Fog Creek Software
Discussion Board




Monty Hall Problem

I'm not sure I agree with Michael's solution to the Monty Hall problem. http://www.techinterview.org/Solutions/fog0000000117.html

His answer actually assumes that Monty is equally likely to pick either one of the 2 wrong doors if you had picked correctly to begin with.

In fact, you can come to a more general formula if you allow Monty's probability of picking the door he picks (given you were right in the first place) to be a variable 'p'.

It turns out by Bayes Theorem, that the probability you should switch = 1 / (1 + p)

If p=0.5, then the probability (by coincidence) happens to be 2/3 that you should switch.

However, let's say p approaches 0.  In other words, Monty would hardly ever pick the door he showed you given a choice, then the probability you should switch approaches 100%.  This goes along with intuition that the only reason he would show that door is if he was forced to (ie. you guessed wrong and he didn't have a choice).

Alternatively, let's say p = 100%.  So Monty always prefers the door he shows you if given a choice.

Now the probability you should switch becomes 1/ (1 + 1) or just 50%.

So, I think, a more correct answer to the question is that you should always switch because the odds of switching being the correct thing to is always at least 50% and can get up to 100% if only you knew in advance Monty's preferences.

Charles Reich
Thursday, April 03, 2003


http://www.dartmouth.edu/~chance/course/topics/Monty_Hall.html

George McBay
Thursday, April 03, 2003

http://discuss.fogcreek.com/techinterview/default.asp?cmd=show&ixPost=738&ixReplies=11

Michael H. Pryor
Friday, April 04, 2003

The way I think about that problem is that you are playing against the moderator; and your odds of winning initially is 1/n and his is (n-1)/n.  When he opens a door to you, since there ALWAYS will be a door that is empty, it doesn't really matter...  Therefore your problem becomes do you switch your initial 1/n chance with the (n-1)/n chance.  (Notice the edge condition is n=3 for the "always will be door that is empty" condition to be true)...

Chang Zhao
Thursday, April 10, 2003

SUCKERS!
There are 4 possible scenarios, NOT THREE. The host has a choice of goat to show you when you choose correctly. Everyone mistakenly combines Case 1 and 2 into a single case.


Case One:
Money--  Goat---  Goat---
Choose  Show

Case Two:
Money--  Goat---  Goat---
Choose                Show

Case Three:
Money--  Goat---  Goat---
              Choose Show

Case Four:
Money--  Goat---  Goat---
              Show    Choose

The Shadow
Thursday, April 24, 2003

Not quite.  There are only 3 scenarios; either the money is behind door #1, #2, or #3.  What door Monty shows you is irelevant; the key is that by switching doors, you effectively get 2 doors.  Let me illustrate:

Say you pick door #1.  If I gave you the option to stick with door #1 or take both doors #2 and #3, what would you do?  Of course you switch; your odds just got better.  You know for a fact that door #2 or #3 has to have a goat behind it since there is only 1 prize, but that doesn't matter in your decision-making.  So just because I show you that goat doesn't change your odds.

Joe Nobody
Friday, April 25, 2003

The simplest thinking is that when you choose the first door you choose the one you DO NOT want to choose (you always switch doors so logically you choose the other TWO). Then Monty shows you which one of the logically chosen two doors contains no money. It is nice from Monty, isn't it?

NoiseEHC
Sunday, May 04, 2003

Never did understand this one! I know what the official answer is, but I don't understand why. My analysis always goes something like this:

1. I pick a door.
2. Monty opens one of the other two, revealing a goat.
3. I can:
a) stay with the door I already picked
b) pick the other door

Since I'm choosing between 2 doors (the one I've got and the other unopened one), that leaves me with 1/2 chance of getting the loot. If Monty opened the 'goat door' before I made any pick, I would have two options from which to choose, one of which would pay off. By opening the 'goat door' after I made my pick and then offering me the chance to switch, he is, in effect, throwing away the results of the first trial, forcing me to make a new, and presumably independent, decision.

This is the first time I've ever had the opportunity to offer my alternate analysis to people who start off disagreeing with that analysis, so I'll be curious to see if I get an explanation I can follow.

Ron Porter
Tuesday, May 13, 2003

Ron,

The thousand-door variation crystalizes it. You pick 1 of 1000 doors. We know that your chance of being right is 0.1%, and your chance of being wrong is 99.9%. Now, Monty opens 998 doors with goats.

Do you believe that your first choice just magically became a 50/50? If so, you're the luckiest person in the world to go picking the right door 50% of the time when faced with 1000 choices! :)

Brad Wilson (dotnetguy.techieswithcats.com)
Tuesday, June 17, 2003

*  Recent Topics

*  Fog Creek Home