Whew! I always hated essay tests 
If you haven’t been worn out, let me try again. I feel like I’m close to a breakthrough.
The combinations that I laid out are equally likely, right? The 12 combos that remain, whatever their probabilities, are equally likely, correct? The combos that aren’t relevant don’t matter.
Here’s an anology. I’m trying to determine how often a 6 or a 4 will come up, relative to each other, if I roll a die. All faces of the die have the same probability: 1/6. But every roll where a non-4 or a non-6 comes up is simply ignored, and I record how often a 4 or a 6 is rolled.
Those outcomes, a 4 or a 6, are still equally likely. The probability just has a different algebra behind it, and I’m certain that over a long haul I will conclude that 4’s come up about 50% of the time, as do 6’s (in rolls that “count”).
So, how is that different from what I’ve laid out? Each of the 27 combos is equally likely, if all events are randomly chosen. Whenever one of 15 comes up, though, I simply ignore it–I only care about outcomes where Monty picks an empty door. I posit that each of the 12 has a 1/12 probability. What’s wrong with that logic?
And when are we going to discuss the great philosophical question, what, exactly, does the phrase “empty door” mean?
Yeah I cringe a bit every time I type “empty door”, but I don’t want to type “door with nothing behind it” five times a paragraph, so I use that sloppy shorthand :p.
Nope - the combos aren’t equally likely. I think I explained that pretty thoroughly in my previous response to you. If you choose a prize door, there are two possible picks Monty can make, but if you choose an empty door [heh] Monty can only make one choice. That is because we know that Monty picks an empty door [sic;)]. So that makes a sequence like CB twice as likely as a sequence like AB (if the prize is behind door A).
Your dice example is not like the Monty Hall problem because it doesn’t have the “x happens, then y happens” setup, so you don’t have to deal with contingent probabilities.
But anyway - the probability of a 4 coming up on a roll is 1/6, also the probability of a 6 coming up is 1/6. If you now say that any other number must be rerolled until a 4 or 6 comes up, the probability of eventually getting a 4 becomes 1/2, and the probability of a 6 also becomes 1/2. Those probabilities are equal, but they are no longer 1/6. The are equal because there is nothing in the re-roll rule that biases toward one outcome or the other.
Now, in the Monty Hall problem, the analog to the re-roll rule is to reject (or replay) any game where Monty opens the prize door. So, that leaves us with the situation where you have an option to switch to the other unopened door. You only have two unopened doors, so one of them must have the prize. When will the prize be behind the door that you didn’t pick? It will be there whenever you picked a door without a prize with your first pick. You know that that will happen 2/3 of the time. So: switch. The game is already biased toward switching because of that 2/3 probability of an initial wrong guess.
You have to work out the probabilities taking into account what you know about the game. You should ask “what can happen first” and “how likely is it”. and then, for each of those, “what can happen next” and “howl likely is that”.
So, there a two things that can happen first. Either you pick the prize door (probability 1/3) or you can pick one of the loser doors (probability 2/3).
Given you picked the right door, Monty can open one loser door (probability 1/2) or the other loser door (probability 1/2).
Given you picked a loser door, Monty can only open the other loser door (probability 1).
Sooo we have
Probability of I-pick-winner, Monty-picks-loser1 = 1/31/2 = 1/6
Probability of I-pick-winner, Monty-picks-loser2 = 1/31/2 = 1/6
Probability of I-pick-loser, Monty-picks-other-loser = 2/3*1 = 2/3
Manduck
if you state from the outset that monty has a 0 probability of selecting the prize door, then you are stating that his choice is non-random by definition, and we’re back to the initial scenario where Monty knows what he’s doing, and a switch is advantageous.
If we’re describing a given instance where Monty has selected a non-prize door through a random process (non-prize probability 2/3), then the two remaining doors have an equal chance of being the prize-door.
The key is that when Monty’s selection is random, he’s not giving you any extra information about the remaining door - he doesn’t have any to give.
Asteroide
Nope - If Monty happens to pick a non-prize door, then the prize must either be behind the door you picked initially, or the other unopened door. You already know that there is a 1/3 chance that the prize is behind your first pick, so the probability that it’s behind the only other alternative has to be 2/3. It doesn’t matter how Monty arrived at his choice, once you know that he did in fact pick a non-prize door.
By that logic if you choose door A and I choose door B and Monty opens door C to reveal nothing each of us has 1/3 of a chance of winning. What happened to the other third?
Manduck, I am beginning to despair of even explaining this to you.
However, you can prove to yourself that the chances are equal by performing a simple test.
Take 3 coins, and put one heads up. That is the “winning coin”.
Now you need some method of choosing randomly. Once you have it, choose a random coin. That is your chosen coin.
