Game Show: Let's Make a Deal & Applicability to Poker

[napawino]

I wanted to post a non-trivial game-theory situation, then ask about it's applicability to poker. I'll follow up in another post with the solution, and I'm interested in how to use this theory in my poker games.

For those people outside the US (or under 30), here is the situation:

Let's Make a Deal - US TV Gameshow in the 1970's
The game is very simple. There are three doors: door #1, door #2, and door #3. Behind one door is a new car. Behind the other two doors are two goats. All you have to do is pick which door you want to open, and you get whatever is behind it. But you only get to open one door. By simple math, then, you obviously have a 1 in 3 chance of picking the correct door and getting a new car.

You pick a door. As soon as you tell Monty (the gameshow host) what door you want to open, he stops and says,
Okay, you've made your choice. Now, I'm going to do what we always do here on this game. I'm going to open one of the other two doors for you that didn't choose.
Sure enough, he opens one of the other doors & the stands a goat. Then he asks, Okay, now, would you like to stay with your original guess, or would you like to switch to the other door that's still closed? You only get one shot, so do you want to stay with your original choice, or switch?

Your choice?

[Dan Mezick]

I've seen this, you have to switch doors because the odds changes when he gives you the new info, ie when he reveals what's behind one of the doors. It goes from 1 in 3 to 1 in 2.

[pudley4]

The answer is: Switch.

If you disagree, you're wrong. Don't post any explanation of why you're right, because you're not. Just search through the old posts for "Monte Hall".

[pudley4]

[ QUOTE ]
I've seen this, you have to switch doors because the odds changes when he gives you the new info, ie when he reveals what's behind one of the doors. It goes from 1 in 3 to 1 in 2.

[/ QUOTE ]

Not quite.

[pudley4]

Test it out here

Notice that you have to set the Host to "Knows" (as Monte obviously knows where the car is). There is a difference based on whether he knows where the car is or not.

[steamboatin]

you guys are smart but in a scary kinda way.

[napawino]

Pudley,
My apologies, but my searches on the forum didn't result in any hits, so I didn't know this had been discussed previously. I'll let everyone flame away before getting a thread started on how to use this in poker tomorrow. (Not that I have a clue how [img]/images/graemlins/confused.gif[/img] , but I want to hear other's thoughts.)

Napawino

[BruceZ]

Since they reorganized the forums, you now have to search the archives to retrieve anything over a few months old. Click on "older archives", then click the link at the top for the recent archives where the latest older stuff is located. Then if you do a search, you will find many Monty Hall threads since they flare up periodically. Here is a thread where I have posted some solutions, both intuitive, and also mathematical by Bayes' theorem:

Monty Hall thread

These are done under the assumption that Monty must always open a door which you did not choose, and one which has a goat. Note that it is not sufficient to simply state that Monty "knows" which door has the car, because if we assume that Monty plays optimal strategy, then he will only allow you to switch when you have picked the right door, and then your correct strategy would be to never switch. This is contrary to the accepted answer which is to always switch. That answer is correct if Monty must always allow you to switch, and he must always open a door with a goat that you didn't pick.