Monty Hall Problem... For Newbies

[anglolawyer]

I voted to change. I have not read the thread. Two times out of three you will choose a goat and in those two cases it will be correct to change your choice once the quizmaster opens the other goat door. One time out of three you will choose the car and it will be correct to stay with that choice. That's it. Now I'll read the thread and see if and where I went wrong.

 

[jiggeryqua]

OK, while you are arguing the numbers, there is something being almost completely ignored: that Monty is playing fair with you.

It did come up almost immediately, as an 'important consideration'. I dismissed it as irrelevant to my lack of understanding (which it was) and nobody else who lacks understanding seems to lack it with regard to that point. It is, nevertheless, an important consideration inasmuch as if he only did it when you'd picked the car, then you know where the car is 100% and there's no problem. So yes, it's important that it is a problem for it to be a problem.

 

[wollery]

I voted to change. I have not read the thread. Two times out of three you will choose a goat and in those two cases it will be correct to change your choice once the quizmaster opens the other goat door. One time out of three you will choose the car and it will be correct to stay with that choice. That's it. Now I'll read the thread and see if and where I went wrong.

You aren't wrong. That's pretty much the most succinct correct description I've seen yet.

 

[pgwenthold]

It did come up almost immediately, as an 'important consideration'. I dismissed it as irrelevant to my lack of understanding (which it was) and nobody else who lacks understanding seems to lack it with regard to that point

Actually, this was a MAJOR source of contention in the original discussion by Marilyn. Many people pointed out that you cannot answer the question without that information. The idea that "Monty's motives" matter was extremely controversial.

And it is a fair criticism. As stated in the OP in this thread, even, the question is unanswerable. You need to make some assumption about the rules of the game in order to derive an answer.

To me, THAT is the real trick to the Monty Hall problem. The math is trivial, and if you can't handle, do the simulation. The real question is about how you interpret the information that is given.

The fact of the matter is that just playing the game doesn't tell you enough about the rules to make this kind of decision. It's more of a cognitive psychology question of how we let biases affect decision making.

 

[GlennB]

Actually, this was a MAJOR source of contention in the original discussion by Marilyn. Many people pointed out that you cannot answer the question without that information. The idea that "Monty's motives" matter was extremely controversial.

Yup. In the original controvery the MH puzzle was poorly phrased. In the o/p here it was set out nicely.

Yet still it leads to hundreds of posts It's a cracker of a puzzle.

eta:

As stated in the OP in this thread, even, the question is unanswerable.

Apologies, I didn't spot this. The o/p avoids this, describing the situation perfectly afaics.

 

[The Greater Fool]

It did come up almost immediately, as an 'important consideration'. I dismissed it as irrelevant to my lack of understanding (which it was) and nobody else who lacks understanding seems to lack it with regard to that point. It is, nevertheless, an important consideration inasmuch as if he only did it when you'd picked the car, then you know where the car is 100% and there's no problem. So yes, it's important that it is a problem for it to be a problem.

You aren't wrong. That's pretty much the most succinct correct description I've seen yet.

+1

 

[hgc]

Actually, this was a MAJOR source of contention in the original discussion by Marilyn. Many people pointed out that you cannot answer the question without that information. The idea that "Monty's motives" matter was extremely controversial.

And it is a fair criticism. As stated in the OP in this thread, even, the question is unanswerable. You need to make some assumption about the rules of the game in order to derive an answer.

To me, THAT is the real trick to the Monty Hall problem. The math is trivial, and if you can't handle, do the simulation. The real question is about how you interpret the information that is given.

The fact of the matter is that just playing the game doesn't tell you enough about the rules to make this kind of decision. It's more of a cognitive psychology question of how we let biases affect decision making.

I don't think this is so complicated. Here's the original description from this thread:

You're on a game-show trying to win a new car. The car is behind one of three doors. Behind the other two doors are goats.

You pick one of the three doors at random and the presenter (who knows which door the car is behind) opens a door that you didn't pick, revealing a goat. (This is standard procedure for the show.)

The presenter then offers you a chance to change your mind and pick the remaining door instead.

It is unambiguous that the presenter knows where the car is and always reveals a goat.

The only choice for the presenter is which goat door he picks if the contestant's first pick is the car. And that makes no difference and might as well be random. In other words, Robot Monty could serve as presenter, and there is no psychology involved whatsoever.

What assumptions are necessary?

 

[Dr. Keith]

Don't see it there are two doors, you have a 50/50 choice. It's a new game after he opens the first door.

I'll be in London this summer. If you want to meet we can play 100 games for a pound per game until you start thinking more critically about it. 1000 games if you like. My greed knows no bounds in the pursuit of knowledge.

 

[jsfisher]

I'll be in London this summer. If you want to meet we can play 100 games for a pound per game until you start thinking more critically about it. 1000 games if you like. My greed knows no bounds in the pursuit of knowledge.

