My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Thanks Tree1Thanks
Reply
 
LinkBack Thread Tools Display Modes
July 22nd, 2016, 08:20 PM   #21
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 391
Thanks: 70

When you picked a door, the probability of the car being behind it was 1/3.

The probability of the car being behind one of the other doors was 2/3.

That probability does not change by the fact that Monty opens a door that does not have the car behind it.

So the probability that the other door that Monty didn't open being the right one is still 2/3.
Timios is offline  
 
July 23rd, 2016, 03:48 AM   #22
Senior Member
 
Joined: Apr 2015
From: Planet Earth

Posts: 141
Thanks: 25

When you picked a door, the probability of the car being behind it was 1/3.

The probability of the car being behind one of the other doors was 2/3.

But that CAN change when Monty opens a door, depending on how he chooses a door to open.

Call the other two doors L, and R, for the one to the left of, and right of, the one you chose. (Wrapping around to the other side of the stage if necessary.)

If the car is behind L, Monty must open R. If the car is behind R, Monty must open L. But if it is behind your door, Monty must choose between L and R. Say he chooses L with probability X, and R with probability 1-X.

So there are two ways Monty can open L, one with probability 1/3, and one with probability X/3. If you see him open L, the probability the car is now behind your door is the probability the car is behind your door AND he opens L, divided by the probability he opens L. That's (X/3)/[(1/3)+(X/3)] = X/(1+X).

Similarly, there are two ways Monty can open R, one with probability 1/3, and one with probability (1-X)/3. If you see him open R, the probability the car is now behind your door is the probability the car is behind your door AND he opens R, divided by the probability he opens R. That's ((1-X)/3)/[(1/3)+((1-X)/3)] = (1-X)/(2-X).

Now, if we don't know how he decides, X=1/2 and both of these probabilities are 1/3. The point is that it doesn't have to be. "That probability does not change by the fact that Monty opens a door that does not have the car behind it" is incorrect logic that by coincidence happens to get the correct conclusion.

Conditional probability is often non-intuitive, and the mistakes that get made are almost always based on assuming something can't change, when it can.
Thanks from manus

Last edited by skipjack; July 23rd, 2016 at 06:26 AM.
JeffJo is offline  
July 23rd, 2016, 04:06 AM   #23
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
Thanks: 1038

Agree...but this is by now quite well known:
just google "Monty Hall problem"...
Denis is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
explain, hall, monty, problem



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Question about the Monty Hall problem Csukar Probability and Statistics 10 July 7th, 2015 05:14 AM
Monty Hall question daigo Algebra 10 July 6th, 2012 09:00 AM
Just launched a Monty Hall problem site jeremy New Users 8 August 17th, 2011 08:08 PM
Question about the Monty Hall problem? giantsteps Algebra 1 July 9th, 2009 01:53 AM





Copyright © 2019 My Math Forum. All rights reserved.