Title Author Date

3 Doors (the Monty Hall show) Simon David, Nesa Nov 02, 2002

In the chapter titled "Probability estimate is often tricky" of Mark Perakh's article Improbable Probabilities the author gives the example of the Monty Hall show, where participants are asked to choose from 3 doors, of which one has a prize behind it.

The author states that the participants double their chances of winning if they change their first choice after the compere opens one of the doors. My mind was quite frankly boggled by this statement. To me it is obvious that the door opened by the compere is irrelevant (after that door has been opened).

The game only really starts after the compere has opened one of the doors, thus taking that door out of the game... and leaving just 2 doors to choose from. One of the doors is a winning door, and the other is a losing door. The probability of winning at this game is thus clearly 50%.

I felt that i couldn't accept your explanation (because i had a nagging feeling that it somehow couldn't be right), so i decided to run a simulation (in my computer). After 8692 iterations, the number of wins to the total number of games is 49.459% (actually it's been fluctuating between 51% & 49%.

What is surprising however, is that the "percentage of games won where the participant changed his/her choice" is 32.881%, but the "percentage of games won where the participant kept his/her choice is 16.578%.

However, it must be made clear that the chance of winning a game is ultimately 50%. The 32% and 16% chances are "after the fact" percentages. A person who keeps first choice has a 50% chance of winning. A person who changes choice also has a 50% chance of winning. It just so happens that 32.881% of persons who win change choice, and 16.678% of persons who win keep first choice (I still don't understand why). But the fact remains, whatever you do, change choice or keep first choice, your chance of winning is 50%.

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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

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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

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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 (N-1) times. Cheers,
Mark

Best wishes, Mark

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