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i agree with the werewolf. the reason that making the decision to switch feels so inherently counter-intuitive is that you're more likely to kick yourself for changing if you had initially picked the right door.

 

BTA has hit the goat on the head. Your original door has a chance of 1/3 of having the car, the second door (the one the host has left closed) has a chance of 2/3 of having the car because the host knows where the car is and would never reveal it

 

Yes, yes I am aware of the maths CL. However, even knowing this I am not sure I would switch. If I do I still have 1 chance in three of being wrong and I would be bloody furious if I were.

If I stuck with my original one I wouldn't be as angry.

In both cases I would eat the goat.

 

Bravo BTA!

Your logic is correct. Since odds are you chose a goat originally, and then having the host eliminate the other...kinda narrows it down.

HOWEVER, you still will have picked the car originally 1 out of every 3 attempts (Statistically speaking anyway.)

 

Me so stupid..

 

If you are ever gonna study maths and/or statistics this is actually the baby babble of the language of logic.

Nice to see so many eager souls though. :thumbs up:

 

Sorry, me still no get it.

 

OK I've come across this example before and it agitated my common sense then. Now I've read the 'proof' provided by Nobleman and I understand about the 2/3 v 1/3 and the example of having a million doors. However, it still feels wrong. So, I've sat down and written out all the possible scenarios to work out the probabilities that way. They suggest that the 50/50 option is indeed right despite what has been said. I admit that it is nearly 20 years since I did my statistics so I am a little rusty but something about only having three options, then two options and degrees of freedom seem to be relevant here.

Anyway, here is the worked example. Please actually read it before responding that I must be stupid if I don't see why changing is the best option. Also don't bother giving me more waffle, as I said, I understand the arguments but they don't support the evidence. If the worked example is wrong please tell me which permutations I have missed, where I have miscounted or how I have misinterpreted the rules.

On that note, I understand the situation to be: There are three doors behind one of which there is a prize. You choose a door. The host then opens one of the doors but this can not be the one you have chosed or the one with the prize. You then choose from the remianing two doors.

I have shown this in my example as follows:

door 1 door 2 door 3
start 0 0 1 - where 1 represents the prize
Pick p 0 1 - p shows which choice is made
remove p x 1 - x shows the door opened and this can not be the door with the prize or the one already chosen
final 0 x p1 - p can move for the last choice but can't be the door that has been opened

Anyway, based on all the permutations I can find, it seems like there is no difference between changing or not changing.

It also occurs to me that this is not really a typical probablity problem. The prize 'is' behind a door and stays there. Our actions do not affect it. It isn't like rolling a dice or those cats that only die when you open the box.

Oh yes, here's the link
http://www.purchase0.freeserve.co.uk/Probability.xls

I look forward to constructive equally fully worked examples rather than opinion or waffle in response.

EDIT - the table has now been modified to take into account some extra possabilities as per the comments from BTA below. It now supports his argument.

[This message has been edited by Slappy (edited March 29, 2002).]

 

I found this link:

http://www.straightdope.com/classics/a3_189.html

It deals with this problem. The "Straight Dope" is a great website by the way. A real time killer.

 

Thing is though, just going on pure statistics, if this is a genuine program, then you're missing half the arguement. If the host knows which is the prize box, then he can always open one with the goat in, effectively rendering it 50/50, as he has no choice as whether he leaves one with a goat in or not. In the event of that being crap, which I guess it is.

Chance of being right first time 1/3. Chance wrong 2/3.

Assuming randon choice as to whether to change or not.

Firstly picking the right box originally. Chance of staying with the right box = 1/3*1/2 = 1/6. Which similarly is the chance of picking the worng box.

Picking the wrong box first. Chance of changing to the right box = 2/3*1/2 = 1/3. Ditto for holding onto the wrong box.

Therefore chance of the right box out of the four possibilities is 1/3+1/6, which is clearly 50%. Therefore it makes no difference to change.

Which makes picking the wrong box and then changing more likely (by a factor of two) than picking and holding the right box. However, as you have no knowledge of whether the first choice was right or not, it's a 50/50 overall.

[This message has been edited by Luan (edited March 28, 2002).]

 

Thanks BA. Yep I had worried about the bias towards starting by picking the right door, especially as , by then, I had appreciated your side of the argument even though it didn't seem to be working. I had put this down to the initial choice affecting the host's options. Of course I can see that there must be an equal chance of picking any of the doors initially even if the subsequent actions are then limited.

Therefore I have amended the list so that everydoor is represented equally. Or put another way, there are four possible outcomes resulting from picking the correct door first so there must also be 4 outcomes considered (to calculate the final probabilities) for each of the other two doors. This now shows the result you would expect as per BTA's argument.

 

Take the last door. If you get a goat all the better. Goats make good pets. They smell, put out large portions of waste, and have hair. If all else fails, you can milk the goat...

 

One reason that the final probability of a goat picking (sounds dirty) is not 50%, is that the host did not say that YOU picked a goat. If the host had said, "You picked a goat on your first try...care to try again?" the probability would be 50/50, as the other doors had been independant of your first step.

However, 66.6% of the time, you WILL pick a goat on your first try. Opening another door to reveal a goat does not level the playing field. If a goat is revealed 100% of the time, and you statistically chose a goat 66.6% of the time...The only door not interracted with will 66.6% of the time be a car.

"Huh?" you say, "Where did the 66.6 come from?"

Recursitivity! Your first pick will be a car 33.3% of the time. Therefore, your second choice will be a car 66.6% of the time. This last part does not work with the huge number of doors example because all but two doors were revealed, but the same principle can be used...or at least seen.

 

Goats have more than one use, especially in Texas.