Please settle an argument for me

[djoyce003]

some friends and I have a probability argument....tell me what the proper decision is here

There are three doors. Our hero has to pick one of the doors. Only one door is the right door. After our hero picks a door, a mystical force then removes one of the two remaining doors from the realm of possible choices. If our hero has chosen wrong, and the correct door still remains the force can never remove the correct door. If our hero has chosen right, it still removes a door

Our hero then has the choice of keeping his original choice, or switching to the door that remains...what should he do?

[gaming_mouse]

switch.

search for "monty hall" for many explanations.

[olavfo]

He should switch.

With 3 doors to choose from he has 1/3 chance of picking the right door, so it's a 2/3 chance he is at the wrong door. This is also true after one of the other doors is removed. Therefore he should switch since this will place him at the right door 2/3 of the times this choice is made.

Let's say he wins $1 if he ends up with the right door and loses $1 when he ends up with a wrong door. He decides to always switch after one door is removed. The EV equation for swithing is

(2/3)*$1 + (1/3)*(-$1) = $0.33

2/3 of the time he switches from wrong to right door and wins $1. 1/3 of the time he had the right door and loses $1 by switching. On average switching wins him $0.33.

olavfo

[djoyce003]

that's what i kept saying but nobody agreed with me.

[AngusThermopyle]

Tell your friends to look at it this way.
They can pick just one door or pick two doors.
Which would they prefer to do?

Pick door #3 only, or pick both doors #1 and #2?

By always switching, they are getting to pick two doors ( ie the two that they do not pick if they only pick one ).

[NiceCatch]

Or try this explanation.

When you pick a door, say door A, it only makes sense not to switch from that door when the prize is behind it. The prize will be behind it 1/3rd of the time. Therefore it will be behind one of the other doors 2/3rds of the time.

Regardless of which door is removed, or is revealed to have the goat behind it, or whatever, the chances that the prize was behind one of the other two doors is 2/3. Therefore you should switch, and the fact that a door was opened to reveal nothing behind it only helps facilitate which of the other two doors to choose.

Interestingly, some fellow actually went back through all the old Let's Make a Deal footage and found that the people who switched were in fact rewarded two-thirds of the time, and the people who didn't, were rewarded one-third of the time.

[NiceCatch]

Here's another way to look at it: empirically.

In this first example, the contestant picks door A, and doesn't change his decision. There are three possible ways the goats and car could be arranged. Obviously, the contestant wants to win the car. I will put an (r) next to the door that Monte opens, though it doesn't really matter, since the contestant decides not to switch.

Door A Door B Door C Result
car goat(r) goat Wins car
goat car goat(r) Loses
goat goat(r) car Loses

Obviously, the contestant wins one in three times. Now lets look at if the contestant switches to the other door (it should be obvious, since one time he'll switch from the car, the other two times he'll switch from the goat; but for the sake of thoroughness). Let's say he initially picks Door A, then switches to whichever door Monte did not open. We'll say (r) is the door Monte reveals, and (s) is the door the contestant switches to.


Door A Door B Door C Result
car goat(r) goat(s) Loses
goat car(s) goat(r) Wins car
goat goat(r) car(s) Wins car

As we guessed, switching doors results in winning the car two out of three times. I'd suspect this is the best way to explain it to your friends, since the concept is seemingly unintuitive.

By the way, sorry about the formatting... the font is not monospaced. Basically there are four columns, and four rows for each table.

[smarterthanyoda]

Several other posters have given the correct answer already, but here is one example that I have used for people who don't really "get" the math:

What if instead of 3 doors, there were fifty doors. You choose one of the fifty, and then 48 more are opened to show they don't have the prize. Would you switch to the one door that they didn't open?

[NiceCatch]

Wow... that's a good one. Nice post.

[djoyce003]

actually i tried that too...except i used 100 doors...but they said that was different...LOL. These guys were amazing, and then they told me they hoped i didn't play poker using my flawed math....one of them was a freakin math major which blows me away.

[Paul2432]

Here is what you should do. Get three playing cards, a queen of spades and two aces. He will deal the three cards face down. You select a card and then he flips up a an ace.

Now make a bet with him. Tell him "I bet you $5 I can find the queen. I am so sure, if I am wrong I will pay you $6." Your friend should jump at this proposition. In fact, you make $1.33 per deal.

Repeat until your friend is broke.

Or you could just do the experiment without the bets. It should be obvious fairly quickly that your friends are wrong.

Paul

Edit: One small caveat. Make sure you randomize which card you pick. For example, if you always picked the first card, your friend would always deal the queen to the first position and you would lose. The best way to do this is to roll a die to pick.