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Comment by LambdaComplex | Hacker News Reader

It's easier if you look at it with 100 doors instead of 3 (and the host opens 98 doors after you pick). It's then easy to see why you should switch: there's only a 1% chance that the door you picked was right, so you should probably go with the other one.

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Intuitively, this sort of makes sense to me but I can't help but think: aren't odds independent for each door? Isn't opening a new door like rolling the dice again every time?

The wikipedia article mentions this:

"the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally"

I don't understand why it doesn't add value to both doors. Just because I picked it doesn't mean it's the wrong choice, does it?

Here's a nice discussion of the Monty Hall problem. (The correct answer depends on the precise wording of the problem). It's full, and clear.

Chapter 18: http://dspace.mit.edu/bitstream/handle/1721.1/104426/6-042j-...

Of this thing: http://dspace.mit.edu/handle/1721.1/104426

Reading it now, thanks!

When you picked your original door, you had a 1% (1 in 100) chance of picking the winner.

Consider the 99% cases that you got the original choice wrong. Then the host has no choice, he must close all doors but the winner. So 99% of the time, the host is forced to leave you the winning door.

The other 1% of the time, you chose the winner, so the host can do whatever he likes.

But 99% of the time (99% of the times you play this game) the host is going to be obligated to close all but the winning door.

Looking at it from the odds of losing side made it much clearer for me. Thanks!

The explanation that made most sense to me is to treat the second choice as a completely independent game (like a coin toss). You're given the choice to stick with your one-in-a-hundred chance, or you can instead go for a coin toss where the chances are obviously much higher.

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