Why the Monty Hall puzzle is categorically 1/2 and not 2/3rds

[CaptainOrange]

So I was browsing the Wikipedia article concerning The Monty Hall Paradox, and I seem to take great issue with the assumption that switching results in a 2/3 probability of winning a car.
(hold on pressed enter by mistake... editing now watch this space)

My reasons are as follows (and I don't believe this has anything to do with intuition bias)

Conditions and assumptions:

* This puzzle uses the word "say" before No.1 and No3... Therefore, it could be equally valid that contestant choose 2, or 3 initially
* Monty Hall must always choose a goat door. If guessed correctly, this would either be 2 of the goat doors (irrespective of number) decided by an a-priori coin toss before the show and not chosen out of the whim of the presenter. Alternatively, the presenter knows what door has the goat and always opens the door with the goat. Therefore the switch door and the door Monty opens can be in variable order.
* I've never watched this show, so I'm going to ignore all subjective behavioural characteristics (e.g. no right/left door bias)
* the presenter must always choose a door with a goat, and always offer a choice to switch or not.
* The contestant has no idea what's behind the door.

The original puzzle:
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?

It is claimed that switching results in a probability of 2/3rds, but it has to be 1/2, factoring all those conditions outlined above.

I didn't study Maths or Physics, but this makes perfect logical sense to me.

 

[Borek]

Have you tried to search the forum?

https://www.physicsforums.com/showthread.php?t=62805&highlight=monty+hall

https://www.physicsforums.com/showthread.php?t=608926&highlight=monty+hall

https://www.physicsforums.com/showthread.php?t=522062&highlight=monty+hall

https://www.physicsforums.com/showthread.php?t=523119&highlight=monty+hall

https://www.physicsforums.com/showthread.php?t=284584&highlight=monty+hall

https://www.physicsforums.com/showthread.php?t=249658&highlight=monty+hall

 

[pwsnafu]

After Monty opens the door, he gives you the option to stay or switch. The probability of the stay strategy being correct is equal to the probability you were correct to begin with. That probability is of course 1/3, because you chose that door when there were three options. So the switch strategy is 2/3. QED.

 

[CaptainOrange]

After Monty opens the door, he gives you the option to stay or switch. The probability of the stay strategy being correct is equal to the probability you were correct to begin with. That probability is of course 1/3, because you chose that door when there were three options. So the switch strategy is 2/3. QED.

Thanks Borek for that link I'm reading those threads to get my head round this...

@pwsnafu

After Monty opens the door, he gives you the option to stay or switch. The probability of the stay strategy being correct is equal to the probability you were correct to begin with. That probability is of course 1/3, because you chose that door when there were three options. So the switch strategy is 2/3. QED.

This is where I'm having trouble grasping. It is true at the start your odds are always 1/3 of winning (before the door is opened and selected at random).

I accept that the odds change when the door is opened...but

(this is how my mind is processing the problem)
1) If Monty has apriori knowledge and always chooses one of the remaining goat doors (100% behaviour). In my head, this leaves 2 doors... with 1 goat, 1 car... Always

Door 1: It could be a goat, it could be a car.
Door 2: It could be a goat, it could be a car.

It no longer becomes 2 goats and 1 car, but 1 car and 1 goat.
I'm going to substitute goat and car with head and tail.

You toss two coins: it could be heads, it could be tails
coin 2: it could be heads, it could be a tails.
50:50

Each time you toss a coin, the odds are always 50:50 for the subsequent coin toss to match the correct guess irrespective if it is heads/head, tails/tail, heads/tail, tail/heads etc.

The decision to switch from the playing perspective is like saying "heads or tail?" before the coin is actually tossed.

Am I misunderstanding the problem?

 

[CaptainOrange]

I found a comment from DaveC426913 which matches my conclusion... I have also included another quote from another poster which concludes with 2/3rds probability and I've highlighting the part that I'm having trouble following.

DaveC426913 wrote:

Monty Hall
A: goat B: goat C: car

1 You choose A
1.1 Monty opens B
1.1.1 You stay with A, You lose
1.1.2 You switch to C, You win

2 You choose B
2.1 Monty opens A
2.1.1 You stay with B, You lose
2.1.2 You switch to C, You win

3 You choose C
3.1 Monty opens A
3.1.1 You stay with C, You win
3.1.2 You switch to B, You lose
or
3.2 Monty opens B
3.2.1 You stay with C, You win
3.2.2 You switch to A, You lose


Conclusion: It makes no difference whether you switch or not. If you switch, you have a 50/50 chance of winning the car. If you do not switch, you have a 50/50 chance of winning the car. The fact that your first choice had a 1/3 chance of winning is a red herring, and is what keeps tripping people up.

