No, it’s now a true 50-50 chance. Call the door that you open #1, the door that Monty opens #2 and the third door #3. The car has an equal chance of being behind any of them. One-third of the time Monty will reveal it and you have already lost. The rest of the time he will not reveal it and the car is equally likely to be behind door #1 or door #3.
This problem is a hardy annual, isn’t it?
I really can’t see why anyone has a problem with this. THe answer IS 2/3. There isn’t any doubt about it, and anyone that says its 50-50 is wrong.
Think of it like a flow chart. You have 3 possible choices at the start- car, goat or goat, although you don’t know what one you picked. It DOES matter which one you pick. If you pick a goat, then the presenter will open the door to the other goat. If you change you win. If you pick the car first off, if you change, you lose.
So, if you pick a goat at the start; changing later means you win. No exception. If you pick the car at the start, changing = lose (duh).
There are 2 goats. There is one car. The chance of a randomly selected door being a goat is 2/3. Therefore, because if you pick a goat, changing= winning, the chance of winning if you change doors is 2/3.
Its not hard. OK?
I’m surprised that nobody else has commented on this. This is absolutely wrong. The law of averages does not say that any given sample must contain equal numbers of boys and girls (assuming that either is equally probable). What it does say is that as your samples get larger, you’re likely to get closer to 50% of each. The ratio of boys to girls approaches 1.
A pitfall here, though: The expected difference between the number of boys and the number of girls does not approach zero. In fact, as you make your sample size larger, the difference between the number of boys and girls is expected to get larger. It just doesn’t get larger as fast as your sample gets larger, so relative to your sample size, the difference is still decreasing.
And if you keep flipping heads, you’re still just as likely to flip another head on the next flip as you are to flip tails (assuming a fair coin). Each coin flip is completely independent of all previous flips. This is not only consistent with the Law of Averages; indeed, the Law of Averages depends on this property.
I can’t confirm either way. I can only point out that this is a new record for a response to one of my questions: almost 4 years from when I posted it. Where the hell were you? One more day and I was going to stop checking. 
So I just made it, then! I figured you hadn’t been able to sleep easy these 4 years, the question gnawing at your mind.
Sadly, despite taking 4 years to compose an answer, I still got it wrong. 
We are given that MH opens an empty door. It doesn’t matter how he decided to open it, whether he knew it was empty or just picked it at random. You still have the situation where there is a 1/3 probability that your original guess was correct. So if you stick with your first guess you have only a 1/3 chance to win; therefore if you switch you must have a 2/3 chance to win, no matter how Monty picked his door.
Ah ha! Now I can use my newfound knowledge.
It does matter how he decided to open it. Here is why:
You pick the wrong door 2/3 of the time, true, but half of those times Monty will open the winning door if he is opening randomly among the two remaining doors!
That means that only 1/3 of the time will you pick the wrong door, AND have the option to switch to the winning door.
You also have a 1/3 chance of picking correctly from the start.
Therefore, if Monty is opening randomly, and he opens a wrong door, your chances are exactly the same whether you stay or switch.
It’s given that he opens an empty door. You have to make your decision whether to switch after seeing that he has opened an empty door. That is part of the puzzle. Given all that, you should switch. The possibility that he might open the prize door is not part of the problem. So given that he opened an empty door, his method for picking that door is irrelevant. In fact you would have no way of knowing whether he picked it at random anyhow, so the correct answer for the random scenario must be the same as the correct answer for the non-random one.
Thank **prokopowicz **
No way of knowing? Why not? Because it doesn’t say so in the problem?
Depends upon which version of the problem you are reading. Generally, if it does not say so explicitly, it is assumed that he does know what is behind the doors, and it is not random.
Of course his method is relevant. Nightime’s example shows why.
I’m still missing why his randomness (or lack thereof) has any bearing at all on the decision, given the fact that Monty opens an empty door. To me it’s the equivalent of Monty showing you one of the sides of a tossed die (but none of the other sides) and asking you the odds that the upper side matches what you guessed before the toss (with me?). Whether he randomly selected to show you a side 4, or whether he randomly selected a side and it turned out to be a 4, you have exactly the same probability calculation in front of you. Why would it be different?
Same in this problem, right?
But you’re saying that in both cases, the choice is random in your two examples. Would he show you the side if it had your number on it? That’s the crux.
