Three doors and 2 goats

Originally posted by TheBloop
Say you pick door #1

If car is behind:
1 - switch and you lose
2 - switch and you win (because host would open #3)
3 - switch and you win (because host would open #2)

The key is that the host knows where the car is, and always opens a losing door.


If you stay with # 1, however:

If car is behind:
1- stay and you win
2 - stay and you lose
...[text shortened]... times.
If you stay with door #1, you only win 1 out of 3 times.


So yes, you should switch.

This is the proof I've always thought of as the most clear.
And if the person you are talking to can't understand it then the only avenue left is to simulate it many many times and let them see what happens.

Originally posted by XanthosNZ
This is the proof I've always thought of as the most clear.
And if the person you are talking to can't understand it then the only avenue left is to simulate it many many times and let them see what happens.

Here in the U.S., there is a weekly feature that appears in "Parade", a magazine that is included in our Sunday newspapers. The column is called "Ask Marilyn", and it is written by Marilyn vos Savant...the column (and her website) claim that she was listed in the Guinness Book of World Records as having the highest IQ ever, for both children and adult scores. Some controversy has arisen about that claim, but that's neither here nor there...

This very question appeared in her column several years ago, and she answered that "yes, you should switch.. " , and as an example, used a game show with a million doors (and one car), and after your guess, the host opens all the doors except # 777,777, and yours... then she concluded "you'd switch to door # 777,777 pretty quickly, wouldn't you?"

She also explained that there was a 2/3 chance of winning if you switched.

You should have seen the controversy that erupted after she wrote that answer... she had math teachers from all levels (including people who claimed to be professors at prestigious colleges) claiming that she didn't know what she was talking about, and that the answer should be 50/50 as to whether it pays to switch.

She then repeated her answer, this time showing a grid of each possible scenario, which clearly showed that if you switch, you win 2/3 of the time.

This convinced a number of people, but not everybody... after another 2 or 3 columns were devoted to this question, she suggested (as you did) that people actually perform the experiment and record the results.

A number of classrooms (at all levels) nationwide did in fact perform the experiment, and finally, most of the people who responded to her were indeed convinced that it does pay to switch your choice of doors, but admitted that they were astonished by the results of their experiment.

In one of the columns devoted to this question, she also pointed out that if, for example, a spaceship landed after the host opened the losing door, that the alien in the spaceship would, in fact, have a 50% chance of winning in this situation, because it would lack the knowledge the original contestant had (that the host opened a losing door on purpose).

I've tried this question on a number of people since that time, but it seems that no one I asked understands the answer until it's demonstrated as I did here (showing all possible scenarios).


A very interesting example of how people think about these things.



EDIT: I just checked, and here is a link to Marilyn's site which discusses this problem. Marilyn herself reprints the question, her response, and responses she printed from her readers.


http://www.marilynvossavant.com/forum/viewtopic.php?t=64

The columns were originally published during 1990 and 1991.

Originally posted by TheBloop
The key is that the host knows where the car is, and always opens a losing door.

Indeed - that is the bit that eventually convinced me when I encountered this problem

http://math.ucsd.edu/~crypto/Monty/monty.html

Anyone interested in the history of this intriguing puzzle might want to check out Monty Hall Problem in Wikipedia.

Originally posted by AThousandYoung
http://math.ucsd.edu/~crypto/Monty/monty.html

I went to that site the other day and played a quick 10 games, and actually won 9 of them (I switched every time).

Today, I decided to play a few more, I played a total of 11, again switching each time. Only won 5 of the 11.

Grand total: 21 games, switching each time, 14 wins, 7 loses, a .667 percentage.

I should have quit while I was ahead...

Originally posted by TheBloop
I went to that site the other day and played a quick 10 games, and actually won 9 of them (I switched every time).

Today, I decided to play a few more, I played a total of 11, again switching each time. Only won 5 of the 11.

Grand total: 21 games, switching each time, 14 wins, 7 loses, a .667 percentage.

I should have quit while I was ahead...

How do you like the car?

Originally posted by HandyAndy
How do you like the car?

Doesn't he have 14 cars by now? And seven goats.

Originally posted by AThousandYoung
Doesn't he have 14 cars by now? And seven goats.

Sounds like a traffic jam in downtown Tijuana.

This was one of those questions that mathematicians got wrong, I read that in a book 😉 The answer is 50% if I'm not mistaken after 1 door is opened.
Doors A B C, lets say A and B have goats. If you pick B and A is opened, there is a 50% chance whether said person would choose C and 50% they would stick with B.
I think that made sense 😛

Originally posted by Tektile
This was one of those questions that mathematicians got wrong, I read that in a book 😉 The answer is 50% if I'm not mistaken after 1 door is opened.
Doors A B C, lets say A and B have goats. If you pick B and A is opened, there is a 50% chance whether said person would choose C and 50% they would stick with B.
I think that made sense 😛

Are you a mathematician?

Originally posted by Tektile
This was one of those questions that mathematicians got wrong, I read that in a book 😉 The answer is 50% if I'm not mistaken after 1 door is opened.
Doors A B C, lets say A and B have goats. If you pick B and A is opened, there is a 50% chance whether said person would choose C and 50% they would stick with B.
I think that made sense 😛

Good for a first try, now try again 🙂

Originally posted by Tektile
This was one of those questions that mathematicians got wrong, I read that in a book 😉 The answer is 50% if I'm not mistaken after 1 door is opened.
Doors A B C, lets say A and B have goats. If you pick B and A is opened, there is a 50% chance whether said person would choose C and 50% they would stick with B.
I think that made sense 😛

Let's phrase it this way. A normal deck of cards has 52 cards and one ace of spades. You draw one card. I then take the cards, look at them, and turn over 50 of them, none of which is the ace of spades. How much you wanna bet you've got the ace?

Originally posted by HandyAndy
A new contestant, with no knowledge of what transpired, is choosing between two doors, one with a goat and one with a car. His/her chance is 50%. If the new contestant witnessed what transpired, his/her chance to win the car by switching doors is 66.6% -- the same as the original contestant.

A new contestant, with no knowledge of what transpired, is choosing between two doors:
Door 1 has goat 2/3, car 1/3
Door 2 has goat 1/3, car 2/3

The only thing is, the new contestant does not know this and cannot deduce it either without knowledge of what happened previously.

you should not switch...
first time: 1/3
second time: 1/3 also...