
The Monty Hall Problem
Hi, I'm still a nonbeliever in the solution to the Monty Hall problem. For those of you not familiar with it, it goes something like this:
Given 3 doors, 2 of them have "goat" prizes behind them, and one of them has a car. You select a door, and then the host shows you one of the other doors  the catch is that other door is ALWAYS a goat. You then have a choice to switch and take the prize behind the other door. Most articles on this problem (including the most accepted answer) say you should switch since it makes the odds of winning 2/3 and the statistics go along to prove this.
Now this makes some assumption that your second pick (i.e. whether to switch or not) is dependent on your first pick. In reality it is not (or does not have to be). I approach the problem like this. I pick a door, and the host shows me a goat. I then randomly choose one of the two remaining doors. Thus, statistically speaking, I now have a 1/2 chance percent of winning.
I'd like to discuss this theory vs the "accepted" theory.
Thanks, Jeff
Re: The Monty Hall Problem
Welcome to the forum!
You're right that randomly picking one of the remaining two doors will give you a half chance of winning, but that doesn't imply anything about the individual probabilities of each door. The statement would still be true even if you knew which door the car was behind.
Why did the vector cross the road? It wanted to be normal.
Re: The Monty Hall Problem
I'm not sure I agree with that. The probability of each door winning is 50%. You can pretty much throw out your first door's pick because it's really irrelevant. Think about this situation. I pick my door and the host shows the goat prize. Let's assume I picked door number 1. Now, someone in the audience says, "I bet door number 1 is correct." He's got a 50% chance that he's correct, but according to the theory of the problem, door number 1 has a 33% chance of winning. So two people with the same exact door have different probabilities  that just doesn't seem to make sense.
 Ricky
 Moderator
Re: The Monty Hall Problem
Rewording the problem makes the solution more apparent.
You are given three doors to pick, A, B, and C. You pick a door. Now you are told you can either stay with the door you picked, or you can get both the other doors. So if you picked A, you could either stay with A or you could go with both B and C.
If you think about it for a bit, this is exactly what the host is doing when he opens a door for you that he knows has a goat behind it. This is because without the host telling you anything, you already know that at least one of the doors you didn't pick contains a goat. Having the host showing you where a goat is exactly the same as being able to pick both doors. Now what would you say your probability is?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 JohnnyReinB
 Power Member
Re: The Monty Hall Problem
Plus, in real life and practice (not just theory), 2/3 does happen. There was a program in the internet (I can't remember where) that shows this.
"There is not a difference between an inlaw and an outlaw, except maybe that an outlaw is wanted" Nisi Quam Primum, Nequequam
Re: The Monty Hall Problem
I'm not really sure why staying with A wouldn't be the same as going with A or B (similar to how switching to C would be the same as B or C).
I understand the math behind the problem and have written some computer simulations to show that if you switch, you do in fact win twice as much. However, if you make your second choice random of the two remaining doors, the odds of winning become 50%.
One thing logically that confuses me is my original statement that if someone else in the audience said, "I think the car is behind door A" after the first goat is shown to you (and let's assume that he was in the bathroom when you picked your original door and the first goat was shown, so he didn't see any of this information). He has a 50% chance of being correct, right? If that's the case, how can I have a 33% chance with door A and he has a 50% chance.
 Ricky
 Moderator
Re: The Monty Hall Problem
I'm not really sure why staying with A wouldn't be the same as going with A or B (similar to how switching to C would be the same as B or C).
I'm not sure what you mean by any of that.
I understand the math behind the problem and have written some computer simulations to show that if you switch, you do in fact win twice as much. However, if you make your second choice random of the two remaining doors, the odds of winning become 50%.
Think about it. You have know the car is behind one of the two doors. Of course picking randomly is going to produce 50%. If I told you the car was behind door A, and you chose a door randomly, you'd still get 50% of getting the car.
One thing logically that confuses me is my original statement that if someone else in the audience said, "I think the car is behind door A" after the first goat is shown to you (and let's assume that he was in the bathroom when you picked your original door and the first goat was shown, so he didn't see any of this information). He has a 50% chance of being correct, right? If that's the case, how can I have a 33% chance with door A and he has a 50% chance.
Because you have more information.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: The Monty Hall Problem
jas511 wrote:One thing logically that confuses me is my original statement that if someone else in the audience said, "I think the car is behind door A" after the first goat is shown to you (and let's assume that he was in the bathroom when you picked your original door and the first goat was shown, so he didn't see any of this information). He has a 50% chance of being correct, right? If that's the case, how can I have a 33% chance with door A and he has a 50% chance.
He actually has a 33% chance, just like anyone else, he just doesn't know that he does.
Why did the vector cross the road? It wanted to be normal.
Re: The Monty Hall Problem
I think the point that is confusing you is how the first pick alters the odds after the first door is opened.
The reason that the odds defy common sense is that the first door that they open and show is is not chosen randomly. They never show you the winning door and then ask if you want to stick with the door you chose first or to change (because what fun would that be?). They will always show you a losing door and then ask you if you want to keep your original choice or change.
