Monty Hall Problem... For Newbies

[jiggeryqua]

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.

Lower case j. More to the point, with the best will in the world, the journey analogy failed in its primary task because ABX and ABY are different journeys. The rationale for combining them is even weaker than in the actual problem.

[Stuff]

Err, yeah. Thanks. RBF already won, a while back, but don't worry about being a late entrant, you wouldn't have won anyway. It didn't address the stated stumbling blocks (thought it may have worked well enough for someone else). Tricky stuff, communication

 

[GlennB]

Lower case j. More to the point, with the best will in the world, the journey analogy failed in its primary task because ABX and ABY are different journeys. The rationale for combining them is even weaker than in the actual problem.

My bolding.

There's no ABX and ABY - I have no idea how you came up with that - just A vs. (BX or BY), where BX and BY are equally likely and their sum = A.

Finally I have to conclude that you're being contrarian for the sheer fun of it. Signing off.

 

[steve s]

Here's a still image from the Mythbusters episode. The person who always switched won 37 out of 49 times. The person who always stayed won about 10 out of 49 (can't tell exactly because part of the board is blocked.)

Still think it's 50/50?

http://i1.ytimg.com/vi/5fSX3brLDjc/maxresdefault.jpg

Steve S

 

[Zax63]

 

[jiggeryqua]

My bolding.

There's no ABX and ABY - I have no idea how you came up with that - just A vs. (BX or BY), where BX and BY are equally likely and their sum = A.

Finally I have to conclude that you're being contrarian for the sheer fun of it. Signing off.

My apologies, I wasn't paying proper attention because it wasn't worth it. There were three routes: A to Whatever, B (with X) to Whatever and B (with Y) to whatever. It's still weaker than just explaining the Monty Hall problem because they are different routes.

While I have been just as interested in the lack of communication skills among mathematicians as in the problem itself, I have never been deliberately contrarian, nor in this case 'deliberately' careless. I just couldn't care that much about an attempt to clarify something by muddying the waters.

 

[boooeee]

My apologies, I wasn't paying proper attention because it wasn't worth it. There were three routes: A to Whatever, B (with X) to Whatever and B (with Y) to whatever. It's still weaker than just explaining the Monty Hall problem because they are different routes.

While I have been just as interested in the lack of communication skills among mathematicians as in the problem itself, I have never been deliberately contrarian, nor in this case 'deliberately' careless. I just couldn't care that much about an attempt to clarify something by muddying the waters.

I'm not sure if I fall in the mathematician camp, but the failure to communicate is going both ways here. I have no idea what point you've been trying to make with your last several posts.

 

[Roboramma]

• Monty opening a door to show a goat (an action that is always available regardless of your choice) cannot affect the result. Whether you picked right or wrong, seeing a goat is always going to happen. Something that is always going to happen cannot be affecting the probabilities?

Right. So, you agree that your first pick has a 1/3 chance of being correct. You also agree that there's a 2/3 chance that the car is behind one of the other doors, yes?

So how does that 2/3 chance that it's behind one of the other doors change when he shows you a goat?

There is still a 2/3 chance that it's behind one of the other doors.

 

[lionking]

Right. So, you agree that your first pick has a 1/3 chance of being correct. You also agree that there's a 2/3 chance that the car is behind one of the other doors, yes?

So how does that 2/3 chance that it's behind one of the other doors change when he shows you a goat?

There is still a 2/3 chance that it's behind one of the other doors.

And that's it. Far too simple an explanation for some, it seems........

 

[This is The End]

I have read the whole thread. I still think it is 50/50

You choose a door he shows you a goat, he is going to show you a goat no matter what is behind your door.
Your next choice will always be 50/50
I think you are all over thinking this.

You're forgetting overall odds which includes repeat trials.

 

[roger]

For those that still doubt this, here is a link to python code which you can execute in the browser that simulates this problem. The code should be readable even for people that don't program.

That is my account I am linking to, I promise there are not viruses.

http://codepad.org/oCYoujgj

 

[Prometheus]

Right. So, you agree that your first pick has a 1/3 chance of being correct. You also agree that there's a 2/3 chance that the car is behind one of the other doors, yes?

So how does that 2/3 chance that it's behind one of the other doors change when he shows you a goat?

There is still a 2/3 chance that it's behind one of the other doors.

Right, so you agree that at the outset, there is a 1/3 chance that it's behind Door A, a 1/3 chance that it's behind Door B, and a 1/3 chance that it's behind door C. You choose A, then Monty shows you a goat behind Door B. So how does that 1/3 chance that it's behind Door B change when he shows you a goat?

There's still a 1/3 chance that it's behind Door B.

EXCEPT... After you see the goat behind Door C, there is now a 1/2 chance that the car is behind Door A, and a 1/2 chance that it's behind Door B, so it doesn't matter whether you switch or stick.

As an individual player, who only gets one shot at this, your odds of becoming one of the contestants who wins the car don't change at any point during the game no matter what you or Monty do. The whole problem, as stated is a red herring.

