Let O be the event that Monty Hall opens door b Now the question is Is PAO PCO

# Let o be the event that monty hall opens door b now

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Let O be the event that Monty Hall opens door b. Now the question is: Is P(A|O) ≤ P(C|O)?

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The Monty Hall problem (5) According to Bayes’ rule (definition of conditional probability), it holds that: 𝑃𝑃 𝐴𝐴 𝑂𝑂 = 𝑃𝑃 𝑂𝑂 𝐴𝐴 ∗ 𝑃𝑃 ( 𝐴𝐴 ) 𝑃𝑃 ( 𝑂𝑂 ) 𝑃𝑃 𝐶𝐶 𝑂𝑂 = 𝑃𝑃 ( 𝑂𝑂𝑂𝐶𝐶 ) ∗ 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝑂𝑂 ) We know that P(A)=P(B)=P(C)=1/3 Hence, all we need to know are the (conditional) probabilities for the event O.
The Monty Hall problem (6) What are the conditional probabilities for the event O (i.e., the event that Monty will open door b)? Remember 1: we assume we have picked door a. P(O|A)=1/2 (if the car is behind a, then Monty opens b or c.) P(O|B)= 0 (if the car is behind b, then Monty never opens b.) P(O|C)=1 (if the car is behind c, then Monty will open b.) What is left is the unconditional probability P(O).

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The Monty Hall problem (7) To calculate P(O), note that the prize is only behind one door (i.e., A, B, C are disjoint events), hence: P(O)= P(O and A) + P(O and B) + P(O and C) The general multiplication rule gives: P(O)=P(A)*P(O|A)+P(B)*P(O|B)+P(C)*P(O|C) P(O)=1/3 * ½ +1/3* 0 +1/3* 1 P(O)=1/2
The Monty Hall Problem (8) Insert the numbers into the definition of conditional probability: 𝑃𝑃 𝐴𝐴 𝑂𝑂 = 𝑃𝑃 𝑂𝑂 𝐴𝐴 ∗ 𝑃𝑃 ( 𝐴𝐴 ) 𝑃𝑃 ( 𝑂𝑂 ) = 1 3 1 2 1 2 = 1/3 𝑃𝑃 𝐶𝐶 𝑂𝑂 = 𝑃𝑃 ( 𝑂𝑂𝑂𝐶𝐶 ) ∗ 𝑃𝑃 ( 𝐶𝐶 ) 𝑃𝑃 ( 𝑂𝑂 ) = 1 3 1 1 2 = 2 3 That’s it! So, when you have picked door #1, always switch to door #3 when door #2 is opened.

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Back to last week: MC-answer guesses Last week, we have guessed MC- answers. By calculating the sample space and calculating the number of possibilities for a certain outcome we were (in principle) able to calculate the probability of an event. E.g. S=4^8=65536 P(X=8) = 1/65536 The probabilities of these events can be calculated in a more efficient way using the binomial setting.
Binomial probabilities (1) Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of n trials. The observations must meet these requirements: The total number of observations n is fixed in advance. The outcomes of all n observations are statistically independent. Each observation falls into just one of two categories: success and failure. All n observations have the same probability of success, p .

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Binomial probabilities (2) The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p: B ( n,p ). The parameter n is the total number of observations. The parameter p is the probability of success on each observation. The count of successes X can be any whole number between 0 and n .
MC-answer guesses (once more) Binomial probabilities can be calculated as follows: So, in the coin toss example of last week: k n k p p n k k X P = = ) 1 ( ) ( 31 . 0 75 . 0 * 25 . 0 * 2 8 ) 2 ( 6 2 = = = X P X 0 1 2 3 4 5 6 7 8 P(X) 0.10 0.27 0.31 0.21 0.09 0.02 0.003 0.0003 0.00002

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