Question: In a variation of the Monty Hall game show, you now have 4 doors and only one has a car behind it. The other 3 doors have goats. This time you choose (without opening them) 2 doors. Monty Hall opens one of the remaining doors and shows that it has a goat. He then asks you “If you keep your two choices and the car is behind one of the two doors you chose, you win. But you can give up your two choices and open the remaining door and win the car if it is there.” In order to maximize your chance of winning the car, should you take Monty’s offer or not?
My Attempt: Initially we have ${{4}\choose{2}} = 6$ ways of choosing 2 doors. 1) We have 3 ways of choosing two doors so that we are on door with car. If we switch, we lose. 2) We have 3 ways of choosing two doors so that both of the doors have goats. If we switch we win.
Since we are equally likely if we switch or do not (in light of the equal number of ways we can choose doors so that we have the car or do not have the car), it does not make any difference, probability-wise to make the switch.
Your solution is correct, and has proceeded exactly the way the demonstration of the normal Monty Hall problem works.
If you want to improve it, a diagram or list of the possible outcomes might make your answer clearer, but it is perfectly clear to me. Very nice!