You'll give him attractive odds, right? How about 60/40 so he'll have a distinct advantage.

It would be only fair.

 

[jiggeryqua]

I'll be in London this summer. If you want to meet we can play 100 games for a pound per game until you start thinking more critically about it. 1000 games if you like. My greed knows no bounds in the pursuit of knowledge.

IANAM but...there were posts earlier suggesting doubters should run a few simulations. I've read Rosencrantz and Guildernstern but I nearly flunked the stats module of my degree, so: how many simulations would you have to run to be sure you weren't just getting a run of flukes? Is it more if you start with a run of 3 or 5 or more change=lose results?

ETA: Offering up cash prizes is a strong persuader - we all know casinos keep an edge, so this new game must have one too. But it's not really a convincer. We know you think it's true, the money merely speaks to your confidence, not the facts of the matter.

ETAA: I'll take 100 games of 'Maybe win a car' for £1 a go. I may still be a little hazy around probabilities, but I have to win once at least, right? Boy oh boy, I hope it's a Bentley.

 

[GlennB]

ETA: Offering up cash prizes is a strong persuader - we all know casinos keep an edge, so this new game must have one too. But it's not really a convincer. We know you think it's true, the money merely speaks to your confidence, not the facts of the matter.

2:1 odds is huge in a series of 100. It's like we roll a die 100 times at £1 a pop, I win if it's 1-4, you win if it's 5-6. It would be freakish if you came out winning.

 

[jiggeryqua]

2:1 odds is huge in a series of 100. It's like we roll a die 100 times at £1 a pop, I win if it's 1-4, you win if it's 5-6. It would be freakish if you came out winning.

Yes. Again, not an idiot. Now, not even conceptually blocked from understanding the MH Problem, as demonstrated. I get that it's strongly persuasive to put your money where your mouth is, but it doesn't convincingly explain why you're right. So you see, we're both right. Let's make some space for people who still don't know what's going on.

 

[Dr. Keith]

IANAM but...there were posts earlier suggesting doubters should run a few simulations. I've read Rosencrantz and Guildernstern but I nearly flunked the stats module of my degree, so: how many simulations would you have to run to be sure you weren't just getting a run of flukes? Is it more if you start with a run of 3 or 5 or more change=lose results?

ETA: Offering up cash prizes is a strong persuader - we all know casinos keep an edge, so this new game must have one too. But it's not really a convincer. We know you think it's true, the money merely speaks to your confidence, not the facts of the matter.

ETAA: I'll take 100 games of 'Maybe win a car' for £1 a go. I may still be a little hazy around probabilities, but I have to win once at least, right? Boy oh boy, I hope it's a Bentley.

I wasn't going to play for a car. Each show up with 100 individual pound notes.

For each game two pound notes are placed together at random by a disinterested third party behind one of three "doors".

The 50/50 believer picks one door. Disinterested third party opens one empty door and I, the changer, get the third door.

Repeat 100 times and I am pretty sure I will walk away with close to 133 pounds. If after 100 trials it is within five pounds of 50/50 I'll give the 50/50 believer 30 pounds. So we both have the same potential upside.

If you don't believe me, James May did something similar on his show Man Lab. But with well shaken beers.

 

[erwinl]

Because it's correct - they're not 'distinct situations', they're different paths stemming from an individual situation. Taking the location of the car and the door you choose as 'the setup' then there are 9 different setups. Each has an outcome, win or lose. You've described 12 setup/outcome combinations.

A parallel case - I can set off down road A or road B to get to work and these are very different routes, but I like a little variety. Route A leaves me no further choices. Route B forks somewhere and I can take fork X or fork Y, both of which take the same time.

Route A takes 40 minutes, route B 50. I set off down A and B equally often. What's my average journey time? 45, obviously.

Route chosen|Fork chosen|Journey time
A|-|40
B|X|50
B|Y|50
Average| |46.66

Nuts, huh? This has happened because I've taken B-X and B-Y as separate events and given them equal weight to A-

Uhm no. And it's because of the highlighted part.
You choose route A and B equally often (let's call this 50%), but you don't say how often on Route B you take X or Y and it doesn't matter in this case.
They're both the same route B, which you choose equally often as route A. Thus making the average travel time 45 minutes.

If on the other hand you would have said that you take each route equally often then you would be correct. It is then on the listener to recognize that route B equals two different routes, with a total of three route to choose from. and thus making the average travel time 46,667 minutes

 

[boooeee]

Uhm no. And it's because of the highlighted part.
You choose route A and B equally often (let's call this 50%), but you don't say how often on Route B you take X or Y and it doesn't matter in this case.
They're both the same route B, which you choose equally often as route A. Thus making the average travel time 45 minutes.