Mister X: 2/3 hypothesis

Dave, I have modified your list to include probabilities. Some of these are conditional probabilities; for "1.1 Monty opens B (1)", (1) means this would always happen given that "1 You choose A"
There is an equation for the probability of each outcome. Winning outcomes are in bold.

1 You choose A (1/3)
1.1 Monty opens B (1)
1.1.1 You stay with A (1), You lose, probability = 1 * 1 * (1/3) = 1/3
1.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

2 You choose B (1/3)
2.1 Monty opens A (1)
2.1.1 You stay with B (1), You lose, probability = 1 * 1 * (1/3) = 1/3
2.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

3 You choose C (1/3)
3.1 Monty opens A (1/2) <----- Why is (3) written as "1/2" and not like (1) and (2) "1"?
3.1.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6 <----Why the change to 1/6th?
3.1.2 You switch to B (1), You lose, win probability = 1 * (1/2) * (1/3) = 1/6<------Ditto
or
3.2 Monty opens B (1/2)
3.2.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6
3.2.2 You switch to A (1), You lose, probability = 1 * (1/2) * (1/3) = 1/6

So, the total probability of winning with the strategy of staying is 1/6 + 1/6 = 1/3.
The total probability of winning with the switch strategy is 1/3 + 1/3 = 2/3.
my comment: Really? surely if the presenter was made to categorically open a door with a goat, doesn't DaveC426913's logic make sense?

(I'm not a maths wizard, but doesn't the terms of the original puzzle change to match the new circumstances?)

 

[micromass]

Consider this: you have 1000 doors, one of which has a car. After you pick a door, the quizmaster opens 998 doors that do not contain the car. You then get the chance of switching. Do you switch in this scenario?

 

[CaptainOrange]

Consider this: you have 1000 doors, one of which has a car. After you pick a door, the quizmaster opens 998 doors that do not contain the car. You then get the chance of switching. Do you switch in this scenario?

Well it's a different puzzle, but.

1000-998 = 2

2 doors.

1 car for 1 door, 1 goat for another.

For the remaining 998 doors if posed this question than this 50% chance assumption is incorrect. When it gets to two possible doors left than it is squarely 50%.

Is this a problem of linguistic semantics and maths (akin to a riddle?). I just don't know why 2/3's assumption of one of the two doors come in... this is like saying the mean average number of children born per couple in the world is 2.33, but this is does not equate to 2 babies, and a baby who is only a 1/3 born?

2 doors left... 50% chosen door is car
3 doors left... 33.33%... chosen door is car
4 doors left... 25% chosen door is car
5 doors left 20% chosen door is car
etcin all 2+ door instances, surely switching presents the same percentage odds? if I switch door if there were 5 left, then there is still 20% chance that the door I switched too is correct. No change?

Bit confused.

 

[CaptainOrange]

Ahh...I sort of get it now but I'm debating whether it will work in a social science context as opposed to a mathematical concept.

I realized if the rules are extended to multiple doors, and the doors chosen by Monty are duds than this can lead a 2/3 interpretation.

However,

I'm wondering (purely from a Pyschologist perspective) If the Monty problem would be congruent with a maths (theory) interpretation.

Assuming 2 rooms...Each room had one participant. Both are blind to each other and no information can be leaked to another (sound proof).

The psychologist is the arbiter... Participant A Selects one of these doors. There are two left.

Participant B is aware of Participant's choice, and must either a) choose the remaining goat or B) select from a possible door to challenge Person A.

Person A is than relayed this information of Person's B choice through a binary receipt inputted by the psychologist. Person A has to make a choice to stay or switch.

I seriously wonder, given the fact that confounding variables are removed and this set up is blind and cannot be leaked. Whether this does actually does lead to the same conclusion, with 100 randomly assigned participants running 50 different tests.

Surely that would be the decider to determine if the internal maths system theory is congruent with happenings observed in the external world.

 

[HallsofIvy]

I'm a bit confused as to why you did not ask why the answer is 2/3, originally, rather than asserting that it is 1/2.

