A little Monday morning math.

I agree with you. Initially, three doors leads to a probability of 1 in 3 for each door. Person chooses “Door 1”, with a probability of 33.3% that the car is behind it. When “Door 3” is opened, to reveal no car, it rules out “Door 3” from the probability problem. Now, in essence, the contestant has two choices: Door 1 and Door 2. The probability is 50% for each door.

This has become known as “The Monty Hall Paradox”. Here’s a good exploration of the problem. Take comfort. Better minds than ours have been thrown by this one

@Les: That’s what I said. But that’s not the case. I was told “You can’t take that door out of the equation just because you know it’s wrong… it’s the law of probability… the third door still factors into the equation.”

Ok, after reading about it (thanks Harp), I completely neglected the fact that the gameshow host’s choice for which door to open is based on the contestant’s choice. If the opening of the door after the contestant chose his door was random, then I think my first statement is still true. But, the game show doesn’t want the player to win the car, therefore their choice of which door to open is based on what the contestant chooses, thereby changing the probability. Guh. I guess math has bested me today. My statistics minor is in shambles. :-(

What if I want the goat? Can I still pick door 3?

Noooooo. Whoever lurved me for my first answer, it’s wrong. I don’t deserve the lurve.

@Les: But as my boyfriend pointed out… it’s a perfectly “logical” mistake to make.

But if your using the process of elimination (knowing door 3 isn’t an option for the car) you know there is a 50–50 chance the car is behind 1 or 2.
This is probably why I failed math.

@elijah: probably… lol

@elijah: But the crux of the problem is the fact that the door the host opens is not a random selection. They don’t want you to win the car. So you pick door one. Let’s say you picked a door with a goat behind it. Now the car is either behind door 2 or 3. The host wants to tempt you some more, so he opens a door with a goat behind it. That probability influences the previous probabilities. So now, you have a better chance of changing doors and getting the car than staying with your door. That wiki article explains it in excruciating detail.

My head hurts now.

There’s a related problem known as “The Beagle Problem” that’s equally maddening:

“A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re both male, both female, or one of each. You tell her that you want only a male, and she telephone’s the guy who’s giving them a bath. “Is at least one a male?” she asks him. “Yes!” she tells you. What is the probability that the other puppy is a male?”

Here’s a link to some pages in a book (by the same woman who famously solved the Monty Hall paradox) where this problem is discussed.

I think the crux of the problem is just that the probability of selecting the right choice doesn’t change just because you know more now. If you have three doors, the probability that the car is behind two of them is still 2/3. If you choose one of them, the probability that it is behind the other two is still 2/3. So, when you find out what’s behind one of those other doors, the chance that the car is behind one of them is still 2/3, but you know it’s not behind one of them, so the 66% chance is now in the choice of the one remaining door, while your original choice of door 1 still only has a 33% chance.

This is a fun question. Everyone needs a little more Monday morning math. Those links were great, Harp.

Yes, Harp, thanks for those links. I’m pretty anxious to read them when I’m not at work and can concentrate. I wish I was good at math! :(

Just for fun I wrote a simulation program:
Here are the results.
Out of 10000 plays…
I won 6663 times.
I lost 3337 times.
I won 66.63 % of the time.

I’ll post the source code (C#) if anyone wants it.

Plus I have taxes to pay on 6663 cars! Yikes!

^5 @Grisson. I did the same thing when I first heard this problem, because I needed to see it actually happen before I could accept the mathematics. :-)

@grisson: How did you execute it? I mean, how did you factor in the door that was revealed? Or did you always pick door one and reveal door 3 but randomize what was behind them?

Another way to think about it is your chances of losing rather than your chances of winning. When you first make your selection, you have a 66% chance of losing. When the host opens the door, he is opening one of the two doors, he already knows that he’s not opening a winning door. There is no new knowledge there. Therefore the math hasn’t changed. Your original door still has a 66% chance of not being correct. So you switch to the only available door, and that gives you a 66% chance of being correct.

