Monty Hall Problem... For Newbies

[Stamuel]

Wrong. Monty's knowledge is irrelevant.





If he opens the door and reveals the car. then:





-- Game over. You lose. Enjoy your goat meat tacos.





The puzzle is only applicable if he reveals a goat. It is the act of revealing the goat that is the key.

It does matter whether Monty is bound to open a goat door, which requires him to know where the car is. There is a proof using Bayes' Theorem earlier in the thread. It shows that if Monty randomly picks which remaining door to open and a goat is revealed, then the probability that the car is behind your original door is 1/2.



Care for a "Here's how to think of it" speech? Ok, then. Here's how to think of it: your initial pick is basically random, in the sense that the location of the car is in no way influenced by which door you pick. Then Monty randomly picks another door to keep closed. Essentially, you end up with two doors picked at random. Then you find out that a goat is behind the other door, which means one of the two left has a goat, and one has a car. Since both were picked at random before the goat was revealed, there is equal probability to find the goat behind either one. So, if you switch it's 1/2 and if you stay it's 1/2. This is why the famous version of the problem requires Monty to know where the car is and to commit to showing a goat.

 

[ynot]

Just for everyone's clarity, we're talking about the difference between "it is known that at least one of the coins landed heads" and "one of the coins is known to have landed heads".
Let's call one coin the blue coin and one the orange coin. "One of the coins is known to have landed heads" would mean (as I interpret it at least) either that we know the blue coin came up heads or that we know the orange coin came up heads. In either case we can remove 1/3 from the Venn diagram and it would become 1/2 and 1/2.

We don’t need to identify the individual coins when the question asked doesn’t refer to a particular coin.

In the "it is known that at least one of the coins landed heads" scenario, we can't look at either coin individually and know how it landed, so we can't get rid of any of the portions from the Venn diagram.

Not true. Since when have we not been able to look at an individual coin and not know how it landed? The looker can look at only one coin and if it’s a head can say “Yes, at least one of the coins landed heads”. If the looker had to look at two coins to establish that one was heads - so what? The asker didn’t ask for that information and doesn’t need it to get the answer to the question asked.

 

[ynot]

ynot, recall that JohnnyG simulated this and confirmed that it is 1/3. You might want to try a simulation of your own.

I think JonnjyG’s simulation was based on the assumption that the question asked relates to the overall odds of a multiple toss. Whether this assumption is true or not is what’s being debated.

 

[ynot]

The fact that two coins were tossed and statement that at least one of the coins was heads (i.e. not specifying which coin) makes it a multiple toss calculation.

I say it’s an individual result from multiple individual tosses. The result is empirical and no calculation is required.

 

[Paul C. Anagnostopoulos]

Largely unargued. But... Monty doesn't offer you both doors, and I can only speak from my conceptual framework, but claiming he's offering you both the goat and the (2/3 car)-door doesn't wash. He's also, at the same time and equally ('50/50') offering you your door plus the goat door. You can change to the other door and see a goat, or keep your door and see a goat. 50/50, right? Wrong - so, with the best will in the world, the explanation must be wrong. Right?

If Monty offered you both doors, certainly you would switch, because you know there is a 2/3 chance that the car is behind one of them.

But you are only allowed to pick one door.

So Monty opens one of the other doors that contains a goat. There is still a 2/3 chance that the car is behind one of the two doors, but now he has given you the ability to take advantage of the 2/3 chance while only picking one door. So you switch.

Does that make sense? If so, how does it differ from your description above?

~~ Paul

 

[ynot]

The fact that "at least one" would make no sense in reference to a set of less than two? "At least one" means "one or more than one".

A set of less than two would make no sense. The question only asks “at least one” not also “more than one”. If the question was interested in “more than one” it would ‘ve been something like “Was one or both tosses heads?”.

 

[Alferd_Packer]

The rules of the game state that Monty MUST reveal a GOAT. NOT a CAR. The game is that the contestant gets two chance at guessing where the car is. If Monty reveals the car before the contestant has had a chance to stick or change the first guess then the game is ruined.

so the game is ruined. so what?

the question is this: if a door is opened, and revealed a goat, what happens to the odds on the remaining two doors?


If Monty didn't know and just guessed and happened to reveal the goat, the odds of the car being behind each of the remaining doors are exactly the same as they would be if he did know.

 

[lionking]

If Monty didn't know and just guessed and happened to reveal the goat, the odds of the car being behind each of the remaining doors are exactly the same as they would be if he did know.

Sure, but you are now talking about a different problem. In the MH problem he always picks a door with a goat because he knows what's behind each door. As ynot said, you are effectively given the choice of two doors rather than one.

 

[ynot]

Why don’t people just play the game as defined by the rules on the box?

 

[ynot]

If Monty offered you both doors, certainly you would switch, because you know there is a 2/3 chance that the car is behind one of them.

But you are only allowed to pick one door.

So Monty opens one of the other doors that contains a goat. There is still a 2/3 chance that the car is behind one of the two doors, but now he has given you the ability to take advantage of the 2/3 chance while only picking one door. So you switch.

Does that make sense? If so, how does it differ from your description above?

