The Monty Hall Problem (Probability Debate)

[Dre89]

On a gameshow, there a three doors. Behind one is a car, behind the other two are goats. The contestant, who doesn't know which one the car is behind, is asked by the host (Monty Hall, who knows where the car is) to chose a door. The contestant chooses a door (let's say door 1 for the sake of the example). MH then opens another door which he knows hides a goat (let's say door 3). He then gives the contestant the opportunity to stay with his original choice (1), or switch to the other remaining door (2). Which is more probable, staying or switching?

The answer provided to the puzzle was switching, despite there only being two choices. The reasoning is that because the only time you shouldn't switch is if you originally chose the car, but because there's only one car and two goats, there was a 2/3 chance you chose a goat, so it is supposedly more probable to switch.

What do you guys think? Do you think the probablity is 50/50, or do you think it is more probable to switch? I did a presentation on this, and I've got a lot more material (both for and against the switching solution) which I'll reveal as/if the debate progresses.

 

[ballin4life]

Always switch (assuming Monty knows where the car is and never accidentally reveals the car). The way you explained it is one way to look at it. Also you can just work out all the conditional probabilities.

 

[blazedaces]

On a gameshow, there a three doors. Behind one is a car, behind the other two are goats. The contestant, who doesn't know which one the car is behind, is asked by the host (Monty Hall, who knows where the car is) to chose a door. The contestant chooses a door (let's say door 1 for the sake of the example). MH then opens another door which he knows hides a goat (let's say door 3). He then gives the contestant the opportunity to stay with his original choice (1), or switch to the other remaining door (2). Which is more probable, staying or switching?

The answer provided to the puzzle was switching, despite there only being two choices. The reasoning is that because the only time you shouldn't switch is if you originally chose the car, but because there's only one car and two goats, there was a 2/3 chance you chose a goat, so it is supposedly more probable to switch.

What do you guys think? Do you think the probablity is 50/50, or do you think it is more probable to switch? I did a presentation on this, and I've got a lot more material (both for and against the switching solution) which I'll reveal as/if the debate progresses.

... While I do enjoy this discussion there is absolutely no debate here. This is not a subjective question of opinion, despite the answer being counter-intuitive. There is one correct answer, and that is switching to the other door.

A simple google search brings up this wikipedia article which explains the problems and solution in great detail: http://en.wikipedia.org/wiki/Monty_Hall_problem

-blazed

 

[Dre89]

Actually this has been heavily debated amongst mathematicians and the like.

Here's one response to the switch propososal. Suppose door 3 is opened by MH, door 1 and door 2 are equally probable. The fact that in this scenario, you don't know which door was originally chosen by the contestant suggests that one and two are equally probable.

The information given here (that door 3 is opened), tells us nothing about which door is more probable, because door three could have been opened regardless of whether the contestant chose door 1 or 2.

 

[blazedaces]

Actually this has been heavily debated amongst mathematicians and the like.

Here's one response to the switch propososal. Suppose door 3 is opened by MH, door 1 and door 2 are equally probable. The fact that in this scenario, you don't know which door was originally chosen by the contestant suggests that one and two are equally probable.

The information given here (that door 3 is opened), tells us nothing about which door is more probable, because door three could have been opened regardless of whether the contestant chose door 1 or 2.

Fine, but now you are changing the parameters of the problem. In each case there is only one correct answer. I don't care much how often it is debated. There's also a flat earth society, but there's no real debate about the shape of the earth...

Even if there were a debate one could always run a simulator of the same conditions some extremely large number of times to see what percentage converges...

-blazed

 

[Dre89]

I think it's a bit of an exaggeration to compare 50/50ers to flat Earthers....

The point is that in an individual case, it is 50/50 because which door the contestant originally chose can have no bearing on which door is opened. Thus there is no distinction between the initial choice and the switch-choice.

 

[-Jumpman-]

If it's hard to understand, try 100 doors instead of 3.

 

[blazedaces]

I think it's a bit of an exaggeration to compare 50/50ers to flat Earthers....

I like taking things to an extreme case because it makes my point rather well.

