## Ask Professor Puzzler

Do you have a question you would like to ask Professor Puzzler? Click here to ask your question!

This blog post is about the "Monty Hall Problem" - which you can find a blog post about here: Ask Professor Puzzler about the Monty Hall Problem, and a game simulation here: Monty Hall Simulation.

Larry from Louisiana writes: "Relating to the 3 door problem the answer given is just plain silly. Consider everything the same except you now have two players, one selects door one and the other selects door two. Now door three is shown to have a goat so, according to the given solution both players switch doors and each now has a 66% chance of winning? This may be possible in the new math but it is not in the old math that I learned."

Hi Larry, thanks for the message.

Before responding, I'd like to mention, in case anyone is confused by the "goat" part of your message, that in different versions of the game either two doors are empty, or they contain a donkey, or a goat, or some other silly prize. In the blog post here on this site I refer to the door as being empty, but to make my answer consistent with Larry's comment, I'll refer to the empty doors as containing a goat.

I'd also like to add that when I'm writing math problems for math competitions, my proofreader and I always comment - half joking (which means half serious!) that we hate probability problems, and wish we didn't have to write them. Probability problems can be very tricky, and it's easy to overlook assumptions we make that completely change our understanding of the problem.

In the case of the Monty Hall problem, there are a couple hidden assumptions that can be overlooked.

- Since the game show host
*always*shows the contestant an empty door, that implies (even though this is not always stated outright) that he*knows*which door contains the prize. His choice is dependent on his own knowledge. - Since only one door has been opened, the game show host
*has a choice*, and it is*always possible*for him to choose an empty door.

It may be helpful to think of the game show host as a "player" or "participant" in the game. If the game show host cannot operate under the same conditions within the game, it is not the same game.

The game you proposed is an entirely different game with conditions that violate the conditions stated above. Specifically, even though in your stated scenario the game show host shows a goat, we need to understand that it is not always possible for him to choose a goat door; if the contestants have each picked a goat door, there is no door left for him to show except the one with the prize. In other words, the game show host's role is completely different. While he still knows which door contains the prize, he no longer has a *choice*, and (just as importantly) it is not always possible for him to choose an empty door. Another way of saying this: in your game, the host is no longer a participant in any meaningful way.

Since your game has conditions that violate the conditions of the Monty Hall Problem, it is not the same game, and we shouldn't expect it to have the same probability analysis. In fact, it doesn't, and you correctly observed that it wouldn't make sense for it to work out to the same probabilities.

I would encourage you to try out the simulation I linked at the top of this page; if you're willing to trust that I haven't either cheated or made a mistake in programming it, you'll see that the probability really does work out as described. And if you don't want to put your trust in my programming (which is fine - I think it's good for people to be skeptical about things they see online!), I'd encourage you to get a friend and run a live simulation. The setup looks like this:

- Take three cards from a deck, and treat one of them as the "prize" (maybe the Ace of Spades?)
- You (as the game show host) spread the three cards face down on the table. Before you do, though, remember that you're going to have to flip one of the cards that
*isn't*the Ace of Spades, which means you need to know which card*is*the Ace of Spades. So before you put the cards on the table, look at them. - Now ask your friend to pick one of the cards (but don't look at it).
- Since you know which card is the Ace of Spades, you know whether he has selected correctly. If he's selected correctly, you need to just randomly pick one of the other two cards to show him. If he hasn't picked the Ace of Spades, you know which one
*is*the Ace of Spades, so you show him the other. - Now ask him if he wants to switch. You'll need to agree beforehand whether he'll always switch or never switch, but he should be consistent and make the same choice each time.
- Do steps 2-5 about 100 times. If your friend always chooses to switch, you'll find that he wins about 2/3 of the time, while if he never switches, he'll win about 1/3 of the time.

I've had classes of high school math students break up into pairs and run this simulation, and the results always come out as described above. Happy simulating!