Fwd: Your version of the Monty Hall Problem

[Somdeb Lahiri]

Dear Friends and Colleagues:

Sharing an interesting paper and a dialogue on the "Monty Hall Problem" with you. As usual, comments would be most welcome.

Regards.

Somdeb.

---------- Forwarded message ---------
From: Keshava PRASADa Halemane <k.pra...@gmail.com>
Date: Mon, Aug 25, 2025 at 10:15 PM
Subject: Re: Your version of the Monty Hall Problem
To: Somdeb Lahiri <somdeb...@gmail.com>

namastE Somdeb Lahiri 
Thank you for your message. 
If you are interested, you may go through my entire paper -
My approach may seem to be somewhat different; 
If you have specific questions about my approach, I may be able to elaborate. 
I am in no mood to comment on your approach. 
All The Best.
KpH

On Mon, Aug 25, 2025 at 9:56 PM Somdeb Lahiri <somdeb...@gmail.com> wrote:

Dear Professor Halemane:

I read your discussion of the Monty Hall problem reported at the link below:

https://engrxiv.org/preprint/view/5102

It appears to me that regardless of who the DM is- the guest or the expert- the problem can be resolved in the following manner.

If y = x, then the guest wins with probability one by sticking to 'y' and hence the guest's optimal choice is 'y'.

If y is not equal to x, then the guest wins with probability 1/2 and loses with probability 1/2 if he sticks to y. If y is not equal to x, then the same win-loss probabilities hold if the guest switches his choice.

Assuming that the guest assigns Probability (y = x) = 1/2 = Probability (y is not equal to x), then the probability that the guest wins if he sticks to y is (1/2) + (1/2)x(1/2) = 1/2 + 1/4 = 3/4; and the probability that the guest wins if he switches his choice is (1/2)x(1/2) = 1/4.

Thus, if the guest wants to maximize his chance of winning, he should stick to y, provided he is neither pessimistic nor optimistic about his initial choice.

Regards.

Sincerely,

Somdeb Lahiri.

--

Somdeb Lahiri
https://sites.google.com/view/somdeblahiri

https://independent.academia.edu/LahiriSomdeb/Teaching-Documents 

https://figshare.com/authors/Somdeb_Lahiri/21580622

"Non-violent non-cooperation" is the most powerful weapon against "zulm"

Lecture Notes: Demand, Demand Curves and Consumer's Surplus

[somdeb...@gmail.com]

Post-Script: It may be more reasonable to assume Probability (y=x) = 1/3 and Probability (y not equal to x) = 2/3.

In that case, probability of winning by sticking to y is 1/3 + (2/3)(1/2) = 2/3 and probability of winning by switching from y is 1/3.

Once again, maximizing probability of winning requires "sticking to y"!!!
S.

[somdeb...@gmail.com]

Apologies for repeated corrections. I think the answer to the problem lies in answering the following two questions:

1) Why did the guest choose y to begin with?

2) Is there a change in the circumstances after the guest has been shown z neither equal to x or y?

The answer to the first question is because he evaluates Probability ( y = x) > Probability (y not equal to x).

The answer to the second question is "No, since he was shown z which is neither equal to y nor is it equal to x.

Thus, after seing z, Probability ( y = x) continues to be strictly greater than Probability (y not equal to x). Seeing z is redundant information.

Thus, if he wants to maximize his probability of winning, he should stick to y.

Thanks. 

[somdeb...@gmail.com]

Here is a link to a "formalization" of what is presented below:

https://drive.google.com/file/d/1p9HiQ72c-YVYIjWXJmyNWYNTWrrUpXME/view?usp=sharing