saw this on myth busters i dont understand any chance someone can explain it please.

there are 3 doors
you select a door (dont get to open it yet) the host opens another door it is empty then he gives you option to STAY with your initial choice (still not opened) or choose the other door remaining.

they said you should always change but i dont understand because theres only 2 doors left staying and changing is the same odds right?

they said if you chance doors your odds become 2/3 from 1/3

thought i would ask the math wizards of 2+2 (:

was season 5 i think

its called the monty hall paradox

am a retard for thinking its all bs?

because saying switching or staying door at the end is just another way of saying left or right door and the probability is equal the initial choice is irrelavant because he exposed a door so now we just have option of 1 or 2 which is the same right?

This 5 minute video is the best explanation I have seen for those who do not understand why switching is correct.

http://www.youtube.com/watch?v=mhlc7peGlGg

The key is (and you have left it out of your problem) that the host KNOWS which doors has the prize and which one's do not. If the door the host decides to open NEVER contains a prize (b/c he intentionally does not choose that one), then switching is best.

Another way people like to think about is to imagine 100 doors with 99 goats and 1 prize. You pick one door. Then the host proceeds to open 98 doors each containing a goat. Do you want to switch now? For what it is worth, the example is useful, but I think not as helpful initially as the nice youtube video.

Look at wikipedia or google it, it's been discussed to death. Yes, you should switch. Short answer as to why: the host has introduced an extra piece of information, because he knows where the prize is. He will NOT open the door that has the prize, he will only pick a non-winning door.

It's easy to work out for yourself - enumerate all the possibilities. That is, number the doors 1 2 and 3. There are 3 places the prize could be, and 3 possible places the winner could be. Make a table, like
C=1 P=1 (chose 1, prize is 1)
C=1 P=2 (chose 1, prize is 1)
for each entry in the table, "open" one of the doors that is NOT a winner and NOT chosen by the player. Could how often you win by switching, and how often you win by staying. If you do this it should be very clear.

thnx sorry for repost was just curious

The repost doesn't really bother me. What would bother me is if you now did not understand why switching it better.

If you did this a million times and always just ignore what the host shows you, staying with your original choice, then obviously your choice will be right 1/3 of the time, and wrong 2/3 of the time. So if you switch every time, you will be right 2/3 of the time and wrong 1/3 of the time.

And there is one other condition that is sometimes left out but must be included to be rigorous. The host must ALWAYS eliminate a door, without exception. He can't choose when to do it. And as you point out, he obviously can never eliminate the prize door, which is the real key.

In case I sounded crotchety in my reply, I'm not bothered by the repost. But there's really not too much to say about it that hasn't already been said. I first saw the problem in 1992 when I got my first access to the internet as a whole (on usenet). So it's been out there for a minimum of 20 years, and I suspect much much longer than that.

Best answer I've ever seen.

This actually isn't that helpful because it misses the common mistake that people make, namely assuming that it doesn't matter if they switch or not (i.e. each one is 1/3 and therefore after the goat has been revealed the odds are now 50:50).

If it was obvious from the problem that "staying = 1/3, therefore switching must be 2/3" it wouldn't be such a famous problem.

Nonsense, of course it's obvious. You'd have to be awfully stupid to not know that your initial choice will be right 1/3 of the time. Most people think that the probabilities have to be 1/2, and they don't see any paradox. My answer shows them that the probabilities clearly have to be 1/3 and 2/3. Then they can think about why their assumption must be wrong.

The majority of people who have never heard of the problem do not reach this conclusion easily, and most people are not awfully stupid. It is a terrific explanation once you understand the problem or have thought it through. But the thought process to get there is more complicated than just "doh 1/3 chance per door herpderp"

I agree with the first half of that and never said otherwise. In fact I stated as much when I said that most people never get past the initial assumption that the doors have the same probability. My explanation gets them to realize that there is even another answer worth considering, and that explanation is compelling even to those with virtually no probability background. The accusation was that the explanation doesn't follow obviously from knowledge of how the game is played, and of course it does.


Well if someone doesn't even know that when they pick 1 door out of 3 that the probability of being right is 1/3, then I doubt that any explanation can help them because it would mean that they have no conception of what probability even means. Then if they're too dumb to understand that it must be behind the other door 2/3 of the time, then perhaps a game of "got your nose" would be more appropriate for them.

This old problem of mine has some similarity to the Monty Hall problem, though it's (I think) more subtle and it's a poker combinatorics problem. David Sklansky answered it correctly, so don't read his if you want to figure it out yourself.