Monty Hall revisited, with a twist- survey

[djptolemy]

Im not sure i understand the whole 'switching' thing. So the way i explained the whole process was like this: because there is an unknown on all of the doors. and monty take away one, leaving still two unknowns. However, one of the other 2 doors is wrong as well, meaning it doesnt matter which of the 2 wrong doors he picked, you are being left with one right and one wrong.
With that said, let us for a moment pretend that a 3 sided solid object exists. On each side is a letter, A, B, C. Respectively, if we toss this figure it will land face down on A B or C 1/3 of the time each, in terms of probability. So now let magic Monty remove on of these sides from our 3 sided solid object leaving us a 2 sided solid object. (Let us also pretend we are betting on which letter will land face down and so monty tosses the 3-sided figure into the air and its supposed to land on the side we want) However, the object, as it is in the air, becomes a 2 sided solid object leaving now only 2 choices. My odds went from 1/3 to 1/2 because he picks to remove a side that I did not bet on and for some reason he knows (although i dont think this matters in the case of the door choosing) it wouldnt land on. Maybe a better example is roulette table with 3 numbers. I place my bet on 1, while the thing is spinning a block of wood falls into the third number slot so the ball cannot land in it leaving 2 places for the ball to land in, 1 and 2. Why should I switch to 2? I would have the same odds choosing either. I think the same principal can be applied to a test with a "right answer" or a door with a monster or woman or the choice of transportation between a car and a donkey.
Im open to being totally wrong here though if the probability between a coin toss does not equate to the probablity where there is a "right answer" pre-determined. Although in a sense, once the coin falls there will always have been a right choice and a wrong one, am i correct?
The other possibility here is that this topic is exhausted and nobody wants to talk about it anymore [img]/forums/images/icons/smile.gif[/img]. Hope this helps someone out.

[DPCondit]

djptolemy wrote: Im not sure i understand the whole 'switching' thing. So the way i explained the whole process was like this: because there is an unknown on all of the doors. and monty take away one, leaving still two unknowns. However, one of the other 2 doors is wrong as well, meaning it doesnt matter which of the 2 wrong doors he picked, you are being left with one right and one wrong.

The point is, each choice initially has 1/3 probability, right?

1/3 + 1/3 + 1/3 = 1

If one door is randomly removed each remaining door now has 1/2 probability.

(1/3 - 1/3) + (1/3 + 1/2(1/3)) + (1/3 + 1/2(1/3)) = 0 + 1/2 + 1/2 = 1/2 + 1/2 = 1 , right?

But, if he is only allowed to choose a wrong answer THAT YOU DID NOT PICK, that changes things quite a bit, you must look at it as two different sets, A= what you chose, always worth 1/3, Monty cannot expose your choice, and B = the 2 remaining choices, whereas

A + B = 1, and A = (1/3) and B = (1/3 + 1/3) = (2/3)

Monty must choose only from B, B could be goat-goat, goat-car, or car-goat, therefore assuming no extra information, B must always be worth 2/3, and A must always be worth 1/3. Because HE MUST ALWAYS CHOOSE FROM B, NOT A, A will always be worth 1/3 when an incorrect answer is shown in set B, and therefore the incorrect answer exposed in set B becomes 1/3 - 1/3, and the remaining unexposed answer in set B becomes 1/3 + 1/3. Removing a choice from set B cannot change the probability of your choice (set A), because he could not choose the one you pick.

The first set in parentheses is A, the second is B

(1/3) + (1/3 + 1/3) = 1
(1/3) + ((1/3 - 1/3) + (1/3 + 1/3)) = 1
(1/3) + (0 + 2/3) = 1
(1/3) + (2/3) = 1
A = 1/3, B = 2/3, you must switch.

One choice did not mysteriously disappear, it was taken only from the two choices that you did not pick, and could not have been taken from your pick, your pick is therefore always worth 1/3.

Given the rules of the problem, there is nothing that takes place here to increase the probability of your original guess from 1/3. A and B MUST ALWAYS BE LOOKED AT AS SEPARATE SETS if the wrong answer can only be shown from set B. If a wrong answer can be exposed from all 3 answers then the previous sentence is no longer true.

Don

[pudley4]

Look at it this way:

There is a 1 in 3 chance of it being behind any of the doors (A, B, C). So there are 3 possible ways the doors could be organized (Y=yes the prize is here, N=no it is not)

<pre><font class="small">code:</font><hr>
Door A Door B Door C
Y N N
N Y N
N N Y</pre><hr>

Each of these three scenarios is equally likely.