Now, consider what would happen if they chose the door to show you completely randomly, meaning that they might show you the winning door. Let's say for the sake of argument that door A is the winner (the probabilities are the same if door B or door C is the winner). What are the odds that they'll show you the winning door before you even get to decide to keep your door or change? There is a 0% chance of this if you choose door A, but a 50% chance if you choose door B or door C, for a total of a 33% chance (0 * 1/3 + 2 * 1/3 * 1/2). That's a 33% chance of losing before you even get to decide to keep your door or switch.
Now, that means that you have a 66% chance of having a meaningful decision after the first door is opened. You now have a 5050 shot of winning, whether you keep your door or change, giving you a 33% chance to win. Note that this is the same chance you have of winning if you keep your door in the original game, but this time you have no chance of improving your odds.
Since you know something about how the door that they open is chosen, you can improve your odds. Basically, that 33% chance to lose right away (if they show you the winning door before asking you to switch) is added to the door that you did not choose. Intuitively it's confusing, but it's what actually happens.
Finally, to address this:
One thing logically that confuses me is my original statement that if someone else in the audience said, "I think the car is behind door A" after the first goat is shown to you (and let's assume that he was in the bathroom when you picked your original door and the first goat was shown, so he didn't see any of this information). He has a 50% chance of being correct, right? If that's the case, how can I have a 33% chance with door A and he has a 50% chance.
There are two points that I want to make here. The first is that while you have a 33% chance of being right with door A while he has a 50% chance, he also has only a 50% chance of being right if he chooses door B (or door C, whichever one wasn't shown to you) while you have a 66% chance. You have the advantage. The second point I want to make is that he has a 5050 chance of being right because he's choosing between two doors. You have a 3366 chance of being right because you're choosing between three doors. Even though one of the doors was shown to you, you're still choosing between all 3 doors because your original choice affects your odds.
I hope that helps.
Wrap it in bacon
Re: The Monty Hall Problem
Contestant 1 Car Goat Goat
Contestant 2 Goat Car Goat
Contestant 3 Goat Goat Car
Let's apply the two differing strategies to the above situation.
Strategy # Choose a door and stick with your original choice. All contestants choose door A As expected only one contestant in three wins.
Strategy @ Choose a door and swap to the only remaining door after the goat is revealed. All contestants choose door A. The swap strategy proves unfortunate for contestant 1; yet notice that both the other two contestants now win. In reality your chances of winning are doubled by using Strategy@. A simple simulation running, say, 300,000 times will give an approximate result of 100,000 wins for Strategy #; and, of course, approximately 200,000 wins for Strategy @.
Last edited by lester_day (20080115 01:48:35)
 Calligar
 Full Member
Re: The Monty Hall Problem
This Monty Hall problem is actually very difficult to see how it works out logically to 1/3 chance of the same door, and 2/3 chance with the door you switch to. It is also difficult to explain for you to understand, I don't know if this will be of much help, but I'll also try....
The main thing happening here is that originally you have 3 doors to choose from. Therefore, assuming it's totally random, you will have 1/3 chance of picking each door. Now, here's where it gets more trickier: Monty Hall eliminates one of the doors that has a goat. When Monty Hall eliminates a door, it is not random anymore, the key is that Monty Hall got rid of the door that has a goat, therefore changing the chances. Only if he eliminated the door randomly it would change the probability down to 1/2 chance of each door. One way of looking at this could be to look at Monty Hall eliminating a door. Monty Hall knows what is behind each door, and depending on what you chose, he has 2/3s chance of eliminating a door that does not have the the car. So in all, by him eliminating that door, there is higher probability that the car is behind the door if you switch rather then stay...
I hope I didn't do that bad of a job of explaining that....
Last edited by Calligar (20111008 05:24:34)
Life isn’t a simple Math: there are always other variables. [unknown]
But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. Aristotle
Re: The Monty Hall Problem
it's just that Monty Hall gets rid of the door that has a goat!
here is a link that explains the whole thing a little better:Monty Hall Problem
The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: The Monty Hall Problem
This game assumes that the car is the prize and the goat is not. If I were a young child playing it, I wouldn't switch if I already had the goat in view, because I like goats, and I would have no use for a car.
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 bob bundy
 Moderator
Re: The Monty Hall Problem
hello everyone,
Would you like to see a diagram?
You would.
Well, I'll show you one then.
Call the door you choose door A. The other two doors are B and C.
When Monty Hall opens a door and shows a goat, you cross that door off from your available choices. (Unless like RYA you like goats ... an excellent point I think.)
The rest is obvious.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: The Monty Hall Problem
I love this problem. So many people get it wrong and many who get it right can't explain it. Here is my stab at it: If your initial choice happened to be the car, then switching would always get you a goat. But if your initial choice was a goat then switching will get you a car. Switching will always get you the opposite of whatever your initial choice was. If you agree with that last statement the solution is obvious. Everyone agrees that your initial choice has only a 1/3 probability of being right and a 2/3 probability of being wrong, so switching would have to be the opposite of that. In other words, would you rather bet that your first choice was a car or bet that it was a goat?