The point is that in an individual case, it is 50/50 because which door the contestant originally chose can have no bearing on which door is opened. Thus there is no distinction between the initial choice and the switch-choice.

Your further clarification of the situation doesn't change anything...

-blazed

 

[SuperBowser]

There's only one answer here. There's not much to debate.

Like jumpman said, if you don't understand the solution, just imagine the same problem with 100 doors. I think most people would understand the statistics more easily then. The important point is that Monty Hall already knows which door is the correct one while you only have a 1 in X chance of choosing the correct door. When Monty Hall knowingly removes all the false doors, there is a greater probability that the remaining door is the correct one, rather than the one you chose by random at the beginning.

You have a 1 in X chance of being correct. Monty Hall has a (X-1)/X chance of leaving you the correct door.

The information given here (that door 3 is opened), tells us nothing about which door is more probable, because door three could have been opened regardless of whether the contestant chose door 1 or 2.

False. This is not your original question; Monty Hall knows where the solution lies. This does affect which door is more probable.

 

[Dre89]

I know of the 100 door example.

And MH knowing where the car is is exactly my point.

The fact that 3 is opened sheds no light on which door hides the car, it doesn't even tell you which door the contestant initially chose. I don't understand what your point about MH knowing where the car is was supposed to demonstrate.

 

[-Jumpman-]

I know of the 100 door example.

And MH knowing where the car is is exactly my point.

The fact that 3 is opened sheds no light on which door hides the car, it doesn't even tell you which door the contestant initially chose. I don't understand what your point about MH knowing where the car is was supposed to demonstrate.

It means he's quoting Wikipedia.

 

[SuperBowser]

Lol I'm not quoting wikipedia. I did this puzzle in school years ago. It's just basic probabilities

I know of the 100 door example.

And MH knowing where the car is is exactly my point.

The fact that 3 is opened sheds no light on which door hides the car, it doesn't even tell you which door the contestant initially chose. I don't understand what your point about MH knowing where the car is was supposed to demonstrate.

Did you even read your own scenario? You did a presentation on it, so you should understand the significance of Monty Hall's prior knowledge. You said in your first post: "The contestant chooses a door (let's say door 1 for the sake of the example). MH then opens another door which he knows hides a goat (let's say door 3)."

Monty Hall can't remove the car because that is the rules of the game (as stated in your own scenario). He therefore only removes goat doors. This matters.

 

[1048576]

Imagine there's 1000 doors, 1 car and 999 goat, and after you pick, Monty reveals 998 goats. Switching or not becomes pretty clear then.

 

[ballin4life]

1/3 chance you are right initially. Monty can then open either of the two doors - it doesn't matter. In this case switching gives a payoff of 0 cars and not switching gives a payoff of 1 car.

2/3 chance you are wrong initially. Monty can then only open only the other door that holds a goat. In this case switching gives a payoff of 1 car and not switching gives a payoff of 0 cars.

So overall switching gives (1/3)*(0 cars) + (2/3)*(1 car) = 2/3 of a car on average.
Not switching gives (2/3)*(0 cars) + (1/3)*(1 car) = 1/3 of a car on average.

Switching is the better option.

 

[SuperBowser]

^Indeed. Just basic probabilities, not a debate.

You have a 1 in X chance of being correct. Monty Hall has a (X-1)/X chance of leaving you the correct door.

 

[Dre89]

I understand the probability behind the solution, I'm not stumped by it or anything. I used alot of the stuff you guys are saying in my presentation. It's just that this was actually heavily debated, so I was trying to provide a different perspective.

Superbowser- Forgive me, but I still don't understand the MH knowledge point. I understand he can't chose the car, which means he can only chose 2/3 doors, but my point is that is the only thing affecting his choice, not which door the contestant initially chose.



Ok, I know this will probably sound like a stupid question, but how does it change if the contestant knows that MH will open a door? If he wants to choose 1, and then knows it's more probable to switch, he'll switch to 2 or 3 depending on which door MH opens. But this is arbitrarily ruling out door 1 based on merely the contestant's decision.

To me, it doesn't make sense to say that a door is less probable simply because the contestant chose it, especially when, for example, MH opens door 3, regardless of whether 1 or 2 is chosen.

