Apparently I'm too stupid to believe in the "Monty Hall problem"

[mreilly]

Call me a blockhead, a pighead, or an idiot, but I remain forever skeptical that switching your choice of doors has any bearing whatsoever on your chance of winning (and yes, the article I reference below does discuss how some people like me refuse to recognize the validity of its claims due to an apparent inability to understand the logic, but I still don't agree with it regardless of how many diagrams and illustrations are thrown at me):

http://en.wikipedia.org/wiki/Monty_Hall_problem

As far as I'm concerned the only thing remotely convincing about this argument is the "Why would someone go to all this trouble to convince me of something if it weren't true?" question....

[LynneSin]

1. Yes but do you subscribe to the Caine-Hackman Theory?

http://en.wikipedia.org/wiki/PCU_(film)#Caine.E2.80.93Hackman_Theory

[Tikki]

5. Here in California it's been following the Schwarzenegger theory..

for the last 5 years or so...mercy on us, I hope it ends quickly with
his departure.


Tikki

[gratuitous]

2. Gullible, thy name is mreilly

You probably think flipping the switch makes the light go on and off, too. Theory of electricity, my sweet Aunt Fanny!

[rug]

3. Any show that rewarded you for dressing like a pickle has lost any appeal to logic.

[Fire Walk With Me]

4. Hanging my head in shame as I put the pickle costume back in my closet.

[rug]

7. Was that you?

[Fire Walk With Me]

8. I won NOTHING.

Yes, they get Gene Wilder to tell me. Highly embarrasing.

[pokerfan]

6. There's an app for that

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html">

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

[zipplewrath]

9. I'm not sure which you are claiming

Do you think their wrong and you just can't prove it?

Or do you think their right and you just can't understand it?

That pretty much would seem to be your only two options.

In the game, there are also only two possible outcomes, you win, you lose. There is no other possible outcome.

Let's start whether you accept that. Do you?

Then, the game has the further reality that they are going to open a door that won't have the winner behind it. They'll pick the one without a winner behind it every single time.

So, when they expose one of the doors, the game is now a new game. In that game there are two doors, the one you picked, and the one you didn't pick. In that game, is there anyway to guess which is the right door? Remember, it's a new game now. There is the door you picked, and the door that isn't open. In essence, the third door is meaningless at this point, there are only two doors.

But here is what you DO know. When the game first started, with 3 doors, you were more likely to pick the wrong door than the right door. There were 3, only one winner, and you could only pick one. So, most likely, you picked the wrong door. That hasn't changed. What has changed is now you don't have 2 choices any more, you only have one.

Pick it. You know you were most likely wrong. Here's your chance to "fix" it.

Will you always win? Nope. But that's the way to bet.

[mreilly]

10. Both, in a way

Keeping with the 50/50 theme, I think either they're wrong and I can't prove it or they may be right and the explanation doesn't make sense to me (and I've spent a good part of the afternoon combing the web - there seem to be TERABYTES worth of posts, diagrams, illustrations, explanations, theories, arguments and analogies about this, none of which has convinced me - it truly seems to be a religious discussion with neither side capable of convincing the other). I do accept that in the game there are only 2 outcomes: win or loss. In fact, that's what seems to fuel my "50/50" argument.

Here's the crux of the issue to me:

-I am told by proponents of the "switching doors gives you a 2/3 chance of winning" theory that there is a 100% chance the prize is behind one of the three doors.

-In this scenario, I pick door #1, which gives me a 1/3 chance of winning.

-Monty opens door #3 to reveal no prize there, which means door #3 now has a 0/3 chance of winning.

-Supporters of the theory claim door #2 therefore has a 2/3 chance of winning since the 1/3 chance represented by door #3 must "go elsewhere" if I am reading this opinion properly (a lot of analogies to 52 cards in a deck, 100 doors being used instead, or even 1,000,000 choices being reduced to 2 come into play here, and again I find these utterly irrelevant).

I don't see how or why that 1/3 "floating" chance has to go to door #2. In my view, Monty took away 1/3 of the doors and the chance that I picked the right door is always going to be 50/50 since it's down to 2 doors.