 

[jsfisher]

Right, so you agree that at the outset, there is a 1/3 chance that it's behind Door A, a 1/3 chance that it's behind Door B, and a 1/3 chance that it's behind door C. You choose A, then Monty shows you a goat behind Door B. So how does that 1/3 chance that it's behind Door B change when he shows you a goat?

There's still a 1/3 chance that it's behind Door B.

EXCEPT... After you see the goat behind Door C, there is now a 1/2 chance that the car is behind Door A, and a 1/2 chance that it's behind Door B, so it doesn't matter whether you switch or stick.

As an individual player, who only gets one shot at this, your odds of becoming one of the contestants who wins the car don't change at any point during the game no matter what you or Monty do. The whole problem, as stated is a red herring.

Wanna bet?

ETA:
It's been said before, but worth repeating: Monty can always open a door with a door behind it, right? So, how does Monty opening one of the non-selected doors give you any information about the door you did select? It doesn't; you door is still a 1/3 chance. Monty didn't provide any information to change that.

The door you picked is a 1/3 chance, and the doors you didn't pick are still a 2/3 chance. However, Monty did give you some information about the doors you didn't pick. By showing you the goat, he eliminated one of the two doors from the 2/3 chance set. Two doors which collectively had a 2/3 chance (1/3 each) now still have a 2/3 collective chance, but the opened door is at 0% and the remaining door, 2/3 for having the car.

!/3 if you stay; 2/3 if you switch.

 

[Roboramma]

Right, so you agree that at the outset, there is a 1/3 chance that it's behind Door A, a 1/3 chance that it's behind Door B, and a 1/3 chance that it's behind door C. You choose A, then Monty shows you a goat behind Door B. So how does that 1/3 chance that it's behind Door B change when he shows you a goat?

There's still a 1/3 chance that it's behind Door B.

How can it be behind door B when he's already shown you a goat behind door B?

I'm sure that was a typo, but it illustrates my point: When you pick door A there's a 1/3 chance that it's behind door A and a 2/3 chance that it's not. At no point does that that change. If you think it changes, please explain how, as Jiggeryqua said, telling you something you already knew (that one of the other doors has a goat) tells you something about whether there is a car behind door A. It doesn't.

Clearly when you see a goat behind door B this means that there's no longer a 1/3 chance that the goat is behind door B. There's 0 chance that it's behind door B. If there is a 2/3 chance that it's behind either door B or door C, and there's 0 chance that it's behind door B, then clearly there's a 2/3 chance that it's behind door C.

Please note: how is this different from if I were to make the same argument with regards to doors A and B? Isn't it also true that there's a 2/3 chance that it's behind doors A and B, and then, it would seem, that we could similarly say that there's now a 2/3 chance that it's behind door A? No. Because Monty couldn't have chosen to open door A, which is why when he opens door B it tells us nothing about A.

 

[lionking]

Right, so you agree that at the outset, there is a 1/3 chance that it's behind Door A, a 1/3 chance that it's behind Door B, and a 1/3 chance that it's behind door C. You choose A, then Monty shows you a goat behind Door B. So how does that 1/3 chance that it's behind Door B change when he shows you a goat?

There's still a 1/3 chance that it's behind Door B.

EXCEPT... After you see the goat behind Door C, there is now a 1/2 chance that the car is behind Door A, and a 1/2 chance that it's behind Door B, so it doesn't matter whether you switch or stick.

As an individual player, who only gets one shot at this, your odds of becoming one of the contestants who wins the car don't change at any point during the game no matter what you or Monty do. The whole problem, as stated is a red herring.

Not correct. Look at the simulations if you don't believe it.

 

[marting]

It seems to be a fairly simple probability question so I suspect there must be some strong psychological factor at work that can cause so many intelligent people to not see it. Anyone run across any Psych papers exploring this?

 

[boooeee]

Not correct. Look at the simulations if you don't believe it.

Or read anglolawyers post a few times. As others have pointed out, it is about as clear and succinct as you can get.

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.

 

[ehcks]

It seems to be a fairly simple probability question so I suspect there must be some strong psychological factor at work that can cause so many intelligent people to not see it. Anyone run across any Psych papers exploring this?

People hate second-guessing themselves.

People also hate other people second-guessing themselves.

 

[bruto]

It seems to be a fairly simple probability question so I suspect there must be some strong psychological factor at work that can cause so many intelligent people to not see it. Anyone run across any Psych papers exploring this?

I've never seen a psychological study, but I think there is a factor at work. When Monty opens the door he is telling us something we already knew, but it seems as if he's providing new information, and when he offers the swap, it refers to the starting choice and its odds, but it seems as if he's offering a new game.

 

[steve s]

I've never seen a psychological study, but I think there is a factor at work.

Dunning-Kruger perhaps?

Steve S

 

[erwinl]

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.



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

Sorry