We can give a complete answer using bayes rule a

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We can give a complete answer using Bayes’ rule. A person’s original choice is car B Monty shows a goat behind one of the other doors 69
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Assuming a fair game, we know that P A 1/3 and P A c 2/3. We want P A | B . To use Bayes’ rule we also need P B | A P Monty shows goat|person intially chose car p P B | A c P Monty shows goat|person intially chose goat q (Note that there is no necessary relationship between p and q .) Then by Bayes’ P A | B p  1/3 p 1/3 q 2/3 p p 2 q 1 1 2 q / p . 70
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If p q 1, so that Monty always shows a goat behind one of the other doors – the initial formulation of the problem – then P A | B 1/3 P A This fits with our intuition: if Monty shows a goat no matter what there is no information, and so it cannot affect the probability that the initial choice was a car. 71
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Here is the important point: Monty’s always showing a goat behind one of the unselected doors does affect the probability that the other unrevealed door has a car. Why? If P A | B 1/3 then P A c | B 2/3. But now one of the two doors we did not select has been eliminated. Therefore, changing to the unselected door that has not been revealed increases the probability of winning the car from 1/3 to 2/3. 72
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Get the same answer if p q for any p 0: if Monty sometimes does not reveal a goat, but does not distinguish between wether the person initially chose the car or goat, then the chance for the inital door is still 1/3, which means the chance for the unselected, unrevealed door is 2/3. So one should switch. But suppose, say, that p 1 and q 1/3: Monty always reveals a goat if the initial choice was a car, but only one-third of the time if it was goat. Then P A | B 3/5 1/2, and so the contestant should stay with the initial choice. One is indifferent if, say, p 1, q 1/2; then P A | B 1/2 P A c | B . 73
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Bayes’ rule has an extension when is partitioned into many events. THEOREM :Let A 1 , A 2 ,... be a partition of . (Could be finite or infinite) and let B be any event with P B 0. Then for each i , P A i | B P B | A i P A i j 1 P B | A j P A j 74
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6 . Independence In the Monty Hall problem when p q , the probabilities P A | B and P A are the same. That is, knowing the outcome on B does not affect the chance that A occurs. Using the definition of conditional probability, P A | B P A is equivalent to P A B P A P B This same condition holds when P B | A P B , too.
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We can give a complete answer using Bayes rule A persons...

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