Now, suppose the GSH reveals a goat behind door 3. Modified Event Tree Representing the Probability GSH will 'R'eveal a Goat Note that sole the modification is that all P( G 1 | C i) become P( R 1 | C i) = 0. Since the GSH is now limited to telling us which of the other two doors contain a goat, we can represent the probability of the GSH revealing ('R') which of the other two doors have a goat with a similar event tree (shown below). Let's assume that we select door 1 in the hopes that a car sits behind it. Note that all conditional probabilities of $ P(G_i|C_i) \equiv 0 $ since the two events are mutually exclusive for the same door i. after the player selects his/her door, the game show host (GSH) will expose one door with a goat that the player has not chosenīelow you can see the event tree of cases, where 'C'=car and 'G'=goat. one door has car, other two doors have goats (three doors total)
The Monty Hall problem outlined in the above video is analogous to our in-class 'Prisoner's Dilemma', where in the Monty Hall case: car = prisoner is released, goat = prisoner dies.įor the following explanation of this 'problem'/'dilemma', I assume two things: Other students are welcome to comment/discuss/point out mistakes/ask questions too! You will receive feedback from your instructor and TA directly on this page. Then use the concept of conditional probability to explain why switching door leads to a probability of success equal to 2/3. Practice Problem on Conditional Probabilityįirst, watch this video explaining the Mounty Hall Problem 1 Practice Problem on Conditional Probability.
0 kommentar(er)
