Finally - A Monty Hall Problem Explanation that makes sense, even to me!

[ParaBul]

I've never encountered this problem before, and I admit my first thought was "50:50".

After reading the above (and a few other discussions from the linked page) I find it easiest to get my head around it this way:

After choosing a door, there is a 2 in 3 chance that the car is behind one of the non-chosen doors. By selectively (NOT randomly) removing one of the non-chosen doors there is still a 2 in 3 chance that the car is within that set - which is now only one door.

I think the key point is that the door that Monty opens (or eliminates from the set) is not chosen randomly. That's why the odds become so much better.

Edit: this is messing with my mind now.

For a new player to now enter and be presented with the two remaining doors, they have a 50:50 chance of randomly picking the car. For me, the odds are different because I have different information, so my choice is biassed - I'm not randomly picking any more.

I remember reading something similar in a novel many years ago. Two men were taking it in turns to call the outcome of a coin flip, and one was having more success than the other. The more successful man had eight coins in his pocket; five were manufactured in one decade, and the other three were manufactured in a different decade. If he pulled one of the five coins, he would call heads; otherwise, he called tails. The coin toss was completely random, but his call wasn't.

Of course, in a novel this works. In real life? I have no idea.

Edit 2: tried playing with the coin-toss game in Excel, using the random number generator. After more than 87000 coin tosses, the non-random call was up by 142 wins. I don't think that's a statistically significant result, so I'd say it doesn't work.

I'll do some drinking now - that should stop me thinking about this.