Monty Hall Problem... For Newbies

[malicus]

I freely admit that I have not the time to read the entire thread and apologize in advance if this has already been shared.

The way I was able to wrap my head around the Monty Hall problem was to put together a quick and dirty Monte Carlo simulation in Excel, which provides tangible and convincing solution. There are also some online simulators available, as below, which some might find helpful.

http://www.grand-illusions.com/simulator/montysim.htm

Oh but you should, it's a cracker

 

[Anders Lindman]

...

ETA: Wait, the chance of picking the wrong door is bigger in the first choice. So there is a dependency. And then in the next choice the bigger chance is the other door. So it's better to choose the other door.

 

[Anders Lindman]

A simple way of looking at it is that in the first choice there is 1/3 chance of selecting the car. After the presenter has opened one of the other doors with a goat behind it, the chance for the first choice is still 1/3 (the presenter will always open a door with a goat behind it and therefore it adds no additional knowledge to the first choice). The chance for the other closed door to be a car, after the presenter has opened a door is 2/3. So the other door has a higher probability of having a car behind it (the presenter has added knowledge to the second choice by revealing which of the remaining doors does not have a car behind it).

 

[Rolfe]

My God are we still at this? I worked this out before I even had a computer. Somebody told me about the puzzle, I think it had been in the Guardian.

Suppose there are 100 doors and one car. And the presenter keeps opening doors and keeps asking you if you want to switch. You keep saying no. Finally you're down to two doors. The one you originally chose and one other.

You still sure you don't want to switch?

Rolfe.

 

[dlorde]

It's deja-vu all over again...

 

[Anders Lindman]

I forgot to explain the 1/3 and 2/3. This has already been explained lots of times in this thread, but I want to present my take on it.

The first choice is easy. One out of three doors has a car behind it, and one choice is first made. The probability of choosing the car in the first choice is therefore p = 1/3.

The other two doors have a combined probability of 2/3 of having a car behind it since that's the probability of picking two doors. After the first choice, the probabilities 1/3 and 2/3 still remain. And after the presenter has opened one of those two doors, the probabilities still remain the same. So when the second choice is made, the probability of p = 2/3 still remains, even though it's now only one door to choose since the presenter has opened the other door.

 

[catsmate]

My God are we still at this? I worked this out before I even had a computer. Somebody told me about the puzzle, I think it had been in the Guardian.

Suppose there are 100 doors and one car. And the presenter keeps opening doors and keeps asking you if you want to switch. You keep saying no. Finally you're down to two doors. The one you originally chose and one other.

You still sure you don't want to switch?

Rolfe.

I remember my first encounter with the idea was in one of Leslie Charteris's Saint short stories about thirty years ago; it intrigued me then but it was pretty simple to understand with a little thought.

 

[Prometheus]

Sheesh! Why doesn't anybody understand that it's QUANTUM!!! You can never know both the goat's postition and velocity at the same time, so no matter which door you choose, you should pick the other one!

 

[The Educator]

Pull the other one (open).

 

[Rolfe]

I remember my first encounter with the idea was in one of Leslie Charteris's Saint short stories about thirty years ago; it intrigued me then but it was pretty simple to understand with a little thought.

It's quite intriguing in that the "right" answer depends on the intent of the host.

• If he's always going to open a door you have not picked, and show a goat, then switch, for sure.
• If he opens one of the doors you didn't pick at random (that is, there is a chance he will show the car and end the game right there) then there's no advantage or disadvantage to switching.
• If he's changing his strategy every time to mess with your head, then the question is meaningless.

Overall, taking all that into account, then the right answer is still to switch since if he's doing the first you are advantaged and if he's doing the second you're not disadvantaged. If he has no consistent strategy then it doesn't matter what you do because he's not predictable.

Rolfe.

 

[dlorde]

It's quite intriguing in that the "right" answer depends on the intent of the host.

• If he's always going to open a door you have not picked, and show a goat, then switch, for sure.
• If he opens one of the doors you didn't pick at random (that is, there is a chance he will show the car and end the game right there) then there's no advantage or disadvantage to switching.
• If he's changing his strategy every time to mess with your head, then the question is meaningless.

But you don't get to choose whether to switch until after he's opened a door. If he opens the door with the car, you lose whether you switch or not; if he opens the door with a goat, it pays to switch... what am I missing?

 

[Meed]

It's quite intriguing in that the "right" answer depends on the intent of the host.

• If he's always going to open a door you have not picked, and show a goat, then switch, for sure.
• If he opens one of the doors you didn't pick at random (that is, there is a chance he will show the car and end the game right there) then there's no advantage or disadvantage to switching.
• If he's changing his strategy every time to mess with your head, then the question is meaningless.

Overall, taking all that into account, then the right answer is still to switch since if he's doing the first you are advantaged and if he's doing the second you're not disadvantaged. If he has no consistent strategy then it doesn't matter what you do because he's not predictable.

Rolfe.

There's one strategy where you should stick. That is if the host only opens a second door when you've initially picked the car as an attempt to trick you into losing the prize. This seems obvious and, of course, asking the best course of action in that situation would be trivial. However, some people seem to get this wrong. Case in point, the movie "21":



"How do you know he's not playing a trick on you, using reverse psychology to try to get you to pick a goat?"

"I wouldn't really care, I mean my answer's based on statistics."

Wrong. The MH problem requires the assumption that MH always follows the same rules. That is he always opens a goat door after the initial pick.

But you don't get to choose whether to switch until after he's opened a door. If he opens the door with the car, you lose whether you switch or not; if he opens the door with a goat, it pays to switch... what am I missing?