Now choose a random coin from the remaining 2 coins. That is the coin “Monty” reveals.
Do this experiment over and over, and keep track of the times that the coin “Monty” reveals is a losing coin.
What you will discover is that half of the times Monty reveals a losing coin, you have already chosen correctly.
I think what you’re asking is, what if I choose a door at random, and you choose one of the other two doors, and Monty reveals that the last door is empty, what are our respective probabilities of winning the prize. If I have that right, here’s how it breaks down.
Lets call the doors A, B, and C and let’s say that the prize is behind door A. The possible outcomes are (first letter is my door, second letter is your door, third letter is Monty’s empty door):
ABC
ACB
BAC
CAB
Note that BCA and CBA don’t appear on the list. That’s because those are the cases that have Monty’s door containing the prize. You have stated that Monty opens an empty door, so BCA and CBA can’t happen.; their probabilities are 0.
So, the probability that I pick A with my guess is 1/3, because I am picking from all three doors. If I pick A, the probability that you will pick B is 1/2, because you have two doors to choose from. Similarly, the probability that you will pick C is also 1/2. In either case Monty has to open the remaining door, so the probability of him doing that is 1. So the ABC and ACB sequences have these probabilities:
ABC: 1/3 * 1/2 * 1 = 1/6
ACB: 1/3 * 1/2 * 1 = 1/6
There is a 1/3 probability that my first pick will be B, If I pick B, the probability that you pick C has to be 0. This is because it’s given that Monty will open an empty door. He can’t do that if you pick the empty door, so you have to leave it for him to pick. Your choice has to be A (probability 1). So:
BAC: 1/3 * 1 * 1 = 1/3
The same logic applies if I pick C:
CAB: 1/3 * 1 * 1 = 1/3
So I win in the cases ABC and ACB, which each have a probability of 1/6; the probability that I will win is 1/3.
You win in cases BAC and CAB; their total probability is 2/3.
So you have an advantage in this game. If that seems counterintuitive, it’s because the restriction that Monty’s door will be empty creates an artificial situation that forces you to pick the winning door if I happen to pick a loser. If you were going to implement a game like this in real life, you would have a do-over if the prize turned out to be behind Monty’s door. You would keep replaying until Monty’s door came up empty.
:eek:
Surely you are not suggesting that the only way for Monty to open a losing door is an “artificial situation”?
Consider that he may have opened it by chance. Nothing artificial about that.
And indeed, that is precisely what the problem says happened.
It wasn’t destiny, or preordained. It just happens that, by chance, he opened a losing door. Thus the remaining doors are equally likely to be the winner.
If angry poltergeist were whispering hints to Monty to guarantee he opens a losing door, then yes, you should switch.
Whether by chance or not, the possibility of Monty picking the prize door is RULED OUT by the statement of the problem. There may be other iterations of the game where he does pick the prize, but the problem only concerns iterations where he does not pick the prize.
So, in don’t ask’s variation, I pick a door, he picks one of the other two doors, Monty picks the remaining door, and Monty’s is reviealed to be empty. In that situation, the probability is 2/3 that don’t ask won the prize.
Before Monty opens his door, the probabilites are indeed 1/3 that I win, 1/3 that don’t ask wins, and 1/3 that neither wins. If Monty now opens the door and reveals that the prize isn’t there, the probabilities become 1/3 that I win and 2/3 that don’t ask wins. That is because we are able to incorporate new information into the analysis: We know that if I picked a losing door, don’t ask must have picked a winning door.
It makes sense when you think about it: The first player to pick has a choice of three doors, so his probability has to be 1/3. The second player does not have a free choice of the remaining two, because he has to leave an empty door for Monty.
Please try the experiment I described two posts ago.
You will be surprised to find that half the times that “Monty” opens a losing door, your original choice is correct!
Seriously. Just try it.
I have to admit… when I first heard about the original question (where Monty does know which door is the winning door) I had to actually do the experiment before it became clear.
Nightime, your coin experiment doesn’t mirror the Monty Hall problem, because your experiment allows Monty to pick the winning ‘coin’, and the original problem does not allow this. The point of the problem is that you are supposed to evaluate the probabilities in the full light of the knowledge that Monty opened a non-winning door.
If you have to make your decision whether to switch before you know what’s behind Monty’s door, then, sure, it doesn’t matter whether you switch or not. But as the problem is formulated, you don’t make your decision until after you have seen that Monty’s door was empty, and you have to incorporate that information into your calculation or you will get the wrong answer. Saying that switching doesn’t help is ignoring information. Switching doesn’t help as long as the possibility exists that Monty picked the prize door, but as soon as you know that he didn’t, switching becomes the best strategy.
You misread the experiment.