 

[micromass]

The answer really is 2/3 instead of 1/2. There are multiple ways to see this fact. If the math doesn't convince you, then maybe a practical experiment will. Do the experiment a large number of time and see how many times you get the right answer by switching the doors.

 

[CaptainOrange]

I'm a bit confused as to why you did not ask why the answer is 2/3, originally, rather than asserting that it is 1/2.

I'm not quite 100% convinced it is 2/3rds. In my head, still debating whether it is a logic trap, akin to St Anselm "Ontological Proof of God" which 'logically' holds but empirically is shaky and can be challenged (from an atheistic/scientific/realist perspective from acquiring evidence outside of the original argument). I know that's a slight jump from the original math puzzle posed but I am now going to look online for evidence from psychology a-posteriori tests with real participants to rest my mind.

Maybe I'm experiencing cognitive dissonance but I'm questioning whether it is a systemic flaw in mathematical modelling or a flaw in my understanding of the original problem. Even a weather man, with a team of meteorologists behind him could be wrong sometimes, and not all models of ecology are predictive of what actually occurs (e.g. humans are not on top of the food chain as they still can be eaten by sharks etc)

 

[arildno]

Remember that MONTY's opening of doors is basically non-random, due to his knowledge of where the car is.
Allowing for his ignorance as well, not just yours (which would add the possibility that he opened the door with the car, giving you a guaranteed loss), switching from your part would never increase your odds.

 

[AlephZero]

I think you are getting sidetracked by talking about modelling here. There is nothing to model here - or, if you prefer, Monty's model of the situation is perfect, because he knows all the relevant facts. And your model of the situation needs to include the fact that Monty is omniscient. The apparent paradox only arises if you ignore Monty's omniscience, and the game situation is set up to disguise that fact - especially if you only play once.

 

[bahamagreen]

Think of a stack of all the tickets for a lottery - one is the winner - you choose one.
Then, except for one all the remaining non-winning tickets are destroyed, so either yours or the one remaining is a winner.
Which ticket do you think is the winner?

Go back to the 1000 doors scenario...

1000 doors... you choose one... what are your chances of picking the one with the car?

998 non-car doors are opened.

There are now just two doors, your original choice, and the other one.

The chance that you originally selected the car with your first choice was a thousand to one.
Now, you know for certain that the car is behind either your first choice or the remaining door... remind yourself how little chance your first choice was of being the car. Think about it... how incredibly lucky you would have had to have been to have chosen correctly the first time.
Now you have a choice between two doors, the one you chose out of a thousand, an the other one...

Substitute any large number for the original number of doors until you see it...
A million doors, a billion doors, a googleplex of doors... your chance of selecting the right one on your first choice approaches infinitesimal.

 

[haruspex]

I found a comment from DaveC426913 which matches my conclusion.

Fwiw, here's where DaveC's logic goes wrong.
He lists four cases and assumes they're equally likely, but they're not. There are three equally likely possibilities: you chose A, B or C. The C case he then splits into two possibilities, but the probabilities are now:
A 1/3
B 1/3
C1 1/6
C2 1/6
So the odds of winning by switching are 2/3 (A+B).

 

[pwsnafu]

I know that's a slight jump from the original math puzzle posed but I am now going to look online for evidence from psychology a-posteriori tests with real participants to rest my mind.

Why are you thinking about psych? You should be looking for computer simulations of the problem.

Maybe I'm experiencing cognitive dissonance but I'm questioning whether it is a systemic flaw in mathematical modelling or a flaw in my understanding of the original problem.

It can't be case one because the problem is purely mathematical to begin with. It is its own mathematical model.

Even a weather man, with a team of meteorologists behind him could be wrong sometimes,

But he is statistically correct, which is what you want.

and not all models of ecology are predictive of what actually occurs (e.g. humans are not on top of the food chain as they still can be eaten by sharks etc)

You are thinking of a food web. A linear sequence along the web from basal to consumer is a food chain. Ecologists are not interested in food chains per se, they are interested in the average over all possible chains in a web. This is called the food chain length.

 

[HallsofIvy]

I'm not quite 100% convinced it is 2/3rds.

That doesn't respond to my question. If you are NOT "100% convinced it is 2/3rds" why did you start by asserting that it is 1/2? Those are quite different statements.