A third way to think about it: When you originally choose, your door has a 33% chance of being right. The other two doors combined have a 66% chance of being right. At least one of those two doors is “wrong”. But that doesn’t matter, there is still a total of 66% chance between them. If you were given the option of switching from your door (33% chance) to BOTH other doors (66% chance), you’d obviously take it, right? Even though you know at least one of them is wrong? So now the host opens one door, that he KNOWS to be wrong. Nothing has changed. That door was always wrong, even when you initially computed the percentages. The 66% chance still applies to the collective of the two other doors. But you know now that one of those two doors has a 0% chance. Therefore the other door has 100% of the original 66% chance that both doors collectively had.

@MrItty: THAT’S IT!! As soon as you put it in terms of chances of losing rather than winning, it makes perfect sense.

@poofandmook Random.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
//using Math;

namespace MontyHall
{
class Program
{
enum Prize
{
Car,
Goat
}
static void Main(string[] args)
{

Random rgen = new Random();
int iterations = 10000;
int win = 0;
int lose = 0;
if (args.Length > 1)
{
Int32.TryParse(args[1], out iterations);
}
for (int idx = 0; idx < iterations; idx++)
{
List<Prize> doors = new List<Prize>();
int carDoor = rgen.Next(0, 3);//Door is behind door#i [Range is 0 – 2]
//Console.Out.WriteLine(“Car:” + carDoor.ToString());
//Put prizes behind the doors:
for (int i = 0; i < 3; i++)
{
if (i == carDoor)
{
doors.Add(Prize.Car);
}
else
{
doors.Add(Prize.Goat);
}
};

int myGuess = rgen.Next(0, 3);//I guess door#j

//Monty selects
int doorRevealed = -1;
if (myGuess != carDoor) //My door is a goat, but I don’t know that yet.
{
//Monty must pick the other goat
for (int i = 0; i < 3; i++)
{
if (doors[i] != Prize.Car)//One that isn’t the car
{
if (i != myGuess) //One that isn’t my door
doorRevealed = i;
}
}
}
else //I chose the car but don’t know it yet.
{
//Monty randomly picks a goat
int goat = rgen.Next(0, 2); // range is 0–1
int goatidx = -1;
for (int i = 0; i < 3; i++)
{
if (doors[i] != Prize.Car) //Don’t show the car which is my guess
{
goatidx++; //Here’s a goat
if (goatidx == goat) //Is it the random goat Monty picked?
doorRevealed = i;
}
}
}
//I picked myGuess;
//Monty revealed doorRevealed
//I change myGuess to the other door.

int myNewGuess = -1;
for (int i = 0; i < 3; i++)
{
if (i != myGuess && i != doorRevealed) //It’s not my old guess and it’s not the door Monty reveald.
{
myNewGuess = i;
}
}
if (myNewGuess == carDoor) win++; //A got it!
else lose++; //Aw shucks!

}
Console.Out.WriteLine(“Out of ” + iterations.ToString() + ” plays…”);
Console.Out.WriteLine(“I won ” + win.ToString() + ” times.”);
Console.Out.WriteLine(“I lost ” + lose.ToString() + ” times.”);
Console.Out.WriteLine(“I won ” + (100.0*win/iterations).ToString() + ” % of the time.”);
Console.In.ReadLine();
}
}
}

//Sorry the tabs don’t do well here.

@poofandmook Whoo! Glad I could help. :-)

@Grisson I’ll try to remember to post my Perl version when I get back to my home computer tonight. :-)

@MrItty Thanks for putting that in terms even a math loser like myself can understand! When I watched that movie I was stumped, too.

Wow, I have been enlightened.

So is it better to trade doors or stick with your original choice?
I haaate math it makes me feel so stupid!
I know I’m doing it wrong, but in my head first there were 3 doors, so I had a 1 in 3 chance of picking the car. Then one option is eliminated. There are only 2 options now, door 1 and door 2. The car is behind one of them. The car isn’t behind the other. I still see it as a 1 in 3 chance that my original door has the car. If I didn’t have the option of picking the other door, my chances are still 1 in 3. But since I do have the option, doesn’t that make it 50–50?

