can choose either... it's back to one in two. The prior one in a billion choice has no bearing on the subsequent one in two choice.


I hope that this helps....

 

[Robin]

Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4. They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

This is exactly the same as a coin flip.
The odds are 1 in 2, or 50%.

Proposition #3 has no bearing only if you ignore the information you got from it and randomly choose between staying and changing.

But you don't make a random choice about this you make an informed choice about this.

Your logic would go "If my initial guess was wrong then the door Monty is offering me must have the car. There is a 0.667 probability of my initial guess being wrong, therefore there is an 0.667 chance of Monty's door having the car."

 

[apoger]

Proposition #3 has no bearing only if you ignore the information you got from it and randomly choose between staying and changing.

What info transfers?
I submit none. Just like coin flips from prior experiments. They already happened and do not influence the current flip.

But you don't make a random choice about this you make an informed choice about this.

Your logic would go "If my initial guess was wrong then the door Monty is offering me must have the car. There is a 0.667 probability of my initial guess being wrong, therefore there is an 0.667 chance of Monty's door having the car."

How do you get the .667 probability? It seems to me that this is the statistic from the first proposition. However new info has been obtained (one goat exposed) and the odds must be recalculated for the new proposition which is just between two choices.

You are making the mistake from the coin example. By your thinking a thrid coin flip would not be 50/50 just becuase the last two experiments came up both heads.

 

[Robin]

You are making the mistake from the coin example. By your thinking a thrid coin flip would not be 50/50 just becuase the last two experiments came up both heads.

No you are making the mistake of assuming the examples are the same. You cannot control how a coin comes down but you can control the decision to stay or switch.

Just tell me, as I asked before, is the following statement true or false?

"If I change my selection and my initial guess was wrong then I will definitely get the car"

If you accept that statement as true then this is the information that transfers from your proposition #3.

If you don't accept that statement then please give an example of a case where your initial guess is wrong, you change your choice and you don't get the car.

 

[rppa]

Let me see if I can add some clarity to all this...

There is no clarity in all this. Monty Hall discussions will go on until the heat death of the universe.

Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4.

This is the fundamental error. Monty's actions depend on yours. Unlike the coin toss, it is not an independent action, but is completely deterministic.

In probability terms, let A = you chose a car on your first trial
~A = you chose a goat
Let B = Monty chooses a car after his actions,
~B = Monty's door is a goat

If your actions and Monty's are independent, then P(B|A) = P(B).
That is, the probability that Monty chooses a goat is independent of what you do. But that's not true.

In fact, the true model is:

P(B|A) = 0
P(B|~A) = 1

If you chose a goat, Monty's door is a car. P(B|~A) = 1.
If you chose a car, Monty's door is a goat. P(B|A) = 0.

Now the question we want to know is, what is P(B)? When all the actions are over, what is the probability that Monty's door holds a car?

P(B) = P(B|A) P(A) + P(B|~A) P(~A)
= 0 * (1/3) + 1 * (2/3)
= 2/3.


They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

Absolutely untrue. The doors Monty is offering you depend on what you chose the first time. Hence the conditional probabilities are different than the unconditioned probability.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

Nope. And it's because Monty's actions are conditioned on yours.

Here are the two possibilities for that fateful 2nd choice, before we open the two remaining doors:

Door 1 2
Goat Car
Car Goat

Now, you are saying that Monty has arranged things so that these are equally likely outcomes. But they aren't. If Door 1 is your original
choice and Door 2 is the one Monty offers after opening the rest, then the probability that the arrangement is (Goat, Car) is (N-1)/N where N = the number of original doors. The probability that the final arrangement is (Goat,Car) is exactly the probability that the first choice was Goat.

The first coin flip was NOT a fair coin. You were NOT equally able to choose Goat or Car on the first choice, and so you aren't equally likely to be based with (Goat, Car) and (Car, Goat) after Monty's move.

If Monty had you choose "your" door after he opened all but 2, then we'd be faced with a 50/50 choice. Then we'd have P(A) = P(B) = 1/2. But that's not the situation. P(A) is not 1/2, it's 1/N.

