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

Question

4. Suppose you change the game in the Monty Hall problem, so that there are 4 closed doors (instead of 3), concealing 1 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 doors, 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

let's say you picked the door with the "goat". you have a 1/4 chance of doing this. in this case, switching gives you a 1/2 chance of winning the prize.
let's say you picked the door with the prize. you also have a 1/4 chance of doing this. in this case, swtiching gives you a 0 chance of winning the prize.
finally, let's say you picked a "empty" door. you have a 1/2 chance of doing this. in this case, switching gives you a 1/2 chance of winning the prize.

so you total chance of winning the prize is: (1/4)(1/2) + (1/4)(0) + (1/2)(1/2) = 3/8.

now let's analyze the "don't switch" strategy:

you pick the door with the "goat (booby prize)". you have a 1/4 chance of doing this, and in this case, you never win.
you pick the door with the prize. you have a 1/4 chance of doing this, and in this case, you always win.
you pick an empty door. you have a 1/2 chance of doing this, and in this case, you never win.

total chance of winning the prize is: (1/4)(0) + (1/4)(1) + (1/2)(0) = 1/4.

it's still better to switch.

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