The Monty Hall problem

[Vorticity]

humber.

Jack said:

Humber, please, just so we can stop going round in pointless circles, is it your view of the 100 doors case that:
If the car is placed randomly, and if you choose a door randomly, then the probability of the car being behind your door is 1/100 and, further, if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

To which you replied:

No. It becomes 1/3, 2/3 ( for the assumed distribution).

What does the "It" refer to in your above statement? Does "It" refer to "the probability of the car being behind your door"?

And yes, you are right. I can neither parse nor understand your posts. A perusal of this thread reveals that I am not alone.

For example, I can neither parse nor understand the following sentence:

What I think he meant, is that if two doors remain, the probability of the remaining closed will not change.

 

[Vorticity]

"Suppose you’re on a game show, and you’re given the choice of
three doors: Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what’s
behind the doors, opens another door, say No. 3, which has a
goat. He then says to you, “Do you want to pick door No. 2?” Is
it to your advantage to switch your choice?"

The accepted answer is swap, because the probability is 2/3 for that door.
Thousands of references you can consult.

Thanks, but I can assure you I'm intimately familiar with the Monty Hall problem.

I am not interested in starting again with the MHP. Just the one point.

What is "the one point"? I've read through this entire thread, and I cannot understand what your one point is.

 

[humber]

What does the "It" refer to in your above statement? Does "It" refer to "the probability of the car being behind your door"?

..then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

No. It becomes 1/3, 2/3 ( for the assumed distribution).

What is "the one point"? I've read through this entire thread, and I cannot understand what your one point is.

OK, we will leave it at that.

 

[Vorticity]

What does the "It" refer to in your above statement? Does "It" refer to "the probability of the car being behind your door"?

..then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

No. It becomes 1/3, 2/3 ( for the assumed distribution).

OK, you've bolded the word "its" in Jack's post, indicating (I assume) that his "it" and your "it" are one. But Jack's "it" referred to the car, so are you saying "The car becomes 1/3, 2/3"?

Since that makes no sense, I still have no idea what you are saying. I can't even tell if you are wrong or not.

Once again, what is "It"?

Fill in the blank in the following sentence with something other than "It":

_______ becomes 1/3, 2/3 (for the assumed distribution).

Thank you.

 

[Dave_46]

How did this get to be all about me?

<snip>

NO. I'm Dave

 

[69dodge]

humber:

You're not explaining yourself well. I have no idea what you're trying to say.

I've read through this entire thread, ...

Have you read post #221?

 

[humber]

OK, you've bolded the word "its" in Jack's post, indicating (I assume) that his "it" and your "it" are one. But Jack's "it" referred to the car, so are you saying "The car becomes 1/3, 2/3"?

Since that makes no sense, I still have no idea what you are saying. I can't even tell if you are wrong or not.

if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

Obviously that one, the above passage, is a bit inscrutable. An enigma.

With the equal distribution you never need to know what the odds are, to improve them.

So easy. We will use the given distribution of 1/3

Probability of car behind contestant's doors p = 1/3
Probability of car behind the remaining closed door 1 - p = 2/3.
That is it.

1 - p is justified because of the Monty's actions. They are fixed, so determined. The only action that is uncontrolled, is the initial action which determines the value p.

ETA:
Is that not clear? To anyone? Trivial. Much ado about nothing. "intuition"...what? A brainteaser?

 

[Vorticity]

Have you read post #221?

I did.

But everyone says that about everyone on the internet, and I've seen far more bizarre statements than humber's asserted with what I am convinced was perfect sincerity...

...but who knows.

ETA:

No, 69dodge is clearly right. See below.

 

[Vorticity]

OK, I think I just understood what you're saying.

You really are trolling, aren't you?

So easy. We will use the given distribution of 1/3

Now I see why you keep saying "For the assumed distribution".

You're assuming that the...

Probability of car behind contestant's doors p = 1/3

...in all cases, no matter how many doors we start out with. In which case...

Probability of car behind the remaining closed door 1 - p = 2/3.

...is trivially true! Even in the 100 doors case!

You've managed to find a way to argue something which is both technically true and utterly unintelligible.

Well played, sir. I salute you. May your bridges be spacious, and your billy goats succulent.