What is interesting is that this “random” line of thinking is what got all those mathematicians into trouble in the first place. They were so used to all the “non-intuitive” solutions to randomized choices, that they objected strongly, and in writing, to Marilyn’s solution. Marilyn, of course, showed them wrong. Now, people have a hard time seeing how the opposing example can be correct, because of the Monte Hall problem.
Maybe we’re talking about two different points. In any single instance where Monty shows you a losing choice (or a die face that isn’t your choice), for that single iteration it matters not whether it was deliberate or not on Monty’s part, you still have the same probability caclulation in making a decision. Period. How could it be otherwise?
Let me put it another way. Say Monty did billion iterations where he deliberately showed you a losing door; let’s collectively call those events Set A. Let’s say he also did a billion+ iterations where he randomly selected a door, and the subset of those where he randomly selected a losing door amounted to a billion iterations (call it Set B). I posit that in either Set A or Set B, it is the same advantage to change your choice on any given iteration (all of them where Monty showed you a loser), and the overall results of those two billion iteration would show this.
Do you agree with this? I’m really just trying to understand if we’re discussing different points, I guess. And, just as I did 4 years ago, I fully allow I may just be completely missing the point.
We don’t appear to be, to me.
I disagree.
Rather than using your example with billions of instances, let’s break it down into counting number of possible outcomes.
With three doors, you can choose any of the three, and they are equally probable. Call them A, B, C, with C being the one with the prize. Each choice has 1/3 chance of occurring. So, Ab would represent you choosing A, and Monty showing B. Notice Ac is not an option, if Monty is not choosing randomly. So, the possibilities are Ab, Ba, Ca, and Cb, but the probability of Ca is only 1/6th, since Ca and Cb can be considered equally probable. Just because there are four possibilities does not mean that they are equally probable. The sum of all the probabilities is 1, of course. In the case of Ab and Ba, you would win if you switched, so the probability of winning is 2/3 if you switch. You don’t win by switching with Ca or Cb. Of course, you should switch.
If Monty is choosing randomly, then all six of Ab, Ac, Ba, Bc, Ca, and Cb are possible, and they all have probability 1/6. Which ones satisfy the terms of our problem? Only Ab, Ba, Ca, and Cb, since the other two would involve revealing the prize, and the problem states that that does not occur. These are the same options as before, but their probabilities are different! In this case, the probability of winning is only 1/2 if you switch. In other words, it doesn’t matter if you switch or not.
The assumptions about how you got to a particular situation do indeed affect the probabilities. Too many mathematicians were treating the problem in the second fashion–but everyone knows that game shows are rigged, right? Well, at least, that Monty Hall knew what he was doing.
Wow, you did ask four years ago, and there wasn’t a real response. I’ll be d*rned.
Remember, the probabilities of all outcomes must add up to one. Since we are given that Ac and Bc did not happen, their probabilities become zero. What then are the probabilities of Ab, Ba, Ca, and Cb?
Here’s the way I look at it: The probability that my first guess was wrong is 2/3. That means that the sum of the probabilities of Ab, Ac, Ba, and Bc must be 2/3. It doesn’t matter how that 2/3 is distributed among those four because I know that if I chose wrong on the first guess I will win if I switch. I also know that there is a 2/3 probability that I was wrong on the first guess. Therefore I should switch.
That depends on how you define them, of course. I defined them two different ways.
[quote]
Here’s the way I look at it: The probability that my first guess was wrong is 2/3. That means that the sum of the probabilities of Ab, Ac, Ba, and Bc must be 2/3. It doesn’t matter how that 2/3 is distributed among those four
[quote]
In my previous post, I showed that it does make a difference.
Not if Monty’s choice is random AND you have thrown out those times where he did choose the prize door–which would be 1/3 of the time.
If the probability is 1/2, it won’t matter if you switch. But there is a situation where you definitely should not switch–Evil Monty. Evil Monty is the guy who knows all about this problem and how everybody has convinced themselves that they should switch. So, what does he do? If you choose the wrong door, he always shows the right door. If you choose the right door, you’ll switch. So you never get the prize.
Good Monty would open all the doors that don’t have prizes–so you know exactly which one has the prize.
Monty’s behavior and motivation definitely changes the odds.
