The Monty Hall Problem

[denny]

Some friends bent my brain yesterday with a thing called 'the Monty Hall problem' - I thought I'd share the sheer confusion with others...

Description of problem, and ensuing discussion (including lots of explanations which may either help or hurt your head, or possibly both):
http://www.livejournal.com/users/duranorak/790534.html?thread=4120070#t4120070

An off-site explanation, also linked in the above discussion but posted separately here out of sheer helpfulness (I still didn't get it after reading this personally):
http://www.io.com/~kmellis/monty.html

Comments

[dennyd.livejournal.com]

No, the odds aren't 1 in 2 after he's opened a door - because they're a function of the original odds.

If you were right, then his choice of door is random, but if you were wrong, then his choice of door is constrained - which means the original odds have bearing on the secondary odds. Something like this: There's a 1 in 3 chance of you having a 1 in 2 chance of winning by sticking with the original door you chose, and there's a 2 in 3 chance of you having a 1 in 1 chance of losing by staying with the original door you chose. The eventual odds are (I think) 1 in 6 if you stay with your original door, and 5 in 6 if you switch. Although I could be wrong.

Bloody weird problem.

[deliberateblank.livejournal.com]

You have a one in three chance of picking the right door to start with. Once you've picked a door, you have that one in three chance of getting the gold. Nothing you later learn changes the odds of this.

To be able to change the odds of that door from 1/3, you need to have extra information about i) that door you chose, or ii) *all* the doors you didn't choose. But the host carefully does not give you that information (in the usual set of assumptions about the game - the 'correct' solution to the problem hinges on what assumptions you make about the host's behaviour. This isn't being cleverclever, you can get different outcomes with different assumptions but you have to state what they are to justify your answer. With the usual assumptions people still intuitively get the wrong answer.)

What the host has actually told you by his choice of door is "*IF* the gold is behind one of the two doors you did not pick, then it's not behind *that* one." He doesn't say anything about whether the gold is behind the other door you didn't pick or not, therefore he's not told you what you would need to know to decide that the combined probability of the gold being behind either of the two other doors has changed away from 2/3.

The probability behind your door stays 1/3, but the two probabilities behind the other doors have changed from 1/3 and 1/3 to 2/3 and 0/3.

[feanelwa.livejournal.com]

Ah, I see. I'd got as far as the logical bit but not as far as putting the numbers into it.

[stuartl.livejournal.com]

Right. This afternoon I set about writing a C program to prove you wrong.

The program plays the part of the player and the host and does 20,000 iterations of the game using 512 doors in about a tenth of a second. Yes, it's quite nippy, and highly unoptimised because I wanted to be 100% sure the logic was right.

However, halfway through the program I realised why you're all correct. I continued writing the program anyway to prove the point (and yes, it does prove it perfectly) but I saw the bit I'd missed before.

This is how I explained it to myself:

After you randomly choose your door there is a (n-1):1 chance that the prize is in the remaining doors. i.e. It's almost definate.

By the host opening the remainder of the doors he's narrowed down the prize location to the one remaining closed door. As someone else said, the chances of it being in that door are still (n-1):1.

QED, I guess. Apologies for being an arse :)

[dennyd.livejournal.com]

Wish I'd thought of doing that... I argued with people about it for ages before I suddenly clicked :)