I give you the Gameshow (aka "Monty Hall"):



Yeah, this has been posted before. But at least this one hasn't seen the ropes as much as the other ones have.

Also: GOD DAMNED POLL TYPO. Actually, wow, how did I wind up typing that... so... horribly. Anyway...

so the goat isn't the prize?

lame

Let's assume the goats are laced with Anthrax.

What do I look like? A mathematician?

No, not really.

No.

Also, we know you don't like higher education, so really, no.

Ha ha!

I like education plenty. My beef is with schooling. But enough derail.

Oh ho, you know there will be plenty of that in this thread.

I just wanted us to start it off, instead of other parties, but I guess you don't like me

You're gonna get a car or a goat. It's 50/50. 1/2 probability....

It's 1/3.... It doesn't matter if you change your mind, you still are only picking 1 door... unless I'm misunderstanding....

I think it's 1/2. Since one of the goat-doors is opened, that leaves 2 doors to choose from. And as far as I can tell from the way the question is worded, you either pick the car-door or the second-goat-door. One choice of two doors. 1/2.

Annnnnnd you're all wrong so far (the people who've posted, not voted).

You made a choice in which 2/3 times, you've chosen the goat. So you're now more likely on a goat. He shows you the other goat, so you are now looking at 2/3 likelyhood that the square you initially didn't choose is the car.

Put it this way: If we had 10 doors, 1 car, and 9 goats. You pick, and you're almost certainly on a goat. The host shows you 8 goats. You'd be stupid to not switch to the other door, because you initially probably picked a goat, and now you're shown where the car is NOT. So it's more than likely in the other position.

It's 2/3. I'm going to draw a diagram in a sec to make it clear.

Simulation that runs n iterations where a switch is made and n iterations where no switch is made and tells the success rate:

Executable doesn't work on my system.

Here's my diagram.

Should work now, recompiled it on an XP machine, apparently the one compiled under Vista only worked under Vista. Guess I should fix that.

But you only have one goat to pick. The other one is completely irrelevant, the way the question is worded.

Maybe I'm misunderstanding.

Oh and Darth, you have to hit enter to quit your program. Not any key.

Wait, I get it now.

Yeah, I noticed that too. I chose to press "1" to end the program, and it didn't end. :saddowns:

Oh well, I just modified some random code I found and threw it together in a couple of minutes. Overlooked that little detail. Too lazy to fix it and recompile it again, lol.

THEN YOU ARE A BAD PROGRAMMER

I didn't comment anything I wrote either.

Zero. "Probabiliry" is not a word, and noone wins at game shows, anyways. ^_^

In a serious note, though, probability is 1/2. First pick does not changes anything. You might pick a car or goat, and host will still have a goat to expose.

Thus, the game is dwindled down to picking between two doors, one with goat and one with car.

No. As a clearer example of the same principle, say you are asked to pick a card out of a deck, and if you draw the ace of spades you win. Once you pick your card, Monty takes fifty cards out of the remaining deck, leaving just one. He asks if you want to take the card he has left instead of your own.

In this case, it's clear that you're much better off taking the other card. Your chance of success with the initial draw was only 1/52, and the fact that Monty removed fifty other cards doesn't change that. Likewise, with the doors, your chance of success was 1/3. Once he opens one of the remaining doors, there's still a 2/3 chance that the car's behind one of the other doors, it's just that there's only one other door remaining.

I disagree. Your example would be valid only in case, if Monty could`ve removed Ace of Spades along with those 50 removed.

First choice does not affects anything, and in second, we get binary yes/no guess, which gives us 1/2 probability.

Well, actually, the outcome of the second choice does depend on the first choice. The answer is 2/3. See my diagram. Also, Darth's program "proves" the fact "experimentally".

Explain.

If I was unclear on this point, Monty removes the 50 cards in the same way he removes the one door in the original problem: by turning them over to reveal that each one is not the ace of spades. In neither case can he choose to remove the winning option.

Pommy, it WOULD`VE depended, if host could "discard" the door with car.

Since goat is always discarded, first choice does not really matters, at all.


Michael - he removes the chances, no?

At first pick, it`s 1/3 of chance to pick a car.

At second pick, it`s 1/2 to pick a car.

Second pick should be considered independantly from first, since the initial set of choices in first and second is not same.

You have to take into account, that host, by removing the obviously wrong choice, alters the initial 1/3 chance into 1/2.

Yes, it does matter. Let me explain this as simply as possible:

There is a 2/3 chance that in your first choice, you will pick the WRONG door. Therefore, there is a 2/3 chance that if you switch you will end up with the CORRECT door.

IF the host had not revealed one of the other doors as wrong, this would not be true because it would not be certain that whatever your second choice is will be the OPPOSITE of what your first choice is. Because the host revealed one of the other doors as wrong, it is certain that IF you switch, if your first choice was right, your second choice will be wrong, and vice-versa. Hence the probability that you will pick the right door for your second choice is simply the probability of picking the wrong door on your first choice.

