I think everyone agrees on this. The only thing people are saying is that if Monty does open winning doors too (which would be pretty stupid, and ruin the game) it wouldn’t matter if you switched or not.
If he opens a winning door, you already lost the game, so sure, in that case it doesn’t matter what you do. But if he is allowed to open a winning door, but happens to open a losing door, then you still improve your chances by switching (because your first guess was probably wrong).
Nope. The reason for this is that opening an empty door randomly adds another element of chance, while opening an empty door that you already know is empty adds no element of chance.
Chances of guess right: 1/3
Chances of randomly opening an empty door after guessing right: 1
(this is a key point… since you already guessed right, Monty will always randomly open an empty door)
1/3 x 1 = 1/3
Chances of guessing wrong: 2/3
Chances of randomly opening an empty door after guessing wrong: 1/2
(since you guessed wrong, there is only a 1/2 chance that Monty will open an empty door)
2/3 x 1/2 = 1/3
So the chances are exactly equal that you will guess right and see an empty door, and that you will guess wrong and see an empty door.
So yes, after seeing an empty door opened randomly, there is no advantage to staying or switching.
You are presented with the option to switch after seeing an empty door opened, so the probability of that empty door being opened isn’t 1/2; it is 1 (because it has actually happened - anything that you know has actually happened has probability 1).
Right.
And since you know the empty door was opened, you know you are in one of two scenarios:
-
You guessed wrong, and Monty opened an empty door.
-
You guessed right, and Monty opened an empty door.
Since these scenarios are equally likely, there is no advantage to staying or switching.
Those scenarios aren’y equally likely. Since you know that Monty opened an empty door, the probability of that event is 1. So the probabilities are:
- You guessed wrong, and Monty opened an empty door: 2/3 * 1 = 2/3
- You guessed right, and Monty opened an empty door: 1/3 * 1 = 1/3
If you calculate the probability of scenario 1 the way you do, i.e. (2/3 * 1/2 = 1/3), you are ignoring information. You are pretending that the probability of Monty opening that empty door is 1/2, when in fact Monty has actually opened that door already, so the probability of that event is 1.
I’ll try a different tactic: What if 2 people were playing?
I pick door number 1.
You pick door number 2.
Monty randomly opens door 3, and it is empty.
According to your logic, both of us should think “I probably chose wrong, so I should switch to the one remaining door”.
But this makes no sense, because we would just be switching with each other.
Play Monty 6 times:
Let’s say you always pick door 1.
1/3 of the time you’re right.
first game Monty picks door 2, it is empty. You swap and lose.
second gameMonty picks door 3, it is empty. You swap and lose.
2/3 of the time you are wrong.
third game Monty picks door 2, it is empty. You swap and win.
fourth gameMonty picks door 3, it is empty. You swap and win.
fifth game Monty picks door 2, it is the prize. Game over.
sixth gameMonty picks door 3, it is the prize. Game over.
There are no other possibilities.
You are wrong with this:
- You guessed wrong, and Monty opened an empty door: 2/3 * 1 = 2/3
As you can see it is 1/3*1 = 1/3 because if he revealed the prize you don’t see the empty door.
If Monty is opening the doors randomly, the probability is 1/2.
So, half the time Monty ruins the game by opening the prize door–and that doesn’t satisfy the original game scenario so it doesn’t allow you to switch, and it reduces your chances to switch to the prize door.
Nightime - your two-person game is quite a bit different from the original problem. Here’s how I see it:
You say that Monty reveals an empty door after we make our picks. That means that one of us must have chosen the right door. Let’s say you pick first. If your first choice is wrong, that means my first choice must be right. We know this because you are saying that the remaining door is not the winner.
So, there is a 2/3 chance that your first guess was wrong, so you should switch. Likewise, there is a 2/3 chance that my first guess was right (because my first guess is right whenever yours is wrong), so I should stay with my first choice.
