I think that you should switch because they didn't open the goat curtain at random, they may have specifically
have chosen the goat, not the car behind the other curtain.  I think I heard this long ago, some question
analogous to this one. 

But it's not about changing your mind, it's what they have done.  They chose not to open the curtain with
the car for suspense.  See what I mean.  I don't know the math behind what I'm saying though.

Google "monty hall problem"

Sorry all you people who think this works. IT DOESN'T!!

Proof

The past will never ever ever determine a greater probability for one posibility. The fact that you made a decision before does not change the probability that you have 2 options, "swap" or "stay" and two posibilities "car" or "goat". The chance is 50/50 (FULLSTOP).

Imagine that you dont choose one to begin with!! Lets play this out. options A B C. They tell you to pick one (BUT YOU DONT). Knowing that they will not pick the "car" they show you that C is a goat. Then they tell you to "stay" or "swap". You dont have a orginal choice. SO WHATS THE PROBABILITY!!! You have to choose A or B. 1 is right, 1is wrong! The probability is 50:50. The same as if they never had a C in the first place.

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

CONCLUSION:
NUMB3RS WAS WRONG!!

100 000 000 times yes!!!

and I did that a few times, so really about 400 000 000 times

and yes "The past will never ever ever determine a greater probability for one posibility"
eg.
if you throw a dice 9 times, and a 6 comes up every single time (highly unlikely, but possible). What would be the probability of a 6 coming up on your tenth throw? ANSWER 1/6. Ive also run this simulation . Even though the chance that of throwing a 6, 10 times in a row is 1 in 60466176 if it has already been thrown 9 times, the next time is 1/6.

Its the same principle

So the probability started at 1-in-3 but then the host revealed one, the probability changes to 50:50.

Last edited by martian (2005-11-22 23:32:01)

Sorry all you people who think this works. IT DOESN'T!!

Proof

The past will never ever ever determine a greater probability for one posibility. The fact that you made a decision before does not change the probability that you have 2 options, "swap" or "stay" and two posibilities "car" or "goat". The chance is 50/50 (FULLSTOP).

Imagine that you dont choose one to begin with!! Lets play this out. options A B C. They tell you to pick one (BUT YOU DONT). Knowing that they will not pick the "car" they show you that C is a goat. Then they tell you to "stay" or "swap". You dont have a orginal choice. SO WHATS THE PROBABILITY!!! You have to choose A or B. 1 is right, 1is wrong! The probability is 50:50. The same as if they never had a C in the first place.

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

CONCLUSION:
NUMB3RS WAS WRONG!!

Something doesn't sound right there. You eliminated the selection. In your example, you said we will not make an official selection and reveal goat C every time. Am I correct? Pardon me if I'm not but If I am, what if you were to pick goat C when actually playing the game? The rules of the game state that when you pick one, one of the goats will be revealed, not the car, and not the one you picked. So if you pick a a door, and he flips over a goat. Now there are three possibilities at this point. You picked goat A so he has to reveal goat B. Or you picked goat B so he has to reveal got A. Or you picked the Car and he randomly flipped over one of the goats. In the first two situations, switching your choice would win you the car since its always more likely he is revealing a remaining goat. In the third situation, if you picked the car on the first try, switching would loose you the car. So in 2 out of 3 situations, switching will win you the car.

I think its because no matter which card you pick, there will always be one remaining goat that is not the door you picked (since there are two goats). So the revelation does not improve your chances at all. So if you pick door B and he reveals door C. It is most likely he revealed door C rather then door A because door A contains the car. Its not that the previous chances count, its that your chances have not improved because there will always be a remaining goat to reveal, and that will always be the one revealed. So your chances have not improved, which means they haven't changed which means they are now what they were then.

I read about computer simulations that proved this concept. Did you program it to make the selection, reveal a remaining goat that is NOT the one you picked, switch the choice, and check whats inside? Thats the only proper way to do it.

Last edited by mikau (2005-11-23 03:41:50)

If You still dont belive me then believe me when I say that Im a computer Scientist and I simulated the whole process in a computer and run the simulation ONE HUNDRED MILLION TIMES!!!!! The simulation showed that it swapping worked 50% of the times.

No, I don't belive in you.
http://en.wikipedia.org/wiki/Monty_Hall_problem

CONCLUSION:
NUMB3RS WAS WRONG!!

No.... Conclusion [ and Advice from a fellow (future) Computer Scientist]:

                    Search the net before wasting time computing solved problems.

------------------------------------------
Wikipedia saves the day (twice)

Last edited by kylekatarn (2005-11-23 04:16:50)

mikau wrote:

Something doesn't sound right there. You eliminated the selection. In your example, you said we will not make an official selection and reveal goat C every time. Am I correct?