If MH opens one of the remaining two doors at random and it happens to be a goat then it doesn't matter what you do. You have a 50% chance of winning if you stick and a 50% chance of winning if you switch. The switching advantage only occurs when MH deliberately opens a goat door.

 

[TheGoldcountry]

delete

 

[Rolfe]

But you don't get to choose whether to switch until after he's opened a door. If he opens the door with the car, you lose whether you switch or not; if he opens the door with a goat, it pays to switch... what am I missing?

If he opens the door with the car, you've lost, game over.

It's about him having a regular strategy. If he is deliberately avoiding the car you should switch. If he isn't there's no advantage to switching, and if he isn't some of these permutations will be occasions when he reveals the car, and you lose right there.

That's what I meant when I said the answer depends on what Monty is thinking. Which is a weird concept.

Rolfe.

 

[Rolfe]

There's one strategy where you should stick. That is if the host only opens a second door when you've initially picked the car as an attempt to trick you into losing the prize. This seems obvious and, of course, asking the best course of action in that situation would be trivial. However, some people seem to get this wrong.

"How do you know he's not playing a trick on you, using reverse psychology to try to get you to pick a goat?"

"I wouldn't really care, I mean my answer's based on statistics."

Your example is one I would include in the third category, that is choosing his strategy to mess with your head. It makes the whole question meaningless.

Wrong. The MH problem requires the assumption that MH always follows the same rules. That is he always opens a goat door after the initial pick.

If MH opens one of the remaining two doors at random and it happens to be a goat then it doesn't matter what you do. You have a 50% chance of winning if you stick and a 50% chance of winning if you switch. The switching advantage only occurs when MH deliberately opens a goat door.

I thought that was what I said.

Rolfe.

 

[Meed]

I thought that was what I said.

Rolfe.

It is. Those two paragraphs from me are in response to the scene from the film 21 and dlorde, respectively. I was reiterating what you said to dlorde, because he/she seemed to be asking for clarification. Perhaps I just repeated what you said in the same way, though, which wouldn't really help anything.

 

[Rolfe]

The more ways that is explained the better, for sure. Don't knock it.

I'm still intrigued by the way the right answer depends on the intent of the host. We're seeing only one iteration, but the answer depends on the system he's using to decide which door he's going to open.

In order to demonstrate the solution you have to do one of two things. One way is to perform numerous repetitions of the puzzle, to show whether switching or not switching gives you the larger number of wins. In this case it soon becomes obvious whether or not Monty is avoiding the door with the car, depending on whether he ever actually shows the car - in other words, it depends on the parameters programmed into the simulation. The other way is to do what I did, as I didn't have a computer. I did a reductio ad absurdum with 100 doors, 99 goats and a car. By the time you're down to two doors remaining, and 98 goats have been revealed, you know that there is only a very small chance he's not deliberately avoiding the car. You realise that if he wasn't, the car would almost certainly have been revealed long since. Again, that it was your choice to have him behave in that way.

The thing that did my head in is that I was arguing about this with a senior lecturer in mathematics and computing, who constantly shouted at me that I was talking nonsense and of course there was no advantage to switching, and then when I suggested it depended on the host's intention, he declared I was barking mad. This all shook my confidence in my own deductions so that I had to check my working quite a few times. In the end I more or less muttered "eppur si muove" under my breath and dropped the subject.

Rolfe.

 

[Meed]

I'm still intrigued by the way the right answer depends on the intent of the host. We're seeing only one iteration, but the answer depends on the system he's using to decide which door he's going to open.

It is rather counter-intuitive that the intent matters. Some people in this thread have insisted (and for all I know continue to insist) that the "Monty randomly selects between the two remaining doors and happens to reveal a goat" situation has the exact same probabilities for switching and sticking that the standard Monty Hall problem has. This despite the fact that I've demonstrated the difference both with Bayes' formula and with a simulation. Others have seemed to think that the difference is real but somehow meaningless for reasons I can't understand (it had something to do with the fact that my simulation didn't incorporate the trials in which Monty revealed the car into the win/loss results for switching and sticking).

The thing that did my head in is that I was arguing about this with a senior lecturer in mathematics and computing, who constantly shouted at me that I was talking nonsense and of course there was no advantage to switching, and then when I suggested it depended on the host's intention, he declared I was barking mad. This all shook my confidence in my own deductions so that I had to check my working quite a few times. In the end I more or less muttered "eppur si muove" under my breath and dropped the subject.

Geez, you'd think a math person would get it. Next time ask a statistics person. We know intent matters.

 

[Rolfe]

It's one of those things I got my head round so long ago I gave up discussing it. Never ceases to amaze me how often it's resurrected though.

It's one of my intellectual boasts, though. Figured out what Peregrinus expectavi pedes meos in cymbalis means, and figured out the Monty Hall problem unaided.

Rolfe.

 

[jiggeryqua]

I freely admit that I have not the time to read the entire thread and apologize in advance if this has already been shared.

The way I was able to wrap my head around the Monty Hall problem was to put together a quick and dirty Monte Carlo simulation in Excel, which provides tangible and convincing solution. There are also some online simulators available, as below, which some might find helpful.

http://www.grand-illusions.com/simulator/montysim.htm

Played your link 10 times, got 7 goats and 3 cars....

If someone wants to scroll up (and click back) there's an authorative (or so I supposed) statement on how often you have to play the game to prove it by the practical method...

ETA: 12 goats, 3 cars
EATA: 20 goats, 3 cars - is your link crooked??
EOATA:30 goats, 3 cars. It did say, when I first clicked the link, that it might not work on my browser (chrome) and I should use IE4 (yeah, right). But that should just mean it failed to work at all, not that it would automatically give you goats in Chrome? Come on, I've only just been convinced that of course you should change, and now it's 10/1 in favour of sticking?