 

[TheIliaD]

This is how I think about it:

The only outcome where it would be beneficial to you to not switch is when you guess correctly with your initial choice. As you know this happens 1/3 of the time. This would mean that the other 2/3 of the time you will get it correct if you switch.

In other words, every time you do not choose the correct door on your first try, switching is guaranteed to give you the correct door since you are currently on one wrong door and the other wrong door has been eliminated. You do not choose the correct door on your first try 2/3 of the time, therefor 2/3 of the time when you switch you will end up on the correct door.

 

[arildno]

This is how I think about it:

The only outcome where it would be beneficial to you to not switch is when you guess correctly with your initial choice. As you know this happens 1/3 of the time. This would mean that the other 2/3 of the time you will get it correct if you switch.

In other words, every time you do not choose the correct door on your first try, switching is guaranteed to give you the correct door since you are currently on one wrong door and the other wrong door has been eliminated. You do not choose the correct door on your first try 2/3 of the time, therefor 2/3 of the time when you switch you will end up on the correct door.

This is incorrect.
If you happen to be on a show with an IGNORANT Monty Hall, who happens to open a door with a goat, then it is irrelevant if you switch or not.
The critical issue is that a rule of the game is that you can rest assured that Monty Hall WILL open a door for the second chance (rather than revealing your first picked goat), and that he ALWAYS will reveal a goat after your first pick.

 

[Integral]

You choose 1 door, so have a 1/3 chance of making the right pick. That means the odds of the car being behind one of the remaining 2 doors is 2/3. That should be clear and obvious, right?

Why does the host opening one of those doors change the odds?

The host shows you which one is NOT the car, there is a 2/3 chance of winning the car if you switch, 1/3 if you don't.

 

[arildno]

Just following Integral:
The reason why the probabilities DON'T change is that you know in beforehand that Monty will reveal at least goat happened to be present in the GROUP of doors you did not pick to begin with. Thus, the Group probability (of containing car) remains unchanged, relative to your initial information after Mobty opens the door.
If you are blind, and are allowed to pick the opened door as well as the closed one, there is no reason to switch..

 

[skiller]

This is incorrect.

Why?

If you happen to be on a show with an IGNORANT Monty Hall, who happens to open a door with a goat, then it is irrelevant if you switch or not.

I don't think anyone's suggested that Monty Hall has ever been ignorant of the contents.

The critical issue is that a rule of the game is that you can rest assured that Monty Hall WILL open a door for the second chance (rather than revealing your first picked goat), and that he ALWAYS will reveal a goat after your first pick.

Precisely, so what TheIliaD said seems pretty correct to me. where does "revealing you first picked goat come" into it?

 

[micromass]

This is incorrect.

I don't see anything wrong with what he said. He clearly said that the wrong door gets eliminated, not a random door.

 

[DonAntonio]

I don't see anything wrong with what he said. He clearly said that the wrong door gets eliminated, not a random door.

This is one of the oldest and, usually, hottest mathematical discussion on the web. I remember it AT LEAST from 1987, when I

was still an undergraduate and the academic net (the nowadays web's daddy) boiled with this stuff.

If mathematicians are involved one can usually expect they will understand more or less pretty quickly why the probability

really is 2/3, but if a greenhorn gets into the discussion...oh, my! In particular, greenhorns with an attitude and lots

of ignorance. Then not only peaceful agreement is NOT reached but things begin to rot pretty soon. Watch it!

DonAntonio

 

[CaptainOrange]

Hmm...

I still don't get it.

Does it depend if the rule changes or stays constant? (e.g. remove door C... left with two other doors. Meaning is the next part of the gameshow somehow unrelated to the first rule?)

honestly struggling to comprehend switching is 2/3rd chance.

You choose 1 door, so have a 1/3 chance of making the right pick. That means the odds of the car being behind one of the remaining 2 doors is 2/3. That should be clear and obvious, right?

Why does the host opening one of those doors change the odds?

The host shows you which one is NOT the car, there is a 2/3 chance of winning the car if you switch, 1/3 if you don't.

Completely agree with the first couple of sentences... but the last sentence "The host shows you which one is not the car" not sure I understand the jump from 1 door being 2/3rds and the other 1/3. I'm not trolling I just don't get it.

 

[lurflurf]

All the talk of door opening just confuses things. The problem is equivalent to this. Suppose there are n>2 doors and one random one is it
Choose one door
Now guess if you were right.
You are wrong more of the time as you picked one door and did not pick more than one.
in particular your are right with probability 1/n and wrong with probability 1-1/n
and
1/n<1-1/n

since n >2

It is better to bet on being wrong.

