Monty Hall Problem... For Newbies

[Meed]

Yes. From before you make your initial pick, we know that the car will wind up either behind the door you choose, or the one Monty doesn't open.

Agreed.

50/50

It's not 50/50. I still don't get why you think it is.

No. Each individual contestant only gets to play once. They don't get 100 trials each so you cannot say that an individual has 99 chances out of 100. Each individual, is only ever going to have 2 doors to choose from, their initial pick, and the one Monty offers them a chance to switch to.

I agree with Roboramma's response to this.

For the same reason that people who play the lottery once don't have a 1/75k chance of winning, but people who play every week do.

My point is that Monty's game is not like the lottery.

If a large population of people each play the lottery once, then 1/175M of them will win.
If one person plays the lottery once then they have a 1/175M shot at winning.
If a large population of people each play the (100 door) Monty Hall game once then 99/100 of them will win.
If one person plays the (100 door) Monty Hall game once then they have a 99/100 at winning.

 

[AdMan]

It's not 50/50. I still don't get why you think it is.

I don't think he does; I think he's just trolling.

It's not as if this is a new problem that hasn't been discussed and analyzed to death here and in many other places for decades.

 

[fromdownunder]

Going out on a limb here, there are two different issues, which are resolved with these scenarios.

Once there are only two doors left, the odds of the car being behind either door are in fact 50/50.

However the odds that the contestant picked the correct door in the first place are completely different and depend totally on how many doors you started with. and can range from 33% (with three doors) 25% with four doors, 20% with five doors...to 1% with 100 doors.

Norm

 

[Modified]

Once there are only two doors left, the odds of the car being behind either door are in fact 50/50.

Only if Monty chose the excluded door from among all three. If he chose it from any random set of two, then the odds are 1/3 for the door he didn't consider and 2/3 for the one of two that he didn't exclude.

We don't need a contestant to choose the first door, it could be chosen at random. As long as Monty was forced to choose the excluded door from a random set of two rather than all three, the odds are not 50/50.

 

[ynot]

Once there are only two doors left, the odds of the car being behind either door are in fact 50/50.

So if you bought a ticket in a lottery with 10 million tickets and, after the draw and before you checked your ticket, the person running the lottery made you the offer of allowing you to keep the ticket you bought or swapping it with a ticket that definitely won if your ticket didn’t, you think the odds of you having the winning ticket or not would be 50/50?

If this offer was made to all ticket holders only one would lose. Without the offer all but one would lose. Switching is a no-brainer.

Two choices don't make the odds 50/50 if one of those choices represents an earlier choice that was made when there were more than two choices.

 

[Prometheus]

I don't think he does; I think he's just trolling.

It's not as if this is a new problem that hasn't been discussed and analyzed to death here and in many other places for decades.

(sorry cornsail, et al -- I'll be good now)

 

[jt512]

Going out on a limb here, there are two different issues, which are resolved with these scenarios.

Once there are only two doors left, the odds of the car being behind either door are in fact 50/50.

That limb just broke.

 

[fromdownunder]

However the odds that the contestant picked the correct door in the first place are completely different and depend totally on how many doors you started with. and can range from 33% (with three doors) 25% with four doors, 20% with five doors...to 1% with 100 doors.

Norm

Why is it that everybody who responded chose to disregard this part of my post? I really would like to know if you are ignoring what people actually say, by ignoring what somebody else actually said.

I will say it again, and put it as simply as possible. With two doors remaining the chance of the car being behind either door is 50/50, and repeat what I said above.

the odds that the contestant picked the correct door in the first place are completely different . I even bolded it so that some people who tend to miss context might see it and actually read what I said

What is it about that statement that you don't understand?

Norm

 

[fromdownunder]

Two choices don't make the odds 50/50 if one of those choices represents an earlier choice that was made when there were more than two choices.

I never mentioned choices.

How about you re read all of my post instead of jumping in like a bull at a gate, and see what I actually said in the second paragraph. We are talking about two different things.

