Question about the movie 21

Just switch the damn doors and you will be right most of the time.

 

But you will never forgive yourself if you were right and switched.

 

If you wanted to redo the bet after the third door is opened, then your odds are changing.

 

I cede to Calc-Boy.

Doesn't know shit elsewhere, but he has this cold.

 

Just like poker. You fold that gutshot straight draw if the pot odds don't figure a call, but you always wince when it hits. But you will be a more consistent winner if you fold.

In this goat problem, yes, you will be wrong a third of the time. But speaking mathematically, you have to switch.

Take three cards: Ace of spades, 2 of hearts and 2 of diamonds.

If you have the Ace of spades at the end of the problem, you win $100.

Now, I shuffle the three cards (without looking at them) and deal you one. You had a 66.67 chance to get a red 2, yes?

Now, I take the remaining two cards, look at them, and show you a red 2 (you haven't looked at your card). You switch because the first time you picked the odds were against you picking the black Ace ... now they are in your favor because you know the other card.

 

The key is that Monte has to give up one of the wrong answers. That's what switches the weight of probability.

 

But how does it make it go to 66 percent if you switch? That's the part that I don't get is having a higher percentage of being right if you switch. What if the car is behind the first door? (And why is it 1 out of three when you already see the door? Why isn't it a 50-50 chance? This is really showing how much I retained from statistics junior year.)

 

Here's how I understand it, correct me if I'm wrong.

Your initial door choice only has a 33% chance of being correct. That means it has a 66% chance of being wrong, meaning there's a 66% percent chance one of Monty's doors has the car. On that we can all agree and understand, yes?

Well, when Monty opens his goat door, that takes that door out of the original equation. Meaning that of the original equation, you had a 33% chance of being right and a 66% chance of being wrong, but now that 66% is stuck onto Monty's one remaining door. So looking at it on a door-by-door basis, your door has a 33% chance and Monty's one remaining door has a 66% chance.

Switching doors, then, is the best idea.

I had never thought about it like that until i read that wikipedia page, but it kinda makes sense. Although I can't imagine who would have had the time or brains to figure it out.

 

Because every time you do it, there is only 1 correct answer (car). It's on Monte's side that he probably has the car since you have only 1/3 chance of picking it right to start with. Then Monte gives up a goat -- the odds, going back to the beginning of the problem, favored Monte, so if you switch to Monte's closed door, you will be right 2 out of 3 times.

 

The graphic and flow chart on the Wikipedia entry helped me understand it the best.

 

Let's just run through the game.
No matter what door you pick, you're going to start with one of three things: Goat 1, Goat 2 or The Car.
So a run through each scenerio, and remember, Monty will always show you a goat. Said another way, he'll never show you a car.

So first, assume you always switch.

1) You start with Goat 1. Monty shows you Goat 2. The Car remains hidden. You switch. SUCCESS.
2) You start with Goat 2. Monty shows you Goat 1. The Car remains hidden. You switch. SUCCESS.
3) You start with The Car. Monty shows you Goat 1 or 2 (doesn't matter). The other goat remains hidden. You switch. FAILURE.

You're successful 2/3 of the time.

Now, if you don't change.

1) You start with Goat 1. Monty shows you Goat 2. The Car remains hidden. You don't change. FAILURE.
2) You start with Goat 2. Monty shows you Goat 1. The Car remains hidden. You don't change. FAILURE.
3) You start with The Car. Monty shows you Goat 1 or 2. The other goat remains hidden. You don't change. SUCCESS.

Since you obviously have no clue what you start with, the smart play is to change. Becuase your odds of winning are 2/3 instead of 1/3 if you keep your first choice.

 

Best explanation I've seen.