You are taking a multiple choices question test. The choices are A, B, C only with only one right answer. Unbeknownst to the examiner, you chose A. The examiner then truthfully reveals that the answer is not B. To increase the probability of getting the answer right, should I switch my answer from A to C?
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Don't you think Answer should be "TRUE" This question is same as Monty Hall problem
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The difference between this problem and the Monty Hall problem is that, in this problem, the examiner does not know your initial choice, meaning that the information he provides is unbiased, while in the Monty Hall problem the host does know your initial choice, meaning that the information provided in this case is biased. It is the bias in the Monty Hall problem that leads one to switch choices in order to improve ones' odds of winning.
The difficulty here for me is, I may have been very certain that A was the right answer to begin with. There should be no assumption that I was guessing randomly!
If that was the point of the question, it should be stated explicitly.
This problem is same as Monty Hall problem. So the answer should be true.
At first, when we are given three choices, then the probability of answer getting right is (1÷3) and of getting wrong is (2÷3). Now I ticked A, examiner came and said B is not right, now I am left with two option A and B.
Now the probability of getting the answer right/wrong is=(1÷2). And I have already ticked B, even if I switch over to C it will hardly make any difference, since both the probability I.e. of getting right and wrong, still remain the same.
However, since the probability of getting it wrong the first time is 3 2 , and the other wrong answer is revealed to you, surely then there is a 3 2 chance of getting it right by switching since the probability of B being correct answer is 0?
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Actually, after you got to know that B is not the correct answer, its probability shifts equally to A and B. The probability, that you are saying is (2÷3), it does not works like this, if the scenario changes, the probability changes too according to the scenario. Like a conditional probability, where the changing the condition will definitely change the probability.
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Ok thanks to you both, I understand now. The title is a twist to the Monty Hall Problem since the teacher does not know what you selected.
The point here is that the teacher didn't you selected A. If the teacher had known, then they would have revealed B with probability 1 if A was also incorrect, and with probability 1/2 if C was incorrect, so it is reduce as likely for A to be incorrect as C. However, in this problem the teacher didn't know, so their choice of B didn't depend on whether A or C was incorrect, so both are equally possible.
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Since B is definitely incorrect , both A , C have the same probability of being correct which is 2 1 . So it doesn't matter which of A , C we choose because they both have same possibility of being correct or wrong.