Monty Hall Logic

Date: 03/09/2001 at 10:35:24
From: George Dinwiddie
Subject: The Monty Hall problem

The Dr. Math FAQ (see The Monty Hall Problem, 
http://mathforum.org/dr.math/faq/faq.monty.hall.html   ), states that 

"We will assume that there is a winning door and that the two 
remaining doors, A and B, both have goats behind them. There are 
three options:

The contestant first chooses the door with the car behind it. She is 
then shown either door A or door B, which reveals a goat. If she 
changes her choice of doors, she loses. If she stays with her 
original choice, she wins.

The contestant first chooses door A. She is then shown door B, which 
has a goat behind it. If she switches to the remaining door, she wins 
the car. Otherwise, she loses.

The contestant first chooses door B. She is then is shown door A, 
which has a goat behind it. If she switches to the remaining door, 
she wins the car. Otherwise, she loses. "

There are, in fact, four options.  The first one described is 
actually two separate options:

1A. the contestant chooses the car, is shown a goat behind A or
1B. the contestant chooses the car, is shown a goat behind B

There are three choice points, not just two.

There is a logical fallacy in suggesting there is any connection 
between the two contestant choice points.

There is a mathematical fallacy in here also. At the first choice, I 
have a 1/3 chance of choosing the right door. Let's presume I choose 
door A. If the prize is behind door C, Monty has a 100% chance of 
opening door B. If the prize is door B, he has a 100% chance of 
opening door C. If the prize is in door A, he has a 50% chance of 
opening B and a 50% chance of opening C. He cannot open either the 
door I've chosen or the door with the prize. This is left out of the 
published "analysis." So, we are left with these possibilities:

      A      B       C
1   prize   open    empty
2   prize   empty   open
3   empty   prize   open
4   empty   open    prize

Hmmm...  I now have a 50% probability of having the right door. The 
mathematical fallacy is in considering the first two possibilities as 
a single possibility when they are, in fact, distinct. There are three 
choice points, not just two.  The host has a choice also, but that 
choice is sometimes 100% (if I've chosen an empty door) and sometimes 
50% (if I've chosen the correct door).

Date: 03/09/2001 at 11:12:53
From: Doctor Anthony
Subject: Re: The Monty Hall problem

Your analysis is interesting but perhaps you would comment on the 
following. There is 2/3 probability that you initially choose a wrong 
door; if you always change, you thereafter must get the correct door.  
Whereas there is 1/3 probability that you initially choose the correct 
door, and if you always stick to your first choice your probability of 
success remains at 1/3.

In the original setting you have initially 1/3 probability of having 
selected the right door and 2/3 probability of having selected the 
wrong door.

The various possibilities are as follows:

  1st choice correct - then changed   Prob of success = 1/3 x 0 = 0
  1st choice correct - did not change       "         = 1/3 x 1 = 1/3

  1st choice incorrect - then changed  Prob of success = 2/3 x 1 = 2/3
  1st choice incorrect - did not change      "         = 2/3 x 0 = 0

Total probability of success if you change = 0 + 2/3  =  2/3
Total probability of success if do not change = 1/3 + 0 = 1/3

So there is a better probability of success if you always change.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   

Date: 03/09/2001 at 11:25:56
From: Doctor Rick
Subject: Re: The Monty Hall problem

Hi, George. I saw Dr. Anthony's response; perhaps I can add to it.

You say yourself that I have a 1/3 probability of choosing the correct 
door first. But then you argue that because Monty can make two choices 
if I have chosen the correct door, and only one choice if I haven't, 
therefore there is a 1/2 probability that I have chosen the correct 
door. You are therefore saying that the probability that I have chosen 
the correct door changes depending on what Monty is going to do after 
I have made the choice!

No, there is a 1/3 probability that I have chosen the correct door, 
regardless of what Monty does next. 

Suppose Monty flips a coin to decide which door to open next if he has 
two choices. The possibilities are then (if the correct door is C):

1. I choose door C, Monty opens door A (probability 1/3 * 1/2 = 1/6)
2. I choose door C, Monty opens door B (probability 1/3 * 1/2 = 1/6)
3. I choose door B, Monty opens door A (probability 1/3)
4. I choose door A, Monty opens door B (probability 1/3)

Analyzing the game in terms of four choices doesn't change the 
remainder of the argument, because the probabilities are the same as 
in the published solution. Your fallacy is in assuming that, if you 
can enumerate N possibilities, the probability of each must be equal.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/