This explanation is basically a description of the diagram I posted above.

You seem to be under the impression that it's a new game once one of the wrong choices is removed. It's not. The car is still behind the same door it was before, and removing a wrong door does not make it more likely that your initial choice was correct. There is still a 2/3 probability that the car is behind one of the doors you didn't choose, even though one of those doors has been shown to be wrong.

Have you looked at Pommy's chart yet? Or run Darth's program? Really, if you doubt this, you can test it yourself. Find three playing cards and a friend, designate one of the cards as a car and the other two as goats, and let your friend play the part of Monty Hall. Stick with your original choice ever time, and if your understanding of the problem is correct, you'll win roughly 50% of the time. But you won't; you'll win only about a third of the time, as the chart and the program show.

That logic is flawed by one simple problem. You pick out of TWO doors the first time too, since one wrong will be discarded in any case.

Think about it this way - when you pick doors first time, for you, there is 1/3 of chance you`ll pick the right one.

BUT! After the discarding of the wrong door, you have to redistribute it`s 1/3 of chance equally between two others.

Or, as an alternative version, keep in mind, that first pick is always wrong, and you are considering the probability of picking right second time.

It`s entirely incorrect to shuttle WHOLE probabilty of wrong door to not-picked one.

No, you don't. Under what statistical reasoning would you do this?

Under what statistical reasoning you would shuttle the whole probability to unpicked door?

No matter, WHAT door we will pick first, we know, that ONE of the remaining ones will ALWAYS be wrong, and discarded. Therefore, pick occurs from two doors, de facto.

Look, let me sum it.

We have three doors. 1, 2 and 3.

Let us assume, that we pick the door 1, and the door 3 is then pointed out as fluke.

At the time, as we pick, each door has 1/3 of chance to have the car, for us. Right?

So, when we pick door 1, we have 1/3 of chance.
1/3 stays with door 2, and 1/3 with door 3.

So, when host points out door 3 to be wrong one, what happens?

We automatically assume, that door 2 now has 2/3 of chance? On WHAT basis?

Entirely correct would be to say, that door 3 is checked against the condition, and it`s 1/3 chance is always a fluke.

Therefore, second pick is to pick right 1/3 out of remaining 2/3 of chance.

Already explained. There's a 2/3 probability that the car is behind one of the two doors you haven't picked. Knowing that one of them contains a goat doesn't change that. Go back to my example with the full deck of cards. Do you mean to tell me that it's also a 50-50 choice in that case?





No, no, no. If we wanted to, we could modify the problem to say that after you choose, Monty opens the correct door and just gives you the prize every single time just because he likes you. While this might affect your motivation to guess correctly, it still wouldn't change the fact that you have a 1/3 chance of getting it right on your own. Nothing he does after the initial choice can change this.

Yes, but if Monty will open the right door, he fullfills the condition. Therefore, no matter, what door you pick, you will get your car with 1 probability.


What I`m trying to say, that when Monty picks the wrong door, he FAILS part of that 2/3 chance for you.

[quote=Ailce Shade, earlier]We automatically assume, that door 2 now has 2/3 of chance? On WHAT basis?[/quote]

We do so on the basis that it's part of a group of two doors that together had a 2/3 chance of containing the prize, and now one of those doors has been eliminated. The eliminated door has a zero chance of containing the prize; 2/3 - 0 = 2/3.



And in what way does this make it a "de facto" pick from two doors?

I have to go report a stolen passport, so in the meantime I'm going to ask you to do what I suggested and test this yourself.

Eh. No. Just NO.

Your suggestion here assumes, that FIRST, we have each door have equal chance, and then one suddenly "gives" it`s chance to the other.

Let`s be consistent here. Either we initially assume, that all doors have 1/3 probability, and one automatically fails, either we take into axiom, that one door is infallably 0 probability, and thus, other two doors share equal 1/2 of probability each.

Clarification:

One door with car, Two doors with goats.

2 possible first-choice doors - he can pick either a door with a car or a door with a goat. We take into account that there are three total doors when we do the simple math below. Firstly, let's explore each choice and its consequences. Note that the fact that the host reveals a door with a goat only serves to simplify the problem by strictly giving the contestant a win or loss solely based on what is behind his first-choice door (prove to yourself that this is so). And so let's now examine the choices:

(1) First choice of Door With Car results in a loss, loss meaning that if he switches, he will get a goat. Probability of picking Door With Car is 1/3.

(2) First choice of Door With Goat results in a win. Probability of picking Door With Goat is 2/3.

Thus, the answer is 2/3. I originally thought it was 1/2 until halfway through trying to prove the "2/3rds" people wrong, I saw the error of my logic. The second pick result is NOT independent of the first pick.