 

[Jack by the hedge]

No. It becomes 1/3, 2/3 ( for the assumed distribution).
That only goes to Savant's support for her argument, as does the matter of if Monty knows or not. That last one is definitely false.
It works for 3 doors. Opening doors until you get there, says nothing other than to support the mistaken interpretation.

The matter of "remaining the same" is about Erdos' objection. I have been unable to find out what he actually meant. Maybe it's subtle, but whatever is was, it has been drowned out by accusations that he is a stubborn bonehead. But, it looks to me that the simulations that are said to have silenced him, are not what they appear to be. That is, he was quite likely to have been given an explanation, that itself was a product of the programme.

What I think he meant, is that if two doors remain, the probability of the remaining closed will not change. He is right, because it doesn't.
It does not switch from being 1/3 when all doors are closed, to 2/3 when the goat is revealed. I can see how he sees that as violating some fundamental ideas of probability.
But, if you say the probability of the chosen door is p, and the remaining (unchosen) closed door is 1- p, there is nothing to discuss, because it always was that way, and the reason can be easily inferred. No question, of course!
Monty is deterministic, so it is not really a matter or probability, but a game in the "game theory" sense.
Conditional probability is IF then, and not WILL then.
Try again
Wiki
Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the (conditional) probability of A, given B" or "the probability of A under the condition B". When in a random experiment the event B is known to have occurred, the possible outcomes of the experiment are reduced to B, and hence the probability of the occurrence of A is changed from the unconditional probability into the conditional probability given B. It can't be said that the probability that something is certain to happen is anything but 1. Writing it as 1 -p makes that clear.

I have highlighted all the parts whose meaning is clear. The initial "No" is interesting. I suspect the following sentence is supposed to convey your belief that if Monty opens 98 doors revealing goats, you think you have a 1/3 or 2/3 chance of winning with the two remaining doors. That's pretty shocking for someone who claims he understands this simple problem.

You can see, I suppose, that for the first choice door to be the winner it must start with the car behind it. And if you say it ends up with a 1/3 chance of winning it must be the winning option in 1/3 of games on average. So it must be the case that 1/3 of the time the first choice door started with the car behind it. So you are claiming that when playing a game with 100 doors, the player will typically guess the right door first time one-out-of-three times. Can you see how absurd that is? 100 doors to choose from and the player guesses right 1/3 of the time.

If you are not already convinced you are wrong, try it yourself with 52 cards: Shuffle and pick a card. Can you get the ace of spades one time in every three? Why not?

 

[Vorticity]

Jack, humber really is trolling. See my previous post.

This is what he has actually been saying all along:

Suppose the probability that the car is behind a given door is not uniform. Say it's 1/3 for one particular door, with the remaining 2/3 of probability distributed somehow amongst the remaining N-1 doors. If my choice of door just happens to coincide with the "special 1/3 door", then it follows trivially that my stick and switch probabilities of winning are indeed 1/3 and 2/3, respectively.

This is why he keeps saying "for the assumed distribution." It's a play on words. In the original three-door problem, the "assumed distribution" is that the picked door has a probability 1/3 of winning (as do all the other doors). See?

What's that you say? It's an absurd and confusing variation on the 100 doors version of the MHP?

Yes it is. That's what makes it clear that this is a classic deliberate troll: Make a statement that is trivially true, but explain it poorly.

 

[Jack by the hedge]

if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?

Obviously that one, the above passage, is a bit inscrutable. An enigma.

You are trolling. Obviously it means:

if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of the car being behind the other door is 99/100?

The meanest intelligence could follow the meaning of the question, and your response to it was "No".

With the equal distribution you never need to know what the odds are, to improve them.

So easy.

Does that mean that you actually meant "yes" when you said "no"? Or does it mean you have no idea what the odds are?

We will use the given distribution of 1/3

That is a different game.

 

[humber]

You've managed to find a way to argue something which is both technically true and utterly unintelligible.

It may appear unintelligible, if following the subject of sentence beyond a few words is a difficulty.

It seems that is the simplest analysis to date. Three lines.
Monty can be expressed with Boolean logic if need be. A doddle.

 

[humber]

You are trolling. Obviously it means:

if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of the car being behind the other door is 99/100?

I said "no". ( the tacit part being..."it does not mean that")

 

[humber]

Jack, humber really is trolling. See my previous post.

