Wednesday, September 10, 2008

Of 21 and Monty Hall

Recently, I saw the movie "21" (with Kevin Spacey and a bunch of young actors I didn't recognize). Early in the movie, there is a Kevin Spacey talks about the classic Monty Hall - a problem that almost everyone gets wrong. It goes like this.

Imagine you are playing the classic TV game show "Let's Make a Deal." You have three doors to pick from. Behind one door is a brand new car. Behind each of the other doors is a goat. Monty let's you pick a door. Not knowing where the car is, you pick a door at random, say door #1. Monty (who knows where the car is) opens door #3, revealing a goat.
He then gives you the option of either keeping your pick or choosing door #2 instead. What should you do? If are like me (and almost everyone else), you would think "It doesn't matter. The car could be behind either door so changing my mind doesn't improve my chances." And like me, you would be wrong. You should ALWAYS choose the other door.

The answer defies common sense. It is completely counter intuitive. It is also correct.

When I saw that scene, I got it wrong as well, but I was so convinced that I was right that I went to the all-knowing Wikipedia for the answer. They have their own explanation as well as few examples to try and explain it. In trying to understand it as well as explain it to some of my friends (who were equally unbelieving), I came up with my own example that, I think, better explains what is going on.

Let's use the same game but with something more familiar. Imagine you are playing a game with three cards - an Ace of Spades, a 2 of Hearts and a 2 of Diamonds. The dealer deals one card face down to you and then deals the other two cards face down to himself. He looks at his two cards and flips over one of them, always a red 2. (The rules state that he must do this.) Then he gives you the option of keeping your card or switching with his remaining down card. What should you do?

Keep in mind that before the cards were dealt:
  • the chances of the dealer getting the Ace were 2 in 3 because he gets 2 of the 3 cards
  • the chances of you getting the Ace were 1 in 3
  • the chances of you getting a 2 were 2 in 3
  • the chances of the dealer getting a 2 were 3 in 3 (there is only 1 ace and since he gets two cards, the other card must be a 2)
Before he flips his card, you already know that one of his cards is going to be a 2. Therefore, when he shows you a 2, it CHANGES NOTHING. The chances of his other card being the Ace ARE STILL 2 IN 3.

So, do you want to change cards now? I thought you would.

If you don't believe that explanation, here's another more rigorous (read "anal retentive") way to prove it. The surest way to explain something in statistics is to list every possible outcome and compare the number of outcomes that succeed vs the number of outcomes that fail. That's not as scary as it sounds. That's exactly what you do when you flip a coin. Think about it.

Here is every possible outcome.

Scenario 1:
Dealer gets an Ace and a 2.
You get a 2.
Dealer flips his 2.
You trade and get his Ace.
You win.

Scenario 2:
Dealer gets a 2 and an Ace.
You get a 2.
Dealer flips his 2.
You trade and get his Ace.
You win.

Scenario 3:
Dealer gets a 2 and a 2.
You get an Ace.
Dealer flips one of his 2's.
You trade and get his other 2.
You lose.

There are only 3 possible scenarios. In 2 of them you win. In 1 of them you lose. It can't be much clearer than that.

1 comment:

Clint Memo said...
This comment has been removed by the author.