 

[DonAntonio]

Hmm...

I still don't get it.

Does it depend if the rule changes or stays constant? (e.g. remove door C... left with two other doors. Meaning is the next part of the gameshow somehow unrelated to the first rule?)

honestly struggling to comprehend switching is 2/3rd chance.



Completely agree with the first couple of sentences... but the last sentence "The host shows you which one is not the car" not sure I understand the jump from 1 door being 2/3rds and the other 1/3. I'm not trolling I just don't get it.

Think of this as follow: if you change your choice after the presenter opened a "bad" door (and yes, he always knows

what door contains the great prize), then:

1) From the beginning you chose first "bad" door, the presenter opens the second "bad" door, you switch then YOU WIN

2) From the beginning you chose second "bad" door, the present opens the first "bad" door, you switch then YOU WIN again

3) From the beginning you chose the "good" door, the present opens either "bad" door, you switch then YOU LOSE.

Be convinced the above exhaust all the possibilities, and thus by switching doors in two of them you win and in one

of them you lose, ergo the probability of winning by switching doors is 2/3. Q.E.D.

DonAntonio

 

[CaptainOrange]

Think of this as follow: if you change your choice after the presenter opened a "bad" door (and yes, he always knows

what door contains the great prize), then:

1) From the beginning you chose first "bad" door, the presenter opens the second "bad" door, you switch then YOU WIN

2) From the beginning you chose second "bad" door, the present opens the first "bad" door, you switch then YOU WIN again

3) From the beginning you chose the "good" door, the present opens either "bad" door, you switch then YOU LOSE.

Be convinced the above exhaust all the possibilities, and thus by switching doors in two of them you win and in one

of them you lose, ergo the probability of winning by switching doors is 2/3. Q.E.D.

DonAntonio

Don, that was so eloquently explained! I don't know why I was stuck at the 50:50 stage (maybe I was imagining the problem differently). After reading that I confess I have eaten humble pie.

 

[DonAntonio]

Don, that was so eloquently explained! I don't know why I was stuck at the 50:50 stage (maybe I was imagining the problem differently). After reading that I confess I have eaten humble pie.

I think we all did when we first meet with this problem. The most important lesson I learned back then: in mathematics, no matter

how good/old/veteran you think you are, always leave a little groove for doubt and humbleness to sit...and wait.

DonAntonio

 

[haruspex]

Be convinced the above exhaust all the possibilities, ..

... and are equally likely ...

 

[DonAntonio]

... and are equally likely ...

Of course, but since this is assumed to be a "choose in a random way something" out of several possibilites, this is

a rather straightforward assumption.

DonAntonio

 

[haruspex]

Of course, but since this is assumed to be a "choose in a random way something" out of several possibilites, this is a rather straightforward assumption.
DonAntonio

Yet that's where DaveC went wrong.

 

[alewisGB]

There are loads of proofs, I think a simple one is the initial probability if 1/3 yes and 2/3 no. The initial probabilery can't change at all so the chance of winning if you stick is always 1/3

 

[haruspex]

There are loads of proofs, I think a simple one is the initial probability if 1/3 yes and 2/3 no. The initial probabilery can't change at all so the chance of winning if you stick is always 1/3

Quite so, but the tricky part is understanding why the probability doesn't change. It is key that MH is always able to open another door without revealing the prize, so the fact that he does so tells you nothing about whether your choice was correct. This breaks down if MH chooses one of the other two randomly, or only knows that a particular door is wrong so can't open a door if you pick that one first.

 

[arildno]

Quite so, but the tricky part is understanding why the probability doesn't change. It is key that MH is always able to open another door without revealing the prize, so the fact that he does so tells you nothing about whether your choice was correct. This breaks down if MH chooses one of the other two randomly, or only knows that a particular door is wrong so can't open a door if you pick that one first.

Prior to choosing the door, we KNOW that at least one of the doors we do NOT open will contain..a goat.
Knowing as well that Monty knows exactly which door, and then opens, adds us no new information.
Had Monty been ignorant, our prediction would become sufficiently skewed towards the "two goats unopened"-scenario (i.e, that we picked the CORRECT door to begin with!) that it would become irrelevant whether we switch or not.