Many people here seem intent on adding bigger numbers, as if that changes anything. It does not. If there are a BILLION doors and I pick one, and only one is a winner, then my odds are one in a billion.

Precisely.

However if God comes by and eliminates all but two choices, one of which is the winner, and I can choose either... it's back to one in two. The prior one in a billion choice has no bearing on the subsequent one in two choice.

Also correct. However, if God lets you choose first, so that there is only a 1 in a bililion chance you've got the car, then there is a 999,999,999 in a billion chance that God's door is the car. Those outcomes are all the ones in which you didn't pick the car first.

I hope that this helps....

It helped me sort out the thinking of the 50-percenters.

 

[apoger]

After rereading all this insanity, suddenly I "got it".
It still seems counter-intuitive... but you (er... all of you) are correct. The first choice does transfer information that can be used to improve the odds in the second pick.

Now the trick will be to figure out a clear way to communicate it.

 

[epepke]

It has nothing to do with what Monty feels like doing as long as he behaves as stated in the problem. I am convinced that all disagreement in this topic is based around misunderstanding the problem.

Here is how I understand it:

If you disagree then please state the problem as you understand it.

As I said, I have empirical evidence that this cannot be explained. It just bounces off people's foreheads. That's why I prefer to win drinks off of them.

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.

But people are just so terribly impressed with themselves for understanding what is essentially the kind of math that should reasonably be expected from a 14-year-old that they work terribly hard to avoid the obvious conclusion. Which means they suck at being skeptics. Which is why I don't feel bad for winning free drinks off of them and humiliating them in front of their peers. Because skeptics should think about these things.

 

[Robin]

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.

Well as I said please state the problem as you understand it or offer a link to it.

I have not seen the version where Monty can decide arbitrarily whether or not to offer the second choice.

If this was the case then you couldn't have a strategy to pick the other curtain 100% of the time because you wouldn' t be offered the choice 100% of the time.

 

[hgc]

Look at it this way...

Each door has a 1/3 chance of being the car. So with the first choice your door has a 1/3 chance and the other 2 doors together have a 2/3 chance. But then one of those other 2 doors is knocked out. But still the other 2 doors have their 2/3 chance. The revealed door went to zero chance and transfered its 1/3 to the switch choice door, giving it 2/3.

 

[Robin]

It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Questions?

I asked epepke and I will ask you - please state the problem or show a link to the version where the host can decide arbitrarily whether or not to open another door and offer to switch.

This is the way I saw it:

Suppose you're on a game show, and you're given the choice of three doors; Behind one door is a car; behind the others, goats. You pick a door, say No.1 and the host, who knows what's behind the doors, opens another door, say No.3, which has a goat. He then says to you, "Do you want to pick door No.2?" Is it to your advantage to switch your choice? -Craig F. Whitaker, Columbia, Md.

Let me emphasise this :

...the host, who knows what's behind the doors, opens another door, say No.3, which has a goat. He then says to you, "Do you want to pick door No.2?"

Revealing the goat and offering a choice is part of the conditions of the problem, and so the probability, if you always change selections, is 0.667.

If the host does not complete these steps then certainly that changes the odds. It also changes the odds if the host does not turn up in the first place and has no goats or cars.

 

[apoger]

And now an attempt to clarify for all those that haven't made the connection...

First thing is we drop Monty, doors, car, and the goats, as they are dragging too much emotional baggage at this point.


Let us say that you and I are sitting at a table.
I place 10 paper cups on the table.
Under one cup I hide a walnut.

I move five cups to the right side of the table.
I move five cups to the left side of the table.

The odds of the walnut being on the left side is 50%. Same goes for the right side. I think we can all see that.

Now I move one cup from left to right. The odds of the walnut are being on the right side is now 60%. The left is down to 40%.

Then I move three more cups from the left to the right. The nine cups on the right represent a 90% chance of having a wlanut. The single cup on the left has a mere 10% chance.

