From: seyyed@wuee1.wustl.edu (Seyyed Mahdavian)
Subject: A general formula for Monty paradox
Organization: Washington University, St. Louis Mo.
Date: Tue, 13 Oct 1992 16:21:46 GMT

In article <1992Oct13.011117.17538@wuecl.wustl.edu> seyyed@wuee1.wustl.edu (Seyyed Mahdavian) writes:

> Common part of the three problems:
> Monty presents three boxes, in one of them a prize. The a priori probability
> of having a prize in a given box is therefore 1/3. You choose one box and 
> put it aside.
> 
> Problem #1:
> Monty opens one of the remaining two boxes randomly. There is no prize in it. 
> He asks whether you want to change your choice.
> Solution: The probability of having a prize in either of the unopened
> boxes is 1/2, therefore it doesn't make any difference. For a proof see
> my previous demonstration.
> 
> Problem #2:
> Monty knows which box contains the prize. He PROMISES to open only the
> box (from the remaining two) which he knows is empty. After opening the
> box (which he doesn't need to open, because he already have promised to
> open the empty box), in other words, after telling you which box from
> the remaining two is for sure empty, he asks whether you want to switch.
> Solution: In this case the probabilities are 1/3 to 2/3, so you better
> switch. I can also prove this using probability theory (not the naive
> way some people think they have done!)
> 
> Problem #3:
> Monty opens one of the remaining boxes. There is no prize in it. However
> you don't assume anything about his strategy of choosing. He might have
> chosen it randomly or always the empty one. He might even have chosen
> the empty with some probability, say 90% or anything. Now would you
> switch?
> Solution: In this case, as Reza Modarres mentioned, there is no correct
> answer, but it can be proved that if you switch, the chance of winning
>         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> is at least 1/2.
  ^^^^^^^^^^^^^^^
 
Sorry, I made a mistake here. I took Reza's word on it, but now I can prove
that in some cases, your chance will be more if you STAY with your first
choice. To do that, I will state "a special case" of problem #3, which is
a somewhat general case by itself. I'll call it problem #4. Problem #4 is
a special case of problem #3, but a general case of problem #1 and #2.
In other words, I will derive a general formula that gives BOTH results
of problem #1 and #2, and furthermore will prove that in some cases, you
better stay with your first choice, rather than switch. I'll also give
an intuitive explanation to this last result.
 
Problem #4:
Monty opens one of the remaining boxes according to a probabilistic
strategy as follows: He looks at the remaining two boxes. If the prize
is in one of them, he chooses the empty box with probability q. If both
are empty, he chooses one of them with probability r. Now he opens the
box that he has chosen and shows that it is empty (meaning that he is
lucky, because there is a non-zero probability that he opens the box
containing the prize). Then he asks whether you want to switch or stay
with your first choice.

Claim: 
In this case, if you stay with your first choice, the chance of
winning is r/(r+q).

Application to problem #1 and #2:
For problem #1, q=1/2 i.e. Monty makes an arbitrary choice. For problem
#2, q=1 i.e. Monty GUARRANTIES to choose the empty box. In both problems,
it is necessary to assume that r=1/2 i.e. when Monty knows that both of the
remaining boxes are empty, he chooses one of them WITH EQUAL PROBABILITY.
So by a simple numerical computation, we obtain the 1/2 probability of
problem #1 and the 1/3 probability of problem #2.

When is it better to STAY?
If r=1/2 and q < 1/2  then from the above formula, it is clear that if you
STAY with your first choice, your chance of winning is > 1/2
Intuitive explanation: Suppose q=0.1 which means that Monty makes a big
gamble, because with probability 0.9, he chooses the box CONTAINING the
prize! He opens the box and you see that it is empty! So you face one of
two possibilities: either the prize is not really in the remaining two
boxes, or Monty is really a very lucky gambler. You better assume that
he doesn't have any special luck!

Proof of the above claim:
The proof will follow in another article.

Seyyed.
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