You only have to keep track of the times that “Monty” reveals a losing coin, because, as you say, the other times are not allowed by the problem.
However, even though you only consider the times where a losing coin is revealed, half of those times you will have already selected the winning coin. Just try it.
Okay, at last I think I see what you’re driving at. Yes, the coin experiment turns out as you predict. But, what that proves is that if you play this game, with the rule that Monty picks his door at random, and that the game is void if Monty picks the prize, then you will probably win about half the non-void games. But that isn’t what the problem asks.
The problem asks, what is the correct strategy in the situation where Monty has opened a door without the prize. When you find yourself in that situation, you have to ask what is the probability that your first guess, in that iteration of the game, was right. It couldn’t be simpler: You had a choice of 3 doors, so that probability has to be 1/3. Since there is only one other door, its probability has to be 2/3.
I think what you’re saying is, that when you see Monty reveal the empty door, that you are in one of those iterations where Monty doesn’t reveal the prize, so that everything must now be modified by the probability that you would go into one of those iterations. But that isn’t how probability works. Once you know an event has happened, its probability is 1. Regardless whether it was something less before it happened.
It’s as if you were playing poker and got dealt a full house. You might think that you had little chance to win, because it’s unlikely that you would be dealt a full house. But you already have the full house, so it’s now irrelevant how unlikely it was for you to get it in the first place.
Here is an illustration of why your theory of probability is wrong:
Suppose a friend gives you a scratch-off lottery ticket.
Given:
-
There is a 1/2 chance that it is a fake your friend made to trick you into thinking you won.
-
There is a 1/2 chance that it is a real ticket, in which case you have a 1/1000000 chance of winning.
You scratch the ticket, and it says you are a winner.
What are the chances that it is a fake?
By your logic, once you see the “winner” on the ticket, the probability of the ticket saying “winner” is 1, and it doesn’t matter how unlikely it was.
So by your logic, there is a 1/2 chance that the ticket is fake.
But I think you can see right away that your method is wrong, because the ticket is almost certainly fake.
Manduck, I hope you don’t mind somebody else joining in this thread. I’m impressed with your stamina, but it looks like you could do with a hand.
Nightime, let’s look at the game with your friend and the lottery ticket a bit closer.
Suppose your friend has two tickets - one is genuine and one is a fake he made. He offers you the choice and you pick one. Don’t scratch it off yet. Just think about it. What is the probability that you have chosen the genuine one? Answer: 1/2.
Now … well, actually, why do we care what happens now? The probability that you have the genuine ticket is 1/2. If you scratch it off and it’s a winner - or if you scratch it off and it’s a loser - or if you decide not to scratch it off at all but to frame it as a memento of the game - the probability that it is genuine is still 1/2. How could it be otherwise?
Let me know if you want the version with the mathematical formulae and stuff.
Manduck,
you seem to be saying that Monty’s opening of a door doesn’t affect the probability of your initial selection being a winner (1/3).
This might seem plausible when Monty picks a loser, but how about when Monty selects the winning door - does your initial pick’s probability of winning remain similarly unaffected ?
In 1/3 of the ‘Random Monty’ plays your initial pick’s probability of winning drops to 0, and in 2/3 of the ‘Random Monty’ plays your initial pick’s probability of winning increases to 1/2.
1/30 + 2/31/2 = 1/3 1/3 being your overall win probability.
Nightime
I’m not sure where you were going with the fake lottery ticket thing - you’re saying that the fact that it’s a winner makes it more likely a fake ? Sounds more like the murky realms of psychology than a probability question !
The crux of the question for me is that when Monty knows which is the winning door, he is in effect creating the probability subset of 2/3 that previous posters have described. The probability of your initial choice being the winner is unaffected because Monty’s selection is not random - he’s knowingly eliminating a losing choice from a subset which has a 2/3 probability of winning.
When he makes a random pick (whatever the outcome) He is essentially playing the same game as you, with the same probability of winning or losing. In those cases where he picks a losing door, he is simply reducing the overall field of possible doors.
But as Manduck has pointed out about 100000000 times, Monty does not select the winning door. That’s a given. The problem as defined states that Monty picks a losing door. So there’s just no point wondering what happens if, in some alternative universe perhaps, he picks a winning door. In this game, he doesn’t. (Well, OK, there may be some value in the exercise from a theoretical point of view. But as dozens of posters to this thread have demonstrated, it doesn’t help answer the problem that was orginally posed!)
I think all that happened was the first time I posted someone chimed in to say that if the original problem was changed and Monty chose randomly there would be no advantage in swapping. This is not the original problem, but everyone accepts the answer to the original problem. It was just an observation that you had to know the circumstances under which Monty chose the empty door to know if it altered the odds. The observation is correct.