RM, I think you are getting lost in the details and missing the big picture. The problem is really simple. There are only two facts you need to consider:
If you chose the wrong door initially, you will win if you switch. This is an inescapable fact because we are given the MH will eliminate a non-prize door, leaving only one other choice.
There is a 2/3 chance you will pick the wrong door at the start, because there are 3 doors and you have no information about them when you make your choice.
The only probability that matters is that 2/3. No matter what Monty’s motivations are, he can’t change that 2/3 because you make that choice before Monty does anything. The only way Monty can affect you chance of winning is by opening the prize door, but we are given that he doesn’t do that, so that is not an event that we should even consider.
It seems to me that back when this problem first came up, a lot of the answers basically said “you have a 2/3 chance of picking wrong from the start, so you should always switch”.
Unfortunately, this is a poor way to answer the question, and leads to difficulty later on when thinking about variations of the problem.
The real reason you should switch in the original problem is not merely because you had a 2/3 chance to be wrong, but rather because 2/3 of the time the winning door will still be available to switch to.
If Monty is opening randomly, then:
1/3 of the time you pick right, and Monty opens an empty door.
1/3 of the time you pick wrong, and Monty opens an empty door.
1/3 of the time you pick wrong, and Monty reveals the winning door.
Since we are told Monty opened an empty door, we know we are in one of the first two scenarios. But as you can see, these scenarios are equally likely. So there is no advantage to switching.
Another illustration of why the chances are equal:
What if TWO people pick doors, and the third door is chosen randomly and turns out to be empty?
By the reasoning that “I had a 2/3 chance to be wrong and should switch”, both people would have a better chance if they switched to the other’s door, but clearly that is impossible.
Nighttime is right. That last post showed it quite clearly.
What Monty does is basically stopping the game (by opening the winning door) half of the times that it would be best to switch doors. So instead of it being 1/3 chance that you will lose and 2/3 chance that you will win it will be 1/3 chance that you will lose, 1/3 chance that you will win and 1/3 chance that Monty will ruin the game by opening to winning door.
Let the doors be a, b, and c. A capital letter signifies that the player selected this door. Underlining means that it’s the door Monty opens randomly. The color red indicates the winning door. Assuming each of these events is random, the following depicts the 27 possible outcomes:
In 9 of the outcomes, Monty reveals the winning door (1, 5, 9, 10, 14, 18, 19, 23, 27). Not counting those already noted, Monty reveals an additional 6 doors that the player selected (2, 3, 13, 15, 25, 26). These 15 are not relevant to the question at hand.
In each of the remaining 12 possibilities, Monty reveals a losing door. In 6 of the 12, the player selected the winning door initially (4, 7, 11, 17, 21, 24). Switching doors would mean the player would lose.
In the other 6, the player initially picked a loser, so switching would win for them (6, 8, 12, 16, 20, 22).
So, in any random scenario where the player is shown a losing door, switching or not has no bearing on winning.
Does this make sense? Did I mess any of this up (I had a couple of beers at dinner, so check me)? If this is right, I still can’t get my head around it. Outcomes 4, 6, 7, 8, 11, 12, 16, 17, 20, 21, 22, and 24 are the 12 possible combinations where Monty could show an empty door, whether he did so randomly or deliberately.
Did somebody already try this? Man, I’m confused. I’m getting another beer.
Again I think people are confusing themselves with irrelevant details. If the probability of picking the wrong door is 2/3 (which I think we all agree on) and it’s a fact that you will win if you pick the wrong door and then switch (which is obvious) then it follows that you have a 2/3 chance of winning if you switch.
A mistake some people are making is to say that the probability of MH opening the winning door is 1/3; but, the probability of him opening the winning door is actually 0, because we are told that he does not open the winning door.
Try to imagine yourself actually playing this game. You have three doors, all alike. You pick one. Then some guy opens one of the others and you see that there is no prize behind it. Maybe he knew that door was empty and opened it deliberately; maybe he picked it at random and it’s a fluke that it was empty - you don’t know. So, do you switch? You know that you should switch if you picked the wrong door at first, right? And how likely was that?
The decision to switch is correct whenever you first choice was wrong. The rightness or wrongness of your first choice is not influenced by whether Monty Hall knows where the prize is, so the decision whether to switch should likewise not depend on what MH knows. You know that your first choice was probably wong,so you know that if you switch you will probably win.