You should all be thankful i dont have to do complex math in real life LOL. No rocket scientist here.

@elijah: Read this explanation by MrItty. It helps a lot. Also, Harp has posted a wiki article explaining it.

@elijahsuicide: It is better to trade.

The flaw in your logic is “one option is eliminated”. No it’s not. All that happens is that one of the two other doors was revealed. You ALWAYS knew that one of the other two doors contained a goat. You just didn’t know which of the other two doors. You know that either A contains a goat, B contains a goat, or A&B both contain goats (ie, you picked the “right” door). When the host opens one door, he’s not opening one randomly. He’s not revealing to you any more information that you didn’t already have. He’s only showing you, “Yup, one of the other two doors has a goat, look”. Therefore with no knew knowledge gained on your part, nothing about the math changes. It’s still 33% chance for your door, and 66% chance for the collective of the other two doors. It’s just that now you obviously won’t pick a certain one of those other two doors.

Is that any clearer?

Yeah I kind of get it Iit makes sense after reading the million door comparison, I mean obviously if there were a million doors the odds of me picking the one with a goat are very high. So if he leaves only 2 doors closed I’m definately going to switch.
It’s just hard for me to understand it on a smaller scale. Even if you apply the math you aren’t guaranteed to win.

Exactly correct. You are not guaranteed to win. You are only twice as likely to win if you switch. There is still a 33% chance that you lose if you switch.

Thanks for taking the time to actually explain it.
I feel better knowing that actual “Nobel physicists” get it wrong (according to the wiki link).

I just found this… it’s fun! Try it :)

I played 60 times… switching 30 and staying 30. When I switched, I won 70% of the time. When I stayed, I won 40% of the time.

Nice, @poofandmook, I like their explanation as well. Maybe I can modify it just a little bit. Instead of it being a car and two goats, let’s make it a car, a goat, and a llama. The car is still the winner, the goat and llama are both losers.

Let’s take a look at every possibility, broken into what you initially picked and whether or not you switch. There are three possibilities – you initially pick the car, you initially pick the goat, or you initially pick the llama.

Initial Pick | Monty Reveals | Switch? | End up with | Win?
——————-+———————-+————-+——————-+——-
Car ________ | Goat ________ | Switch_ | Llama _____ | Lose
Car ________ | Goat ________ | Stay __ | Car _______ | Win
Car ________ | Llama _______ | Switch_ | Goat ______ | Lose
Car ________ | Llama _______ | Stay __ | Car _______ | Win
——————-+———————-+————-+——————-+——-
Goat _______ | Llama _______ | Switch_ | Car _______ | Win
Goat _______ | Llama _______ | Stay __ | Goat ______ | Lose
——————-+———————-+————-+——————-+——-
Llama ______ | Goat ________ | Switch_ | Car _______ | Win
Llama ______ | Goat ________ | Stay __ | Llama _____ | Lose

As you can see, in 2 of the three original possibilities, if you switch, you win. In the other possibility (you originally chose the car), if you switch, you lose. Now all you have to remember is that there is no difference between Goat and Llama in the actual show.

@MrItty – great explanation. I like how you explained it in so many different ways. Thanks for putting so much work into it. Lurve :)

@Grisson many luffles, when I read about the problem a long time ago I wanted to write a program for it but never got the chance, maybe I can use this now to make something in Python. Thanks :D

There’s another way to think about it that makes it clear why switching gives you better odds.

In the original problem the game show host will always show you a goat. Regardless of what’s behind your door.

Suppose that there are 1000 doors. Lets say there is one car, and 999 goats. The first choice you make will most likely be a goat. Then the game show host goes through and opens all but one door, and shows you a goat each time. Then you’re left with the door you first chose and another closed door. Would you switch?

Of course you would.

The odds of you picking the car the first time is 1 in 1000. That means that one in one thousand times, you’d lose if you switched. The rest of the time you’d win the car.

The same logic follows for 3 doors.

Yeah, I always explain it Perchik’s way. Instead of having to go through all the math it seems to make sense to most people by increasing the amount of doors.