In: Economics
consider the Monte Hall problem as discussed in lecture. Recall that a game show host(Monte Hall) gives a contest a chance to choose from three doors of which one is a new car and the other two are goats. After the contestant chooses a door, the game show host who knows what is behind all of the doors decides to open another door behind which a goat sits. The games show host. then, offers the contestant an opportunity to switch doors. For the purposes of this question, note the following:
1. The contestant initially selects door B
2.Game show host (Monte Hall) open door C behind which a goat sits
also note the following notation for events
1. Car A= the car is actually behind the door A
2. car B= the car is actually behind the door B
3. car C= the car is actually behind the door C
4. A= the host reveals a goat behind Door A
5.B=the host reveals a goat to be behind Door B
6.C= the host reveals a goat to be behind Door C
Given that the host reveals a goat behind Door C, what is the (conditional)probability that the car is actually behind door A [P(Car A| C)]?
The contestant selects Door B and there is a probability of the car being behind any of the doors of 1/3.
However, the probability of a goat behind each of the doors is 2/3
Now the door C is opened and a Goat is revealed.
Initial Choice | Prize Behind | Door Opened | Switch | Don't Switch |
A | A | B or C | Lose | Win |
A | B | C | Win | Lose |
A | C | B | Win | Lose |
B | A | C | Win | Lose |
B | B | A or C | Lose | Win |
B | C | A | Win | Lose |
C | A | B | Win | Lose |
C | B | A | Win | Lose |
C | C | A | Lose | Win |
From the above table we see that the number of cases when the contestant chooses door B and door C is opened is 2.
Out of those 2 cases, the car is behind A in only of the cases.
So the probability of the car being behind A while not being behind C is 1/2.
In general, if the contestant switches, he wins 6 out of 9 times or 66.67% times and if the contestant does not switch he wins 33.33% times as per the above table.
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