Let's say you always choose door A. So 1/3 of the time you are right, but the other 2/3 you are wrong. If you always switch, you'll be wrong 1/3 of the time, but right 2/3 of the time.

If you randomly pick a door, you will be right 1/3 of the time, and wrong 2/3. Again, switch doors and you'll be right 2/3 of the time.

Last way to look at it. We're going to play the game 6 times. We'll keep the setup the same each time - Door A is the winner (but you don't know that). You'll pick each door 2 times.

<pre><font class="small">code:</font><hr>

Try # You pick Monty eliminates You stay (W/L)
1 A B W
2 A C W
3 B C L
4 B C L
5 C B L
6 C B L
</pre><hr>

Again, you win only 1/3 of the time if you stay, but 2/3 if you switch.

[Cyrus]

A dismissive comment elsewhere about the worth of Philosophy itself, got me thinking about your enquiry, ie the relation between Philosophy and Mathematics. Others may disagree but personally, I find it impossible to overrate the effect that Goedel's Theorem had on Philosophy (and not just Philosophy of Mathematics but the pure thing proper.) The realization (which amusingly enough came by way of math'matical proof) that a full understanding of the human condition and getting the answers that philosohphers have posed might be impossible, is a monumental event. So, forget trivialities like the Monty Hall, here's the real thing.

I'm not suggesting that this was a novel thought in early 20th century or that this had not been posited before. But Goedel proved it "on the blackboard" and not just by logic. A rarity for Philosophy.

--Cyrus

PS : BruceZ had a couple of posts up about Goedel somewhere. The latest one chronologically is the one to read. But I believe that a web search would provide anyone so inclinded with a wealth of information about Goedel's work. Have your anti-depressant handy.

[Erudito]

I agree with Marilyn's answer. The reason is you are writing a test, and you should have studied for the exam, and your choice of selecting the answer is based on the information which you studied before taking the test. Therefore, when the professor tells you which one is the wrong answer, it won't matter because, if you know what is the right answer, based on your hard studying before taking the test, you won't switch to the wrong answer. The fact the professor tells you the wrong answer, it only increases your chances that your first pick was the right answer. Of course, if you are taking a test on a subject which you have absolute no knowledge, than your choice is random. Taking an exam for a grade [img]/forums/images/icons/tongue.gif[/img] is not random [img]/forums/images/icons/smirk.gif[/img] .

[Cyrus]

You are in front of a thousand closed doors. When you'll pick one, I have promised to open 998 doors showing a goat. You choose door #74. True to my promise, I switch open 998 doors and they all show a goat behind 'em.

One door is left closed, #569. So now there are only 2 doors closed, #74 and #569. I ask you to choose again. You stick to your original choice.

The result is_____________

..OK, we do this again. The prize is moved around a different door -- maybe. Now you choose door #6. I open 998 doors and they all show a goat but door #991 I leave closed.

Again you stick to your original choice.

The result is______________

...We do this again. And again.

The question is no longer where is the prize. It's How many times do you have to go through the above to realize that you gotta be switching ev'ry time??

[pudley4]

The solution to this problem does not depend on the fact that you are taking an exam, or whether or not you have studied for it.

The difference between this problem and the Monty Hall problem is that the teacher will tell you one of the wrong answers without knowing which answer you selected. So it's possible he will tell you your answer is wrong.

Let's look at 6 examples again.

<pre><font class="small">code:</font><hr>
Problem # A B C You pick Teacher eliminates You switch
1 Y N N A B L
2 Y N N A C L
3 N Y N A A 1/2 W
4 N Y N A C W
5 N N Y A A 1/2 W
6 N N Y A B W
</pre><hr>

So you'll end up with 2 guaranteed correct, 2 guaranteed wrong, and a 1 in 2 chance at the other 2. On average you'll end up with 3 correct answers if you always switch, and 3 correct answer if you never switch (except when the teacher tells you you are wrong)

[djptolemy]

[punkass]

Wow, I totally forgot about this post.

Your scenarios, the 3 sided solid figure changing into 2, or 3 numbers on a roulette table with a block blocking one, are not analogous to Monty's situation. In your scenarios, when one of the choices is blocked, you introduce an element of chance. If Monty didn't know which door had the car, then yes, if he chose a door with a sack of potatoes, then it doesn't matter whether you switch or not, your chances went up to 50% from 33%.

However, Monty knows where the car is. That seemingly minor fact changes the scenario. Because he will always choose a door with nothing to show you is why you should switch.