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Title 
Author 
Date 
3 Doors (the Monty Hall show) 
Perakh, Mark 
Nov 02, 2002

Dear Nesa Simon David: Thanks for your letter wherein you doubt the assertion made in my article on probabilities regarding the well known Monty Hall game. If it is a consolation to you, you are not alone as many people before you, including some well qualified mathematicians, had difficulty in figuring out the correct solution. The problem, though, has been solved beyond doubt and I gave the correct answer in my article. What is odd in your message is that you don’t believe your own data obtained by means of a computer simulation of the game in question. You have obtained precisely the results predicted by my explanation – in about 2/3 of games the change of the original choice led to winning while only in about 1/3 of games the preservation of the original choice led to winning. Your attempt to explain away this result by referring to the difference between prior and posterior probabilities is obviously contrived ad hoc and invalid. In this case Bayesian approach is irrelevant while a frequentist one is fully justified. Btw, a simulation with a much larger number of iterations was conducted by a group of mathematicians and computer experts at the IBM Research center and resulted in exactly the same result – 2/3 wins if the choice is changed and 1/3 if it is not changed. You wrote that you don’t understand why you got such a result. If, though, you wish to follow the facts, try to erase from your mind all preconceived notions and listen to explanation. I know of at least four different ways to explain the result. In my article I offered my own explanation which, in my view, is the simplest, but there are other explanations as well. I’ll repeat here my explanation once again. In the first step of the game you have a choice among three doors of which one is winning and two are losing. Obviously the probability of choosing a winning door is 1/3 and of choosing a losing door is 2/3. Hence, if the game is repeated many times, in about 2/3 of games the door chosen on the first step will be a losing one, and in 1/3 of all games it will be the winning door chosen. Now at the second step there are two possible situations, to wit: either the originally chosen door is the winning one (the probability of that is 1/3) or it is a losing one (with the probability of 2/3).
Take a deep breath. If at the first step you have chosen the winning door (the probability is 1/3) then to win you have to keep your choice (which happens in 1/3 of games). If, though, at the first step you happened to choose the losing door (probability 2/3) to win you need to change your choice (and this happen in 2/3 of games). Hence, in terms of probabilities, changing the choice doubles the probability of winning, QED.
I hope this will clarify for you the problem although you will have to figure out yourself where you went astray in your thinking (relying on a gut feeling you refer to is not always the best way to understand the problem).
Best Wishes

Related Articles: 
Improbable Probabilities

Title 
Author 
Date 
3 Doors (the Monty Hall show) 
Simon David, Nesa 
Nov 06, 2002

Hi Mark, thanks for replying...
I still had doubts, so i ran 2 more simulations... one where first choice was always kept.. and the other simulation where the choice was always changed. In the first simulation, the chance of winning was just 33% approx, and in the second simulation, the chance of winning was 66% approx. So i guess you're right... the data has spoken..
After that i've been thinking about the problem, and now understand why it is so... it's MUCH easier to understand this problem, if instead of 3 doors, we extrapolate and use say... 10 doors. (Participant first chooses a door, then the compere opens 8 doors.... the remaining door is almost sure to be the winning door, therefore participant has almost 100% chance of winning if the first choice is changed.) This 10 door game better helps to understand the odds involved in this type of game. If you use this example, i'm sure ppl who doubt will see why the game works the way it does...
Thanks & best regards... :)
Simon David

Related Articles: 
Improbable Probabilities

Title 
Author 
Date 
3 Doors (the Monty Hall show) 
Perakh, Mark 
Nov 06, 2002

Hi, Simon David. Thanks for your message. I am glad that you have figured
it
out. Of course, there is also a rigorous mathematical way to prove it, but
since my article was written for a general audience, I did not use it. Here
is a brief rendition of a formal proof. Denote the doors A, B, and C. P(X)
is probability of X being the winning door. Obviously P(A)=P(B)=P(C)=1/3 and
P(A)+P(B)+P(C)=1. Assume A was chosen. Then P(A)=1/3 and
P(~A)=P(B)+P(C)=2/3; Assume B is opened and found empty. Now P(B)=0, hence
P(A)+P(C)=1. Since P(A)=1/3, P(C)=2/3. QED. Instead of 3, any number N of
doors can be used, all of the above remains valid by replacing 3 with N, so
changing the choice increases probability of winning (N1) times. Cheers,
Mark
Best wishes, Mark

Related Articles: 
Improbable Probabilities