 

[SuperBowser]

You might have known the probabilities for your presentation but you clearly didn't understand them. The point is that the contestant is never faced with a 50/50 decision. Read below and try reading the puzzle again.

When the contestant chooses a door he is making a 1 in 3 decision. I hope we can agree upon this - there are 3 doors present and the contestant chooses a door by random.

Monty Hall is then left with 2 out of 3 doors. I hope we can agree there is a 2 in 3 chance the car lies within these 2 doors. Again, this is basic probability. Now the key point is that Monty Hall removes a goat door from his options and leaves you one remaining door. However, this doesn't change the fact that he sorted through 2 doors and there was a 2 in 3 chance the car lay within his options. Even though Monty Hall only presents you 1 door from his group, he has actually sorted through 2. This is why you should always switch to the door Monty Hall presents you with.

 

[rvkevin]

The only way to get away from the answer given is to change the assumptions that they are operating on. If you want to change the scenario by saying that MH is forgetful and forgot where the car was and opened up a random door, then it would be 50/50 when given the opportunity to switch. However, him knowing where the car is and purposely choosing the goat changes the payoff matrix which makes it in your favor to switch.

 

[ballin4life]

Here's a better one:

Scenario A: You're walking down the street and see a friend walking his dog. You stop to chat, and your friend says he has been very busy raising his 2 children. He mentions that he is buying a My Little Pony DVD later that day. What is the probability that both of your friend's children are girls?

Scenario B: This time, your friend is walking his dog along with a little girl. You stop to chat, and your friend says he has been very busy raising his 2 children. He introduces the girl to you as his daughter. What is the probability that both of your friend's children are girls?

 

[Battlecow]

100% in both cases. I don't see your point

 

[T-block]

Ok, I know this will probably sound like a stupid question, but how does it change if the contestant knows that MH will open a door? If he wants to choose 1, and then knows it's more probable to switch, he'll switch to 2 or 3 depending on which door MH opens. But this is arbitrarily ruling out door 1 based on merely the contestant's decision.

To me, it doesn't make sense to say that a door is less probable simply because the contestant chose it, especially when, for example, MH opens door 3, regardless of whether 1 or 2 is chosen.

It's not arbitrarily ruling out anything. The contestant should realize that by sticking with his choice, he has a 1/3 chance of winning the prize. By switching, he is essentially able to say "I bet the prize is in either this one or this one", except that the either-or is actually solved before the reveal, giving him a 2/3 chance. If the prize is revealed to be behind the original choice, the contestant shouldn't be going "wtf happened this is the best strategy" - he should acknowledge that there is still a 1/3 chance of the original choice being correct.

I'm really confused at what you're trying to do, Dre o.o I believe that you understand the probabilities, but there's no other perspective to show...

Another interesting approach to the problem is to consider the overall probabilities of being correct. A contestant who always stays with his choice will have a 1/3 chance of winning the prize. Since there is only one choice with two options to be made, a contest who always performs the other action, switching, must have 1-(1/3) = 2/3 chance of winning.

 

[jaswa]

Lets hypothesise that I walk onto the show not knowing what's going on and take Dre's place. I see 2 doors and am told to choose between them. What's the chance that I pick the one with the prize?

 

[eschemat]

50%....?

What's the point you're trying to make?

 

[jaswa]

50%....?

What's the point you're trying to make?

My point is: is the probability absolute? or relative?

 

[ciaza]

It's 50/50, but what I think you're forgetting is that you don't know what Dre.'s initial decision was. If you did know it, the probabilities would change.

Everyone play the game here!

 

[Dre89]

Ok, but suppose I choose door 1, and Jaswa, watching it on TV and deciding to play it for himself, chooses door 2. If MH opens 3, then which door is most probable? If both of us switched, we'd be saying different doors are probable.

It doesn't matter whether Jaswa or I is the contestant, because MH could have opened 3 if either 1 or 2 was chosen.

So you're either saying that 2 is more probable than 1 for me, and vice versa for Jaswa, or one of us shouldn't switch. If one of us doesn't switch, then switching isn't more probable for one of us, and how would you decide who switches?