If on the other hand you would have said that you take each route equally often then you would be correct. It is then on the listener to recognize that route B equals two different routes, with a total of three route to choose from. and thus making the average travel time 46,667 minutes

I think that was GlennB's point (i.e. I don't think you're disagreeing with him).

 

[malicus]

Don't see it there are two doors,

The game started with three doors. You made your pick under those conditions. Your pick has a 1 in 3 chance of being correct.

you have a 50/50 choice.

The game started with three doors. You made your pick under those conditions. Your pick has a 1 in 3 chance of being correct.

It's a new game after he opens the first door.

It is not a new game, the car has not been shuffled behind the two remaining doors. Remember, and it is worth repeating not for snark but because this is an essential piece of information for this problem, the game started with three doors. You made your pick under those conditions. Your pick has a 1 in 3 chance of being correct.

And now the kicker, when you made your initial pick, the odds of the car being behind the other two doors, in total, are 2 out of 3. But dastardly MH only allowed you to pick one door. Picking two doors out of three would have been awesome right?

Let's plug those odds in to the mantra:
The game started with three doors. You made you pick under those conditions. The remaining doors have a 2 out of 3 chance of hiding the car.

Monty is offering you those two cards. Showing you a wrong door, 100% of the time, is pure misdirection. Showmanship.

Because, remember...

.....the game started with three doors, you made your pick under those conditions.

 

[Modified]

The game started with three doors. You made your pick under those conditions. Your pick has a 1 in 3 chance of being correct.

Unless there is new relevant information. That seems to be the confusion for many. Monty knowingly showing you a goat is new information, but it does not tell you anything about the odds that your original guess was correct. You already knew there was a goat behind at least one of those doors.

I think a lot of people will have a hard time intuitively understanding that Monty knowingly showing you a goat does not change the odds that you were originally correct, but him randomly showing you a door which happens to have a goat behind it does change them.

 

[Bindamel]

IANAM but...there were posts earlier suggesting doubters should run a few simulations. I've read Rosencrantz and Guildernstern but I nearly flunked the stats module of my degree, so: how many simulations would you have to run to be sure you weren't just getting a run of flukes? Is it more if you start with a run of 3 or 5 or more change=lose results?

At 30 simulations, there is <5% chance that you will achieve 50/50 by coincidence.

At 50, you're down around 1%

At 80, < 1 / 1000

At 190, you're in the neighborhood of 1 in a million chance of breaking even.

 

[GlennB]

Uhm no. And it's because of the highlighted part.
You choose route A and B equally often (let's call this 50%), but you don't say how often on Route B you take X or Y and it doesn't matter in this case.

Correct, which is exactly what I said. I start down A or B equally often and the possible fork on route B is irrelevant to any analysis of time.

They're both the same route B, which you choose equally often as route A. Thus making the average travel time 45 minutes.

Yes. This is exactly what I said.

Relating it back to the MH puzzle, it's as if choosing fork X or Y on route B makes a difference to the average journey time. It's the same route B and variations on B shouldn't be considered as separate cases in terms of working out average times, which is exactly what Jiggeryqua's table did when turning the 9 possible starting MH scenarios into 12 possible 'courses of action' and lending each occurence of a 'fork' equal weight to each journey that started on A.

You know, I really, really tried to phrase that "route A vs. route B" analogy as precisely as humanly possible. Failed <sob> <extra sob>

eta: I guess you took that 46.77 average time to be my conclusion. My bad - that was meant to be irony, which is in itself a kind of failure. Just because it was clear to me doesn't mean my meaning was clear. I should have stuck a in there

 

[Brian-M]

Why does deciding on one door affect the mathematics? If there's probability to be shifted, why to the door you haven't picked? Your picking it hasn't affected the material world nor, surely, has it affected the mathematics underpinning the material world.

Let's put it this way...

When you pick a door, there's a 2/3 chance that you're wrong.

Even after he reveals the goat there's still a 2/3 chance that you're wrong, because he was always going to reveal a goat no matter what door you chose. It doesn't affect the odds of the car being behind the door you picked in the slightest.

But now there's only one alternate door remaining... only one door the car could possibly be behind if your original choice is wrong.

So if there's a 2/3 chance of the door you picked being the wrong one, then logically there must be a 2/3 chance of the only other unopened door being the right one.

I understand that you'd all like me to shift the whole of the 'spare' probability to the other door, but what distinguishes the two available doors??

Because picking a door constrains Hall's actions. He's not free to open the door you picked, so whenever your door has a goat behind it (as it does 2/3 of the time) there's only ever one door that Hall is allowed to open, and the car must be behind the remaining door.

Whenever a goat is behind your door, the car is always behind the door Hall didn't open. Because a goat is behind your door 2/3 of the time, then the car must also be behind the door Hall didn't open 2/3 of the time.

(Of course, the 1/3 of the time the car is behind the door you picked it doesn't make the slightest difference which of the remaining two doors Hall opens, because the car isn't behind either of them.)