RM- It’s given that Monty does not open the prize door, so that event cannot have a non-zero probability. If you do assign it a non-zero probability, it would mean that Monty might not open an empty door, but we already know (because it is given) that he does open an empty door. So the probability of him opening the prize door has to be 0 and the probability of him opening a non-prize door has to be 1.
What extraneous details are included in what I lined up?
I itemized all the possible outcomes of 3 events: The player’s initial selection, the winning door, and the door Monty reveals. There are 27 possibilities, each of them equally likely.
I eliminate from this group all combinations where Monty reveals the winning door and/or the player’s initial selection, since those outcomes aren’t relevant. The 27 is reduced to 12, 12 combinations where Monty is revealing an empty door. If I am thinking of this clearly, there are 12, and only 12, combinations of these 3 events where Monty reveals a loser.
Within those 12, there are 6 where the player originally selected correctly and 6 where he didn’t. Put another way, there are 6 where switching is the right move and 6 where it isn’t.
Again, what is wrong with what I outlined? (There may well be something, but I can’t see what.)
Neither of us knows the other is playing, and we choose at precisely the same time.
Yet there is a 2/3 chance that your original guess is right?
Maybe if you are psychic. Otherwise our chances are equal.
Just as the chances are always equal for the two remaining doors, if Monty randomly opens an empty door.
1/3 of the time you guess right, and Monty opens an empty door.
1/3 of the time you guess wrong, and Monty opens an empty door.
1/3 of the time you guess wrong, and Monty opens the winning door.
The fact that Monty opened an empty door tells us that we are in one of the first 2 scenarios. They are equally likely.
It’s not a matter of assigning probabilities. It’s a matter of computing probabilities. You have to make sure that your assumptions are consistent.
Let me get this straight. Your assumption is that Monty is faced with two doors, one of which is the prize door, and he is choosing randomly between them, and you say the probability of him choosing the prize door is zero?
Try playing this game six times, with Monty picking randomly, with all six possible outcomes. What do you do with the situations where Monty picks the prize door? (If he is picking randomly, he will sometimes pick the prize door.) How do you propose to incorporate them into the calculations? Do you think it affects the relative probabilities of the other events?
Your list looks good to me, Stratocaster. It shows why the doors are equally likely to be the winner if Monty chooses randomly.
True. But if Monty knows what is behind the doors, he can change the probability of each of these outcomes.
Look at outcomes 6 and 9 on your list. If Monty knows what is behind the doors, outcome 9 will be replaced by another instance of outcome 6, thus making outcome 6 twice as likely as if Monty is picking randomly.
The same happens to all the outcomes in which you guess wrong. Each will occur twice as often, thus giving you a 2/3 chance if you switch.
That last line should say:
“The same happens to all the outcomes in which you guess wrong AND Monty opens an empty door. Each will occur twice as often, thus giving you a 2/3 chance if you switch.”
Listen here:
The chance of you picking the wrong door at first is 2/3. But half of the times you start by picking the wrong door and thus would win, Monty ruins the game for you instead. So that outcome that happens 2/3 of the time is split into a 1/3 where you will win and a 1/3 where Monty will ruin the game.
You guys are wearing me out {:^P
I’ll try to get to everybody over the next few posts.
First, Stratocaster:
The mistake your are making is eliminating several of your 27 outcomes and assuming that the probabilities of the others remain equal. In fact, some outcomes are more likely than others.
Let’s say that the prize is behind door A. The possible sequences of my pick/Monty’s pick are:
AB
AC
BA
BC
CA
CB
(Actually BA and CA aren’t possible, but bear with me for a moment.)
So, what is the probability of AB? The probability of A is 1/3 (because I’m choosing among 3 doors). Once I have chosen A, the probabilty of Monty choosing B is 1/2 (because Monty could choose either B or C). So the probability of AB is:
AB: 1/3 * 1/2 = 1/6
By the same argument, the probability of AC is
AC: 1/3 * 1/2 = 1/6
Now we come to BA. The probability that I pick B is 1/3; so, given that I have picked B, what is the probability that Monty picks A? The prize is behind door A, and we are given that Monty does not reveal the prize, so the probability that Monty picks A must be 0. I can’t stress that enough. That is the key to the problem. We know that Monty doesn’t pick A, so we have to assign it a 0 probability. So the probability of BA is:
BA: 1/3 * 0 = 0
So, if I pick B, and we know that Monty doesn’t pick A, then he must pick C because that is the only remaining choice. That means that the probability that he picks C is 1. So
BC: 1/3 * 1 = 1/3.