 

[ciaza]

You should still both switch based on probabilities. All it means is that one of you got unlucky by initially choosing the door with the car. This is the difference, one of you would have chosen the door with the car. However neither of you knew that. Of course after the game is over then it would have been apparent that the loser should have stayed, but it would have been a dumb move since they would have only had a 1/3 chance of being right.

Ultimately it's still down to being a game of luck. **** happens.

 

[Dre89]

The point is that one box is favoured based on a subjective arbitrary reason ie. the the person's initial choice.

 

[ciaza]

AKA luck. I'm not disagreeing. I think I get what you're getting at. If you did that 10 times there is a loser and a winner 50/50 (probably) of the time. The variable that you're taking out however is that if you both chose a door it restricts MH's choice of door. For the game to work there has to be a maximum of 1 contestant.

 

[blazedaces]

Ok, but suppose I choose door 1, and Jaswa, watching it on TV and deciding to play it for himself, chooses door 2. If MH opens 3, then which door is most probable? If both of us switched, we'd be saying different doors are probable.

It doesn't matter whether Jaswa or I is the contestant, because MH could have opened 3 if either 1 or 2 was chosen.

So you're either saying that 2 is more probable than 1 for me, and vice versa for Jaswa, or one of us shouldn't switch. If one of us doesn't switch, then switching isn't more probable for one of us, and how would you decide who switches?

This is a different game now because what happens if MH opens door 1 (the one you chose)? Do you see how playing while another plays changes things drastically?

Changing the parameters of the game can change the probability of the outcome. There's not much more to debate though.

We can go through 100 manipulations of the MH problem. There's plenty of people who find it very enjoyable to do so and discuss... but it's not a debate.

-blazed

 

[ballin4life]

You should still both switch based on probabilities. All it means is that one of you got unlucky by initially choosing the door with the car. This is the difference, one of you would have chosen the door with the car. However neither of you knew that. Of course after the game is over then it would have been apparent that the loser should have stayed, but it would have been a dumb move since they would have only had a 1/3 chance of being right.

Ultimately it's still down to being a game of luck. **** happens.

No you are wrong.

The person on the show should switch, and the person at home should not switch.

This is because Monty is only aware of which door was guessed by the person on the show.

 

[ciaza]

I assumed he did know however, otherwise if MH opened the door jaswa chose initially the game falls apart. Dre. sort of stated that as well. Unless I'm interpreting wrong.

 

[A1lion835]

If you're really fixated on debating probability, the Sleeping Beauty Problem would've been a better choice of topic.

 

[blazedaces]

If you're really fixated on debating probability, the Sleeping Beauty Problem would've been a better choice of topic.

Very interesting problem. I'd never heard of it before.

-blazed

 

[T-block]

No you are wrong.

The person on the show should switch, and the person at home should not switch.

This is because Monty is only aware of which door was guessed by the person on the show.

This. In Dre's example, when he picks door 1, there is a 2/3 chance of the car being in door 2 or door 3. When Monty opens door 3, he moves that 2/3 chance all into door 2. The person playing at home should realize this =\

Sleeping Beauty problem is interesting... I'm pretty sure it's 1/3 chance though.

St. Petersburg paradox might be a fun one to debate too.

 

[ciaza]

Well then I apologise with the utmost sincerity.

Spoiler

HA HA! A mature apology is apparently all I need to get +1'd by some members

 

[1048576]

I like the traveller's dilemma. http://en.wikipedia.org/wiki/Traveler's_dilemma

 

[Dre89]

If anyone talks about the solution to the Sleeping Beauty problem, I will have you killed, because I'm going through a puzzle book and haven't got to it yet.

I know that's totes selfish but whatevs, I live by my own rules...

 

[ballin4life]

I like the traveller's dilemma. http://en.wikipedia.org/wiki/Traveler's_dilemma

BAH. simple game theory.

At least go with the one where the pirates are dividing up the gold.

 

[Battlecow]

I remember you talked about that one on the 64 boards one time and then I went and made decision trees with my dad for like 4 hours. **** you, man.