Here comes the trick...
I now turn over eight cups that have no walnut. All eight are on the right side. Now we have new information. If the 90% chance of the walnut being on the right side is valid, then there is a 100% chance that the walnut is under the last cup on the right side. However this does not change the 90% chance the the walnut is on the right side. Thus there is a 90% chance of the right cup having the walnut and a 10% chance of the left cup having it.

Given a choice between the two cups, you would of course gamble on the right cups 90% chance.


This visualization helped me focus on the probem. I hope it's simplicity has helped someone else.

 

[TeaBag420]

Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here.

But I'm convinced from empirical evidence that it's not possible to explain this to people. Fortunately, it is possible and a lot more fun to win free drinks off of them.

I think that your use of the word "'tards" is offensive.

 

[Robin]

And now an attempt to clarify for all those that haven't made the connection...

First thing is we drop Monty, doors, car, and the goats, as they are dragging too much emotional baggage at this point.


Let us say that you and I are sitting at a table.
I place 10 paper cups on the table.
Under one cup I hide a walnut.

I move five cups to the right side of the table.
I move five cups to the left side of the table.

The odds of the walnut being on the left side is 50%. Same goes for the right side. I think we can all see that.

Now I move one cup from left to right. The odds of the walnut are being on the right side is now 60%. The left is down to 40%.

Then I move three more cups from the left to the right. The nine cups on the right represent a 90% chance of having a wlanut. The single cup on the left has a mere 10% chance.

Here comes the trick...
I now turn over eight cups that have no walnut. All eight are on the right side. Now we have new information. If the 90% chance of the walnut being on the right side is valid, then there is a 100% chance that the walnut is under the last cup on the right side. However this does not change the 90% chance the the walnut is on the right side. Thus there is a 90% chance of the right cup having the walnut and a 10% chance of the left cup having it.

Given a choice between the two cups, you would of course gamble on the right cups 90% chance.


This visualization helped me foucus on the probem. I hope it's simplicity has helped someone else.

I think that is a very good way of putting it. When I first heard the problem I said 50% and it took me quite a long time to change my mind.

 

[Robin]

Originally posted by TeaBag420 in response to epepke
I think that your use of the word "'tards" is offensive.

Not only that but inaccurate because the result has nothing to do with how the host feels.

The host can be malicious, fair or helpful but it does not affect the outcome one iota.

So it would be offensive if he was right. If he is wrong then it blows right back in his face and is rather amusing.

Epepke, feel free to answer my last post and state the version of the problem where the host has the option of not offering the second choice.

 

[TeaBag420]

As I said, I have empirical evidence that this cannot be explained. It just bounces off people's foreheads. That's why I prefer to win drinks off of them.

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.

But people are just..... a 14-year-old..... terribly hard to avoid ....... suck .....Which is why I don't feel bad for ..... humiliating them in front ...... think about these things.

And if Monty followed the strategy outlined above, most contestants would go home winners, because they wouldn't switch...WHETHER THEY KNEW ABOUT THE RESTRICTION OR NOT.

Why don't you back up your assertion about the original question by citing your source?

 

[69dodge]

Originally posted by DaveW
This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtml

Here's an excerpt from that web page:<blockquote>What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?</blockquote>If you write your initial door choice on a secret ballot and don't tell Monty what it is, he might end up opening the same door you chose. If you use a secret ballot, 50/50 is indeed the correct answer, even in those instances where Monty happens not to open your door.

If Monty is guaranteed to open a goat door, and if he's guaranteed not to open your door, then the extra car odds go to the door that he might have opened but in fact didn't. This makes perfect sense: probably, the reason he didn't open that door is precisely that it has a car behind it. On the other hand, you knew from the beginning that he wasn't going to open your door, so the fact that he didn't open it doesn't tell you anything you didn't already know about what's behind it.

It's not seeing the goat that's important; it's seeing Monty's door choice. In particular, it's seeing which door he chooses to open when he is constrained (a) to open one door, (b) not to open your door, and (c) not to open the car door. If he is not so constrained, the odds will be different, even if you happen to end up seeing the same goat behind the same door as before.