October 11th, 2015, 01:01 PM  #1 
Newbie Joined: Oct 2015 From: York Posts: 1 Thanks: 0  Can someone explain the Monty Hall problem to me?
After one door was opened and thus eliminated, why exactly isn't there a 50% chance for the car to be behind either door? I know for a fact it's not behind the opened door so only 2 are left and I have no more information about which of these 2 doors has the car. How can that be? And why do the solutions always continue taking 3 doors and chances out of 3 into account when there are only 2 doors left? 
October 11th, 2015, 03:52 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
Which would you rather have  what's behind one door or what's behind two doors?

October 11th, 2015, 03:59 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,852 Thanks: 743 
Try the problem with 100 doors. You pick one and the host opens 98 to show nothing. What do you do?

October 12th, 2015, 08:05 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Suppose you chose one door and were then given the choice of switching to both of the two other doors. You'd take that, right? If so, why are things different when one of those doors is opened?

October 12th, 2015, 01:37 PM  #5  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25  Quote:
Call the door you originally chose C (for Contestant). Call the door to its right R, and the door to its left L. (Assume they are in a circle, so the door to the "right" of Door #3 is Door #1.) If the car is behind R, Monty must open L. If the car is behind L, Monty must open R. If the car is behind C, he can open either R or L. But he can never open C. Let's assume he opens L (you can reverse R and L in the following argument in he opens R). What confuses you is thinking that all the information you have is that L does not have the prize. But you know more than that: you know that Monty choose to open L. He must open R if the car was behind L, so you eliminate all those games. But he can also open R if the car is behind C, so you also should eliminate half of the games where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN R. What's left is the 1/3 where the car is behind R, and the 1/6 where the car is behind L BUT MONTY HALL WOULD CHOOSE TO OPEN L. Since there are twice as many cases left where the car is behind R, than cases where it is behind C, you should switch. Quote:
Technical: Say something random happens. You can make a list of all the possible outcomes, and the probability of each. Those probabilities will add up to 100%, right? An example is rolling two sixsided dice, and adding them. You can get a sum of {2,3,4,5,6,7,8,9,10,11,12}, with probabilities {1/36,2/36,3/36/,4/36,5/36,6/36,5/36,4/36,3/36,2/36,1/36}.The point is that you can only do this correctly if you start with all of the possibilities that existed before you learned anything. That is, all 3 doors, or all 36 combinations of 2 dice. The mistake people make is eliminating only those cases that couldn't produce the information you have, when they need to also eliminate those that wouldn't. In Monty Hall, eliminating "couldn't" only takes care of the cases where the car is behind L so he had to open R. Eliminating "wouldn't" also takes care those cases where the car is behind C and Monty Hall chose to open R.  
October 13th, 2015, 08:58 AM  #6  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25  Quote:
EXAMPLE: Monty always asks if you want to switch to both doors, and then opens the closest door to him that he can. If he walks past a door to open the furthest door, you know the prize is behind the door he passed. If he opens the one that actually is closest, your chances are 50% with either of the other two. Now, I know this doesn't happen in the problem as it is stated. But the fact that it could requires more of an explanation than you gave. +++++ But I really am replying to correct a typo in my reply: What confuses you is thinking that all the information you have is that L does not have the prize. But you know more than that: you know that Monty choose to open L. He must open R if the car was behind L, so you eliminate all those games. But he can also open R if the car is behind C, so you also should eliminate half of the games where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN R. What's left is the 1/3 where the car is behind R, and the 1/6 where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN L. Since there are twice as many cases left where the car is behind R, than cases where it is behind C, you should switch.  
October 13th, 2015, 10:55 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326 
By "Monty chooses at random", do you mean he may open the door you want to choose? One can list all the cases to obtain a correct answer, but some people don't find that sufficiently clear either. 
October 13th, 2015, 12:26 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  
October 13th, 2015, 02:36 PM  #9  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25  Quote:
That leaves one, or two, choices. A correct solution has to use the probability that he would open the door he did when you picked correctly (and he had a choice of two). If that probability is Q, the probability you win by switching is 1/(1+Q). Of course, if you aren't told how he decides, the best assumption is Q=1/2 and so a 2/3 probability you win if you switch. But some people refuse to acknowledge that this assumption is being made here, or in similar problems. Quote:
Quote:  
October 13th, 2015, 04:52 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 21,110 Thanks: 2326  That's incorrect. If you picked correctly, you lose by switching, and if you picked incorrectly you win by switching, regardless of the value of Q. Hence the probability that you win by switching is independent of the value of Q.


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