1. The odds of which door the car contains: With two doors it is 50/50

2. The odds that the contestant chose the correct door. This is, as I clearly stated in my post completely dependent on the number of doors. Do you need me to repeat it. I can if you are not willing to reread my post.

Do people on JREF sometimes read 1/5th of a post then jump in and base conclusions on what was said in the first sentence, and totally ignore context and the rest of the post? It would seem so.

Norm

 

[Daald]

I never mentioned choices.

How about you re read all of my post instead of jumping in like a bull at a gate, and see what I actually said in the second paragraph. We are talking about two different things.

1. The odds of which door the car contains: With two doors it is 50/50

2. The odds that the contestant chose the correct door. This is, as I clearly stated in my post completely dependent on the number of doors. Do you need me to repeat it. I can if you are not willing to reread my post.

Do people on JREF sometimes read 1/5th of a post then jump in and base conclusions on what was said in the first sentence, and totally ignore context and the rest of the post? It would seem so.

Norm

They read it and quoted the part they disagreed with. That is point number 1.

Point number 1 is true only if the events are independent which they are not. That is the whole point of the monty hall problem and the point where most people go wrong.

The points you made are only true if.

Person A comes in and picks 1 door out of three and then leaves.

Person B comes in and picks 1 door out of the remaining 2 and then leaves.

That is not the Monty Hall problem.

Person A persists and the agent exposing wrong choices has knowledge of the system. That changes the odds so you can no longer say when you get to step 2 it is 50/50.

The system basically allows a person to pick twice (with the I will always switch strategy) thus the odds of success are 2/3.

 

[fromdownunder]

Only if Monty chose the excluded door from among all three. If he chose it from any random set of two, then the odds are 1/3 for the door he didn't consider and 2/3 for the one of two that he didn't exclude.

We don't need a contestant to choose the first door, it could be chosen at random. As long as Monty was forced to choose the excluded door from a random set of two rather than all three, the odds are not 50/50.

People seem to be missing or not understanding what I said. So I need to say it again and even more simply. Using three doors with one open, the odds of the car being behind either door to an independent observer is 50/50. The odds that the contestant who picked the first door are 33/67.

Using 100 doors with 98 open, the odds of the car being behind either door to an independent observer is 50/50. The odds that the contestant who picked the first door are 1/100

How is this that so hard to understand?

Norm

 

[fromdownunder]

They read it and quoted the part they disagreed with. That is point number 1.

Point number 1 is true only if the events are independent which they are not. That is the whole point of the monty hall problem and the point where most people go wrong.

The points you made are only true if.

Person A comes in and picks 1 door out of three and then leaves.

Person B comes in and picks 1 door out of the remaining 2 and then leaves.

That is not the Monty Hall problem.

Person A persists and the agent exposing wrong choices has knowledge of the system. That changes the odds so you can no longer say when you get to step 2 it is 50/50.

The system basically allows a person to pick twice (with the I will always switch strategy) thus the odds of success are 2/3.

I agree that my first paragraph was not directly related to the Monty Hall problem in its pure form. I was discussing some of the points argued on this thread, and perhaps I should have quoted parts of the threads, that directly claim a 50/50 argument. The 50/50 is true enough to an independent observer, but does not apply to the contestant, who must always change.

Norm

 

[jt512]

Why is it that everybody who responded chose to disregard this part of my post?

Obviously because that wasn't the part of your post that was wrong.

I will say it again, and put it as simply as possible. With two doors remaining the chance of the car being behind either door is 50/50...

And, no matter how many times you repeat that, it will still be wrong. With two doors remaining, the probability that the car is behind the door that the contestant picked is 1/n, where n is the total number of doors that the contestant had to choose among. Then it is necessarily the case that the probability that the car is behind the remaining door is 1–1/n.

the odds that the contestant picked the correct door in the first place are completely different .

They are completely different only because you were wrong about the odds being 50/50.

I even bolded it so that some people who tend to miss context might see it and actually read what I said

Thanks, but that wasn't necessary. To most of us it was perfectly clear where your error was.

