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Comment by oliv__ | Hacker News Reader

In the gameshow contestant "problem", the sentence concludes like so:

>So the first choice the contestant made had a 1/3 chance of concealing the car, but after the host had thrown a lemon away, the remaining door had a 2/3 chance of concealing the car.

Why doesn't the remaining door have a 1/2 chance of concealing the car? Doesn't the act of revealing a lemon change the odds here? Now there are only two unopened doors, one containing a car and the other a lemon.

Am I missing something here?

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It is 1/2 if the host just randomly opens one of the doors, irrespective of the things behind them.

It is 2/3 if the host always opens one of the remaining doors with a lemon.

It is 0 if the host only opens a door with a lemon if he can't choose a door with a car.

In my mind, when you take out one lemon, you end up in a new situation with two doors: one with a lemon, the other with a car.

I still can't see how having picked one of those two doors in a previous setting changes the current situation of two doors, one pick.

The host's choice depends both on your choice and the things behind the doors. Therefore he is revealing information about them.

Even if you didn't pick which doors the host must choose between; if he only ever opens a door with a lemon, you should choose one of the doors he had a chance to open but didn't.

I get it. Framed this way, it makes more sense.

Thanks for bearing with me!

Some presentations of the problem leave it ambiguous, but this article makes it clear that the host always opens a door to reveal a lemon. In this correctly stated version of the Monty Hall problem, switching always improves your chances.

The description in the article is

> The contestant must pick a door, and then the host opens one of the two doors the contestant didn't pick, to reveal a lemon.

I'm not a native English speaker, but I think this is ambiguous between "to reveal a lemon" specifying the reason or the outcome. At least I seem to remember constructions like "They tried again and again, only to fail each time.", which obviously aren't about motivation.

I retract my statement that it's unambiguous. The article leaves the possibility that the host didn't know where the car was and revealed a lemon only by chance. In this case there's no reason to switch doors. The host has to know where the car is and always reveal a lemon.

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.

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.

Try the same problem with 100 doors instead of 3. You pick a door, the host closes all other doors except one. Then ask yourself would you really stay on the door you originally picked?

This is called the Monte Hall problem and the conclusion is correct.

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