This is what he has actually been saying all along:

Suppose the probability that the car is behind a given door is not uniform. Say it's 1/3 for one particular door, with the remaining 2/3 of probability distributed somehow amongst the remaining N-1 doors. If my choice of door just happens to coincide with the "special 1/3 door", then it follows trivially that my stick and switch probabilities of winning are indeed 1/3 and 2/3, respectively.

This is why he keeps saying "for the assumed distribution." It's a play on words. In the original three-door problem, the "assumed distribution" is that the picked door has a probability 1/3 of winning (as do all the other doors). See?

What's that you say? It's an absurd and confusing variation on the 100 doors version of the MHP?

Yes it is. That's what makes it clear that this is a classic deliberate troll: Make a statement that is trivially true, but explain it poorly.

You missed.
It is not trivial, but succinct. None of the window dressing of the other attempts, no Bayes..nothing. Covers all occasions.
Easy and clear.

 

[Jack by the hedge]

I said "no". ( the tacit part being..."it does not mean that")

Why then consider this, humber;

I asked if you agreed that: "If the car is placed randomly, and if you choose a door randomly, then the probability of the car being behind your door is 1/100 and, further, if Monty, knowing where the car is, opens 98 other doors which do not reveal the car, then the probability of the car being behind your door remains 1/100 throughout and the probability of its being behind the other door is 99/100?"

You replied: "No". And you are wrong.

Just because you wanted to answer a different question does not make your wrong anwer right. I imagine you found that out at school. Eventually.

 

[GlennB]

Jack, humber really is trolling. See my previous post.

I would respectfully disagree.

humber is not trolling, humber is determined to find some way - any way - to be wrong. Quite a difference

 

[humber]

I have highlighted all the parts whose meaning is clear. The initial "No" is interesting. I suspect the following sentence is supposed to convey your belief that if Monty opens 98 doors revealing goats, you think you have a 1/3 or 2/3 chance of winning with the two remaining doors. That's pretty shocking for someone who claims he understands this simple problem.

You can see, I suppose, that for the first choice door to be the winner it must start with the car behind it. And if you say it ends up with a 1/3 chance of winning it must be the winning option in 1/3 of games on average. So it must be the case that 1/3 of the time the first choice door started with the car behind it. So you are claiming that when playing a game with 100 doors, the player will typically guess the right door first time one-out-of-three times. Can you see how absurd that is? 100 doors to choose from and the player guesses right 1/3 of the time.

If you are not already convinced you are wrong, try it yourself with 52 cards: Shuffle and pick a card. Can you get the ace of spades one time in every three? Why not?

Nothing to it
Monty always shows a goat, but not the car. That allows a simple Boolean statement to describe his actions. He is determined.

The result, for the even distribution of the car amongst the three doors is:

Chosen door has car p= 1/3
The remaining closed door has car p = 1 - 1/3


Generalize for the distribution, p

Chosen door has car = p
The remaining closed door has car = 1 - p

Done.

 

[Vorticity]

I know I'll regret this, but...

You missed.
It is not trivial, but succinct. None of the window dressing of the other attempts, no Bayes..nothing. Covers all occasions.
Easy and clear.

OK, on the off chance you're not trolling:

Do you agree with the following:

1) Start with N doors. If I pick a door, and the probability that my door hides the car happens to be p, then the probability that I'll win if I stick with my door is p and the probability that I'll win if I switch to the one door remaining after Monty has revealed N-2 goats is 1 - p.

2) If we have N = 100 doors, and each has an equal probability of hiding the car, then each door has a p = 1/N = 1/100 probability of hiding the car.

3) If I pick one of the doors, then my door has a p = 1/100 probability of hiding the car.

4) Combining 1 and 3: If I stick with my door, I have a p = 1/100 probability of getting the car. If I switch to the one door remaining after Monty has revealed 98 goats, I have a 1 - p = 1 - 1/100 = 99/100 probability of getting the car.

Which, if any, of these do you disagree with?

 

[Vorticity]

The result, for the even distribution of the car amongst the three doors is:

Chosen door has car p= 1/3
The remaining closed door has car p = 1 - 1/3

Correct.

What about for the even distribution amongst 100 doors?