C is just simply one that is not the car.

mikau wrote:

Pardon me if I'm not but If I am, what if you were to pick goat C when actually playing the game? The rules of the game state that when you pick one, one of the goats will be revealed, not the car, and not the one you picked.

Ok, lets just say that B "is" the car. Just to clarify
If I picked A then they would reviel C
If I picked C then they would reveil A

So you see it didn't matter whether I picked A or C in the first place. Im still in the same predicament. And obviously, if Id have picked B in the first place, then then I swap, I'd be wrong.

mikau wrote:

I read about computer simulations that proved this concept. Did you program it to make the selection, reveal a remaining goat that is NOT the one you picked, switch the choice, and check whats inside? Thats the only proper way to do it.

Thats exactly what I did. By all means,please, check it yourself.

import java.util.Random;

class  NumbersThing
{
    public static void main(String[] args) 
    {
        Random r = new Random();
        int worked = 0;
        int didnt = 0;
        for (int trys = 0; trys<100000000;trys++ ){
            int answer = r.nextInt(3);
            int guess = r.nextInt(3);
            int takeaway=guess;
            //Ensures the reveiled wrong answer is not the one you guessed
            while (takeaway==guess){
                takeaway = r.nextInt(3);
            }
            //Makes sure the one reveiled was not the answer, if it was, the whole thing is discarded.
            if (takeaway!=answer){
                //Continue
                int swap = -1;
                for (int i = 0;i<3;i++){
                    if (i!=guess && i!=takeaway)
                        swap=i;
                }
                if (swap==answer)
                    //It worked
                    worked++;
                else
                    //Didn't Work
                    didnt++;

                //JUST TO MAKE SURE!
                if (swap == guess){
                    System.out.println("ERROR");
                }
            }
        }
        double ratio = (double)worked/didnt;
        System.out.println("Swapping worked: " + worked + " times");
        System.out.println("Swapping didn't work: " + didnt + " times");
        System.out.println("Ratio: " + ratio);
    }
}

-output-

Swapping worked: 33331990 times
Swapping didn't work: 33335219 times
Ratio: 0.999903135479626

-another output-

Swapping worked: 33330446 times
Swapping didn't work: 33338379 times
Ratio: 0.9997620460190941

You said it yourself.

If you pick A, the host would reveal C, give you the choice between A and B and swapping would win.
If you pick C, the host would reveal A, give you the choice between B and C and swapping would win.

Picking A or C means you have to swap, swapping gives a 2/3 chance.

I'm not a program writer, so I can't check the programming for your simulator, but I would guess that you've made a wrong assumption somewhere along the line, and so the simulation is inaccurate.

If you still don't believe me, then look at some of the many websites concerning this problem. They are unanimous that switching has a better chance.

I know why 1/3 of the answers are missing, and its cause I was lazy. Rather than finding a door which wasn't the guess and which wasn't the answer, I simply picked a random number and if it wasn't the guess and wasn't the answer then I started from the top at the next itteration

shouldn't make a difference, but I try and go back and be more through if you like. I be back with modified code

The idea was that the points didn't go to anyone

But now, I have changed the code

import java.util.Random;

class  NumbersThing
{
    public static void main(String[] args) 
    {
        Random r = new Random();
        int worked = 0;
        int didnt = 0;
        for (int trys = 0; trys<100000000;trys++ ){
            int answer = r.nextInt(3);
            int guess = r.nextInt(3);
            int takeaway=guess;
            //Ensures the reveiled wrong answer is not 
            //the one you guessed 
            //or the right answer
            while (takeaway==guess || takeaway==answer){
                takeaway = r.nextInt(3);
            }
            int swap = -1;
            for (int i = 0;i<3;i++){
                if (i!=guess && i!=takeaway)
                    swap=i;
            }
            //System.out.println("Swap:     " + swap);
            if (swap==answer)
                //It worked
                worked++;
            else
                //Didn't Work
                didnt++;
        }

        double ratio = (double)worked/didnt;
        System.out.println("Swapping worked: " + worked + " times");
        System.out.println("Swapping didn't work: " + didnt + " times");
        System.out.println("Ratio: " + ratio);
    }
}

Well the conclusion is, Ive been wrong. Im still confused though lol!

--output--

Swapping worked: 66668889 times
Swapping didn't work: 33331111 times
Ratio: 2.0002000233355557

-----------

So there you go.

Im deffinatly willing to accept my being wrong now!

The thing that I couldn't get my head around was, if you hadn't choosen one in first place. I couldn't work out what was the difference. Now I realise the difference is that you can choose one card that you dont want to be eliminated as being a goat. And by restricting the host to what he can remove thus it increses you chances on the swap.

This is a very good discussion. It has really opened my eyes...

thanks, and thanks putting up with my ignorance...lol ahaha