The same logic that applies to BA and CA also applies to CA and CB, so their probabilities are
CA: 1/3 * 0 = 0
CB: 1/3 * 1 = 1/3
So the possible outcomes and their probabilties are
AB: 1/6
AC: 1/6
BA: 0
BC: 1/3
CA: 0
CB: 1/3
The cases where I will win if I switch are BC and CB. Their probabilities add up to 2/3. so I should switch.
Where your analysis goes wrong is that you say that each outcome has 1/27 probability, then you eliminate 15 of them as being impossible, but not reexamining the probabilities of remaining 12 outcomes. So the probabilities of all possible outcomes in your analysis was adding up to 12/27, which can’t be right. The total probabilities of all possible outcomes has to be 1. Also, when you eliminate a combination, you can’t assume that that doesn’t affect the distribution of probabilities among the other combinations. You can see that in my analysis above, where eliminating BA made BC more likely, but didn’t afftect the probabilities of AB or BC.
In your description of the game, you said that you pick a door, I pick a door, and the remaining door is opened by Monty. From that I inferred that the game didn’t allow us to both open the same door. If we are allowed to choose the same door, that changes everything. But that still leaves the situation where we both pick the two empty doors, leaving only the prize door for Monty. You description of the game says that Monty opens an empty door, so that situation can’t arise, so IF we pick different doors, one of them must be the winner. IF we aren’t allowed to pick the same door, then the player who goes first should switch and the other player shouldn’t.
If I was correct, that we can’t pick the same door, then presumably the game would have some mechanism to enforce this. I don’t know, you made up the rules.
Not true, because it is given that Monty doesn’t open the prize door. If the prize door is one of the two remaining, then the probability of him opening that one is 0 and the probability of him opening the other is 1. Anything else would violate the premises of the problem. See my response to Stratocaster above for the mind-numbing details.
They are not equally likely withing the problem domain. The problem says that scenario three doesn’t happen. Therefore its probability is 0, and the 1/3 that was being assigned to it must be distributed among the other two scenarios. Clearly it belongs with scenario two, because that is the only other one that has me picking the wrong door.
Whether you call it assigning or computing, every event has to have a probability. An event that you know does not happen has to have 0 as its probability. An event that you know does happen has to have 1 as its probability. We are given that Monty opens an empty door, so that event has to have probability 1, and the event that he opens the prize door has to have probability 0. We are given this in the statement of the problem; we have no choice about those probabilities.
That’s not my assumption; that is given. We are told that Monty opened and empty door. Given that, he can’t open a prize door, because that would take us into a situation that this problem isn’t about.
If you play this game and Monty picks the prize door, then that particular iteration of the game isn’t one that we are being asked about in this problem. The problem asks “What should you do, given that Monty opens an empty door and gives you the option to switch”.
Again, imagine you are playing this game. There are 3 doors. A guy tells you there is a prize behind one of them, and invites you to pick one. So you pick one. The guy then opens one of the other doors, and you see that there’s nothing behind it. The guy asks if you want to switch. Let’s stop here for a second.
Maybe if you played the game another time, Monty would open the prize door and say “ha ha you lose. sucker!” But should that affect your answer in the game you’re playing right now, in which you know that he opened an empty door? The prize is either behind the door you picked first, or the other unopened door. There is a 1/3 chance you picked the correct door first. That is all you need to figure out your answer. It doesn’t matter what might have happened in some other iteration of the game, because you know that in this iteration, it didn’t happen.
Not really, because the statement of the problem says that Monty doesn’t pick the prize, so that event is out of play. The last two paragraphs of my previous post explain it (I hope).