[pokerfan]

12. If the door was opened at random

ala Deal or No Deal, then there would be no advantage to switching. Of course, if the door is opened at random, then there's also the chance (1 in 3 probability) that the prize is revealed and the game is over. In the DoND case, you have a 1/3 chance of being right, a 1/3 chance the host reveals the prize and a 1/3 chance the last door has the prize. So providing that the random reveal didn't expose the prize, then yes, your odds at that point would have improved to 50:50.

But in Let's Make a Deal, the host 1) knows the location of the prize and 2) will always reveal one of the other doors. By always eliminating a losing door (there's always going to be at least one unchosen losing door) he's not really improving your odds of your being right with your first choice.

Look at it this way. You choose a door and then the host gives you the option of either keeping your door or switching to both of the other doors. You would probably switch as those two doors have double the chance of containing the prize. The "reveal" essentially accomplishes the same thing except the probability of the doors is condensed into the door that wasn't revealed. The odds of you being wrong with your first choice remains 2/3.

Or this way: We'll use a deck of cards and the Ace of Spades will represent the prize. You will pick a card and set aside face down. I will go through the rest of the deck and reveal all the cards except one. At this point there are two cards left, your original choice and mine. Are you sure you don't want to switch?

[zipplewrath]

14. It doesn't "have" to go "elsewhere"

Remember, they DON'T open the door with money behind it. The ONLY open a loser door. When you pick one, the combined probability of it being ONE of the two "other" doors is 2/3. The knowledgeable person picking that door knows which one it is. So the probability of it being the other door is STILL 2/3. The only chance that it is NOT the other door is if it is the one you picked, which STILL has a 1/3 chance.

The trick here is the "knowledgeable observer". When they expose one door, they don't really tell you anything about the "other" door. The knowledgeable observer WON'T expose the winning door, so he hasn't yet told you anything about the other door. Therefore, the 2/3 chance "still" exists for that combination of two doors.

[Dr. Strange]

11. It has bearing...

because IF you switch, then you are taking advantage of extra information (namely the information that Monty Hall is giving you by revealing one of the losing doors).

To make it more intuitive, imagine it this way: suppose there are 100 doors, and you pick one (say #20). Monty then opens 98 doors (say #1 - 19, 21- 48, and 50 - 100), and asks if you want to switch (which means switch to door #49). Should you switch?
Don't switch--you're betting that #20 is the one with the prize, which has a 1% chance of being right.
Switch--you're betting that the prize is not in #20, which has a 99% chance of being right.

[cemaphonic]

13. It wouldn't be such a classic puzzle if it wasn't so counterintuitive.

And it is often presented incompletely (usually leaving out that Monty knows the contents of each door and is *required* to open an empty one), which changes the answer. I think the best way to understand it has been posted already - when you make your initial choice, you have 1/3 chance to get it right. Monty is essentially giving you a choice to either stick with your original door, or open *both* of the other doors, for a 2/3 chance.

[zipplewrath]

16. That's pretty near perfect

"Monty is essentially giving you a choice to either stick with your original door, or open *both* of the other doors, for a 2/3 chance."

This is the most perfect explanation I've seen so far. Better than any of mine ever were. The only addition I'd ever make is to remind everyone that in the end, you can STILL lose, because there is a chance you've already picked the "right" door.

(In fact, in some arguments, folks have made the assertion that Monty would only offer you this option if you HAD already picked the right door, so you should NEVER switch. I think the historical record reflects that Monty wasn't that consistent.)

[begin_within]

15. The solution is to bribe Carol Merrill to tell you which door.

[zipplewrath]

17. Ha anyone done this for Deal or No Deal?

It's a far more complicated version, but it is basically the same thing. They open doors and you have greater and greater knowledge about what remains. What always occurs to me is that you have roughly the same problem. There are some "prizes" that will be "good" to win. The odds are pretty low thought that your one case will be that one (well somewhere around 20% or so depending upon what you believe to be a "good" number. It would seem that at some point, enough of those are "gone" to justify taking any "deal" that's offered in that range of "good" numbers.

[pokerfan]

18. see post 12. (nt)

The key difference is that in DoND, the "doors" are opened at random and in LMAD, the host opens the door. That makes all the difference.