# For those of you who have never heard of the Monty Hall Problem

For those of you who are familiar with the problem and/or saw the movie 21, pop up some popcorn and just watch until the answer is revealed.

Here goes...

This is an apparently simple problem of statistical and logical analysis and all you have to do is consider a very easily explained problem and answer a rather straightforward question. Here's the setup:

Suppose you're on a game show hosted by famous game show host Monty Hall, and you're given the choice of three doors. Behind one door is a brand new Lamborghini and behind the other two doors are ugly and smelly goats. You win whatever is behind the door you choose. You have no basis for choosing so you go ahead and just pick door No. 1. Monty, who knows what's behind the doors, tries to make things simpler by opening door No. 3, revealing a goat. He then says to you, "Do you want to stay with door No. 1 or do you want to switch to door No. 2?"

Should you switch?

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### Replies to This Discussion

Even if one door is revealed nothing has changed the original odds of one chance in three.

The fallacy lies in thinking that it becomes a case of 1 in 2, which may be how it appears, but the 1/3 bit of information which has been discovered apparently doesn't go away, it becomes part of the remaining problem, so that one door is 1/3 probable and the other 2/3. I'm not a statistician and I probably don't talk like one. LOL

This is why I like Matt's kind of example.

Imagine not 3 doors, but 10. Pick one... you have 1 chance in 10. Now Monty opens 8 other doors, and you feel fortunate to have not picked any of those. Are the odds really still 1 in 10 for the door you didn't choose, and changed in the door you did choose?

Arghh! I can't word this properly without numerous re-write. I can see the best answer, I just can't explain it, yet.

I got it. The thing I need to remember is, the odds never rise above 50%, no matter what choice one makes. Now, if the first choice had a 10% chance of being the right choice, that won't change just because the other door now has a 50% chance of being the right one.

Dang, that's not an accurate explanation either, but at least it gets me beyond the merely intuitive part of the understanding.

I give up now. And I even did well in college statistics.

Don't beat yourself up. Lot's of world class physicists are dummies when it comes to statistics. I think it takes a special kind of (sick) mind to be a competent statistician. LOL

If you switch, the only way you can lose is if you picked the right door in the first place, and there's only a 1/3 chance of picking the right door. So if you switch you have a 1/3 chance of losing, which leaves a 2/3 chance of winning.

If there are 10 doors, you pick one and monty opens 8, if you switch you have a 9/10 chance of winning? You have a 1/10 chance of losing by picking the right door on the first try, which leaves 9/10 chance of winning, as long as monty only leaves you with 2 doors to choose from. Is that right?

I think that must be right.

The funny thing is, it only feels intuitively right if I simulate it further in my head multiple times... then I can accept that "on average", the win must occur 9/10 of the time after switching from the first, 1/10 choice.

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