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I'm solving a single-choice question in the exam. The question has $4$ options: A,B,C,D. These are my steps:

$1$. I didn't understand any choice of the four, so I randomly selected one: A.

$2$. After $1$, I figured out that D is wrong, according to the conclusion of Monty Hall problem, I change my answer, randomly chose one between B and C, suppose B. I should change, right? :) to increase probability from $1/4$ to $3/8$.

$3$. After $2$, I figured out that C is wrong, so should I change back to A? And what's the probability of each one? What if I figured out A instead of C wrong?

Thanks!

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No, your deductions about other wrong answers do not change the probability that your original randomly selected answer is correct.

A crucial piece of information about the Monty Hall problem is Monty's algorithm, which is assumed to be "open a door without the prize behind it". In an alternate universe, imagine you are the contestant who has just selected Door #1. Then, by looking in a well-placed mirror, you figure out that there's no prize behind Door #3. In this situation, there is no value for you switching from Door #1 to Door #2: both are equally likely to contain the prize.

If a game show host had been obligated to tell you that D was a wrong answer, then yes, you should switch from A to B or C. But if you simply figured out that D was wrong, then there's no value in you switching.

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    $\begingroup$ Indeed. In fact, the Monty Hall situation requires "unbiassedly openning a door without the prize behind it, but also certainly not openning the door you are selecting even if there is no prize behind it". $\endgroup$ – Graham Kemp Jul 11 '19 at 10:17
  • $\begingroup$ Thanks ! I think I get it. The key factor is the independence. My deduction about the wrong answer was not affected by the first random choice, while the TV host choose the door with goat among the doors the guest didn't choose. $\endgroup$ – Charing Lau Jul 12 '19 at 7:01

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