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.

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    Are you just trying to verify your solution? – The Count Feb 17 '17 at 17:48
  • Sort of. I think this may be the solution, though I feel I have something going missing. – user278039 Feb 17 '17 at 17:54
  • Does Monty Hall know where the car is? Did Monty Hall open the door closest to him? – Bram28 Feb 17 '17 at 17:58
  • @Bram28 That's a good point, but we seem to be assuming that there is always a possibility that switching could win us the car, ergo they always reveal a goat before we make our choice. – The Count Feb 17 '17 at 18:00
  • @TheCount But what if Monty Hall is lazy and always opens the door without the prize closest to him? – Bram28 Feb 17 '17 at 18:02
  • That won't change anything. As long he doesn't reveal the car, ruining the premise of the problem, we still have exactly the same odds, regardless. – The Count Feb 17 '17 at 18:03
  • @Bram28 in class we assumed Monty hall knew the layout of what is where – user278039 Feb 17 '17 at 18:04
  • @user278039 You can simplify your reasoning in terms of counting: in picking 2 doors out of 4, the probability of getting the car with your initial choice is 0.5. – Théophile Feb 17 '17 at 18:26
  • @TheCount Wait, that doesn;t sound right. Let's assume that Monty Hall does indeed always open the door closest to him that does not have a prize. That is, let's supose we have doors A, B, C, and D in that order and that A is closest to Monty. Let's also assume that you picked doors B and D. Now, if at this point Monty open door C, then we immediately know that the prize is behind door A, otherwise Monty would have opened door A. But if Monty opens door A, all other doors now have a 1/3 chance. So if I knew Monty is lazy and he does open the door closest to him I would stick with the original. – Bram28 Feb 17 '17 at 18:32
  • Oh, I see what you mean @Bram. Sorry I was misinterpreting. Yes, we need to assume that Monty randomly chooses a door containing a goat. My mistake. – The Count Feb 17 '17 at 18:37
  • I meant that it doesn't matter if Monty knows, but just that *someone* does and tells Monty what to do in a random way, if the car has been chosen already. – The Count Feb 17 '17 at 18:39
  • @TheCount Yes, that's a good way of putting it! – Bram28 Feb 17 '17 at 18:43
  • @Bram28 Haha, I thought you meant that Monty had to be trying to trick the contestants. But all sorted out now. Good teamwork. – The Count Feb 17 '17 at 18:47
  • @TheCount No, just the opposite almost: he's just lazy! Much more realistic! :) – Bram28 Feb 17 '17 at 19:07
  • @Bram28 I can certainly relate to Monty, then. – The Count Feb 17 '17 at 19:12

1 Answers1


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!

The Count
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