~~ Paul

Monty giving you the goat door is essentially the same as Monty offering you both doors. Realising that fact is a good way of highlighting the advantage of switching.

 

[boooeee]

so the game is ruined. so what?

the question is this: if a door is opened, and revealed a goat, what happens to the odds on the remaining two doors?


If Monty didn't know and just guessed and happened to reveal the goat, the odds of the car being behind each of the remaining doors are exactly the same as they would be if he did know.

As an illustration why Monty's knowledge is relevant. What if he operated under the following rule: He always shows the car if you didn't pick it, otherwise he shows a goat.

Now it's time to play the game. You make your pick and then Monty shows you a goat. Do you still think that the probabilities are 1/3 for your door and 2/3 for the remaining door?

If not, then you need to explain why the other two "Monty scenarios" lead to the same probabilities, but this one does not.

 

[Brian-M]

It doesn't matter which coin is heads, as long as one of them is. The probability is then dependent entirely on the other coin. 50/50 heads or tails.

There are two possible scenarios here, each with different odds. The description of what's going on is ambiguous, it could apply for either situation. The answers people give going to depend on which scenario they assume is being talked about.

Let's say X represents either heads or tails.

Scenario A:

You pick a value for X at random
You flip a pair of coins until at least one coin lands X
You then say "At least one coin is X"
The odds of both coins being X is 1/3​

Scenario B:

You flip a pair of coins.
You set X as the value of an arbitrarily selected coin
You then say "At least one coin is X"
The odds of both coins being X is 1/2​

 

[Stamuel]

Why don’t people just play the game as defined by the rules on the box?

If I understand correctly, Alferd_Packer is claiming that Monty's knowledge is an irrelevant detail that is included merely to make the problem more colorful. In essence his claim is that objecting "But Monty knows where the car is in the original Monty Hall problem!" is the same as objecting to someone who calls the host Morgan Henry. The host's name is completely irrelevant, and objecting to using the wrong name would be absolutely ridiculous, pathetic and stupid. Alferd thinks that Monty's knowledge in the classic problem is just like Monty's name, totally irrelevant. As has been shown, this is not true. Monty's knowledge actually makes a difference when it comes to the probabilities.

 

[Meed]

Not true. Since when have we not been able to look at an individual coin and not know how it landed?

I said that we can't look at a coin and know how it landed, not we can't look at a coin and not know how it landed.

The looker can look at only one coin and if it’s a head can say “Yes, at least one of the coins landed heads”. If the looker had to look at two coins to establish that one was heads - so what? The asker didn’t ask for that information and doesn’t need it to get the answer to the question asked.

It doesn't matter how many coins the looker looks at, so long as he is able to accurately assess whether at least one of the two coin tosses came up heads.

I don't understand what part you are disagreeing with.

I think JonnjyG’s simulation was based on the assumption that the question asked relates to the overall odds of a multiple toss. Whether this assumption is true or not is what’s being debated.

It seemed like Johnny's simulation represented the question perfectly accurately to me.

There is no assumption necessary. It's in the wording of the question that it's about a multiple toss (e.g. "at least one", "both", "two").

A set of less than two would make no sense. The question only asks “at least one” not also “more than one”. If the question was interested in “more than one” it would ‘ve been something like “Was one or both tosses heads?”.

"At least one" means "one or more than one".

 

[ynot]

s
the question is this: if a door is opened, and revealed a goat, what happens to the odds on the remaining two doors?

The odds of the first choice door don’t change but the odds of the door hiding the car does as it become the only alternate choice as opposed to previously being one of two alternate choices.

 

[Meed]

There are two possible scenarios here, each with different odds. The description of what's going on is ambiguous, it could apply for either situation. The answers people give going to depend on which scenario they assume is being talked about.

Let's say X represents either heads or tails.

Scenario A:

You pick a value for X at random
You flip a pair of coins until at least one coin lands X
You then say "At least one coin is X"
The odds of both coins being X is 1/3​

Scenario B:

You flip a pair of coins.
You set X as the value of an arbitrarily selected coin
You then say "At least one coin is X"
The odds of both coins being X is 1/2​

That is true. I have basically been assuming a random sample from the set of double tosses in which at least one is H.

 

[This is The End]

There's a difference between "it is known that at least one of the coins landed heads" and "one of the coins is known to have landed heads".

In the former the odds of 2 heads is 1/3, in the latter it is 1/2.

Please explain what the difference is (I can't see any).

In the former you do not know which of the 2 coins landed heads. The important phrase there is "at least one"; it is not specifically talking about just one coin. That still leaves open the possibility of either of them being the coin to being known as having landed heads.

In the latter you do know which coin landed heads. The important phrase there is "one of the coins is known"; it is specifically talking about just 1 coin. At that point you are merely asking for the odds of the other coin, which of course is 50/50.

 

[ynot]

I said that we can't look at a coin and know how it landed, not we can't look at a coin and not know how it landed.

Is this better for you? - "Since when have we not been able to look at an individual coin and not (being able to) . . . "know how it landed"?

 

[Meed]

Is this better for you? - "Since when have we not been able to look at an individual coin and not (being able to) . . . "know how it landed"?

?

Again, I didn't say that. You have an extra "not" there.

 

[Meed]

Am I getting trolled again?