Suppose you change the game in the Monty Hall problem, so that there are 4 close

Question

Suppose you change the game in the Monty Hall problem, so that there are 4 closed doors (instead of 3), concealing l car and 3 goats. As before, you select one door initially. Now the host opens two of the other doors, revealing a goat behind each one, i.e. there still two doors closed, including the one you initially selected. If you switch your choice of doo what is the probability of getting the car? Explain your thinking. [If you need some experimental inspiration, you can simulate the 4-door version at http://onlinestatbook.com/2/probability/monty hall demo.html]

Explanation / Answer

Suppose the contestant opts for door 1, then following are the possibilities.

If the Car is behibd the door1, then contestants wins the car if he/she does not switch.

If the Car is behind the door2, door3 or door4, then contestants wins the car if he/she switches the option of door 1 as the car is not behind the door1.

So, out of 4 possibilities, there are 3 number of times, where the contestant wins the car if he/she switch the options of the doors.

So, the probability of getting the car in switching the options = 3/4

Behind door 1 Behind door 2 Behind door 3 Behind door 4 Result if staying at door #1 Result if switching to the door offered Car Goat Goat Goat Wins Car Wins Goat Goat Car Goat Goat Wins Goat Wins Car Goat Goat Car Goat Wins Goat Wins Car Goat Goat Goat Car Wins Goat Wins Car

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