 

[Roboramma]

I agree that my first paragraph was not directly related to the Monty Hall problem in its pure form. I was discussing some of the points argued on this thread, and perhaps I should have quoted parts of the threads, that directly claim a 50/50 argument. The 50/50 is true enough to an independent observer, but does not apply to the contestant, who must always change.

Norm

Wait, you were trying to point out that if someone came in half-way through and only knew that there were two doors and behind one of them is a car, then the odds would be 50/50?

Obviously. Why did you think that was worth pointing out?

 

[fromdownunder]

Wait, you were trying to point out that if someone came in half-way through and only knew that there were two doors and behind one of them is a car, then the odds would be 50/50?

Obviously. Why did you think that was worth pointing out?

Because other people on this thread had already suggested that the final choice was always a 50/50. I was trying to refute this, but apparently my point is invisible to some people.

Norm

 

[fromdownunder]

With two doors remaining, the probability that the car is behind the door that the contestant picked is 1/n, where n is the total number of doors that the contestant had to choose among. Then it is necessarily the case that the probability that the car is behind the remaining door is 1–1/n.

You might want to find anywhere where I said that I even mentioned that the contestant had anything to do with the odds where the final two doors had odds less than 50/50. As I said earlier, people have not even bothered to read my posts for context.[/quote]

they are completely different only because you were wrong about the odds being 50/50.

No. the remaining two doors are 50/50 to anybody.

Thanks, but that wasn't necessary. To most of us it was perfectly clear where your error was.

How is this an error?

However the odds that the contestant picked the correct door in the first place are completely different and depend totally on how many doors you started with. and can range from 33% (with three doors) 25% with four doors, 20% with five doors...to 1% with 100 doors.

Norm

 

[Brian-M]

People seem to be missing or not understanding what I said. So I need to say it again and even more simply. Using three doors with one open, the odds of the car being behind either door to an independent observer is 50/50. The odds that the contestant who picked the first door are 33/67.

I think it would be better described this way...

The contestant picks which door the host cannot open.
The host then (intentionally) opens one door to reveal a goat.

At this point...

The odds of the car being behind the unopenable door are 1 in 3.
The odds of the car being behind the remaining door are 2 in 3.

And...

The odds of an observer who has no idea which door the host wasn't allowed to open guessing which of the two remaining doors has the car behind it is 1 in 2. (Because this observer is effectively just flipping a coin to decide.)

 

[Roboramma]

Because other people on this thread had already suggested that the final choice was always a 50/50. I was trying to refute this, but apparently my point is invisible to some people.

Norm

I see. I can only say that the way that you wrote it didn't seem to me to refute it, so I didn't find it very clear. I understand your point now, though, and if you'd made it clearly it probably would have been a useful way to make clear the mistake of those who think it is 50/50.

 

[Daald]

I was thinking about the 50/50 last night and I came up with another way to look at it.

The 50/50 does actually exist but it is not the odds of the doors.

The 50/50 is the choice you have between the two strategies if you are ignorant of them.

So you have a 50/50 chance to go with "always change strategy" and "never change strategy".

"always change" is 2/3 success rate and "never change" is 1/3. The whole point of the problem is that you should go with the "always change" strategy.

 

[s4zando]

And that's an interesting way to explain a problem by merely inserting the numbers you're trying to justify. My own formula (G+?=G+?) jokingly addressed an earlier attempt to explain the problem by reasoning that when a goat is shown, the 'other' door is now worth more than the other door. I made that deliberately confusing because really, so was the explanation. It didn't account for why two indistinguishable doors are made different by an unconnected, inevitable, irrelevant goat. It merely claims that whichever door you pick, it's more likely to be the other one. Explaining that 'six of one' is the same as 'half a dozen of the other' doesn't help the innumerate.

I am certainly over my head here with the math. From following the thread, I thought it might be beneficial to try to frame an explaination in terms of 'G+?', in such a way that might help to shed light on the solution. It seems most in this thread have a preferred explaination that 'works' for them. I don't expect you to accept what I had proposed, but I thought I would take a stab at it.