I’m not sure I understand your answer here. If Monty doesn’t know, then you are indifferent between switching or not. One-third of the time you were right in your first choice. One third of the time Monty picked the right door (and assuming your not allowed to switch to this you lose) and one third of the time neither of you picked the right door. Once he shows you the zonk behind his door, your odds are therefore (1/3)/(1/3 + 1/3) = 1/2 whether you switch or not (or 33/66 as you put it. But the result is definitely not the same as if he knew, where your odds favor switching 2-1.
You sure about that?
I don’t agree. If we are given that Monty has already rejected a blank door (after you’ve made your choice), then it doesn’t matter what his policy was for selecting it. You will win 2/3 of the time — 2/3 of those times, when the given condition is true — by switching to the remaining door.
If Monty ends up rejecting the prize door, you obviously have no winning strategy in that case. (B*stard!) That’s probably why the game is never played that way in practice. The audience would revolt. (As if they weren’t revolting enough.)
I don’t think that Cecil ever addressed the argument that it becomes a new game when one door is opened. He simply said, “No, it’s not.”
Since my first analogy apparently didn’t do the trick, let’s try this:
Suppose instead of me picking one of the three doors, a computer picks it for me, completely randomly, before I even arrive at the studio. Monty tells me, when it’s my turn to play, "The computer has randomly chosen door 1 for you. Now it’s a 1 in 3 chance that the computer picked the prize door.
Monty knows which door has the prize. He chooses one zonk to open, based on a randomized formula: he either opens the highest numbered zonk or the lowest. In this case, it must be a high day, because he has opened number three for me. Now he wants to know if I want to switch.
At least that’s how it’s supposed to work. But my alarm didn’t go off, and I missed my bus. By the time I got there, Monty had already told the audience what number door I had, and opened number three, having revealed it to be a zonk. When I arrive, out of breath and apologizing, an enormous fat woman from makeup arrives and starts dabbing my forehead. She’s so enormous that she completely obscures door number 3. I can choose to keep door number 1 or give it up for door number two. One of them has the prize. One does not.
So my question: How is that not 1:1 odds that door 1 is the prize door?
Because the door you were randomly assigned has a 1/3 chance of being the right door. The other one has a 2/3 chance of being the right door. Nothing in your scenerio changes those odds.
Nope, if Monty opens a door randomly, and happens to select the zonk, the choice you are left with is between two equally probable doors, so it’s 50/50.
This gets back to what I keep pointing out every time this comes up. Monty’s intentions do matter! Here are the different scenarios for what his intentions might be:
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Monty always reveals a zonk door. You should switch, because switching will let you win 2/3 of the time.
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Monty reveals the zonk door only when you’ve chosen correctly on your first guess. You should never switch, obviously.
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Monty picks a door at random, and this time has happened to pick a zonk. Your intuition is right in this case - it’s 50/50.
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Monty plays some strategy to make the game show exciting, and knows which door has the prize, but doesn’t always show a zonk and offer the switch. The probability of winning by switching can be anywhere from 0% to 100%, depending on his mood.
I’m not entirely sure I completely understand your analogy, saoirse, so please correct me if I’m mistaken.
As I see it, you’re saying that you arrive at the studio, are informed that door number 1 has been chosen for you, and you may either keep it or switch to door number 2. At no point (here’s where I may be misunderstanding something) are you made aware of the existance of door number three. You then proceed to conclude that you have a 50/50 chance with either door 1 or 2. The problem is that you are playing with false assumptions: namely that there are two doors, one of which has a prize behind it, and that you may pick one of the two (i.e., stick with door 1 or switch to 2).
The catch is that with a false assumption, you can draw a false conclusion. Suppose, for example, we tweak the rules slightly. Let us allow Monty to occasionally open a door with the prize behind it. Let’s also assume that today, door number 3 was opened, and had the prize. Now, continue with the scenario as before – that when you arrive, you don’t know about door 3. Now Monty is a bit upset with you for arriving late, so he decides to play a joke. He tells you that the computer has chosen door number 1, and you may either keep it, or switch to 2. (At no point does he mention door 3, or that the prize has already been revealed.) You’re operating under the same assumptions as you did in original scenario in the paragraph above, so you figure you’ve got a 50/50 shot. Of course, this isn’t true. No matter which door you take, you’ll never win the prize, since it was behind door 3. The problem was that your initial assumptions didn’t represent the actual situation, so the conclusion you drew wasn’t reliable.
I also want to second the suggestion that you try it for yourself. Find a friend, and have them put a ball under one of 3 cups. Pick one, and have your friend reveal a dud. Choose to stay or switch, and record whether or not you made the right decision. It becomes pretty obvious pretty quickly that switching is the better option about 2/3 of the time. I’ll also point out that one of the reasons this question tends to draw controversy is that often (at least, when I’ve seen it discussed), all the factors aren’t explicitly listed: whether or not Monty always opens a door, whether or not he knows where the prize is, whether or not the opened door can contain the prize, etc. Changing those factors can change the outcome, so it’s very important to make sure everything is stated at the beginning.
Here is where I get stuck. If Monty didn’t know which door had the prize, and could reveal it, then showing door 3 as a zonk would be useful information to me, and could get me to change. Monty can only show a zonk to me, though, so essentially he’s always going to remove one of the three. My odds of winning were never really 1/3 to begin with, because one of the wrong answers was always going to be eliminated. Once that door is shown to be wrong, I have two doors, one right and one wrong. Asking me to keep or switch is really no different than asking me to pick one or the other.
Right. And Marilyn’s answer only works IF we make the same assumptions about why & when Monty opens another door. As Cece puts it: "only if we make several assumptions: (1) Monty Hall knows which door conceals the prize; (2) he only opens doors that do NOT conceal the prize; and (3) he always opens a door. Assumptions #1 and #2 are reasonable. #3 is not. Monty Hall is not stupid. He knows, empirically at least, that if he always opens one of the doors without a prize behind it, the odds greatly favor contestants who switch to the remaining door. He also knows the contestants…will tumble to this eventually. To make the game more interesting, therefore, a reasonable strategy for him would be to open a door only when the contestant has guessed right in the first place. In that case the contestant would be a fool to change his pick.
But that’s absurd, you say. If Monty only opened a door when you’d chosen correctly in the first place, no one would ever switch. Exactly–so it’s likely Monty adds one last twist. Most of the time he only opens a door when you’ve chosen correctly–but not always. In other words, he tries to bluff the contestants, then counterbluff them."
I think this is the problem. It doesn’t matter to your initial choice that one wrong door will be eliminated. Suppose you enter the studio and are told to pick a door. (Let’s say you choose door 1.) That door is then opened and if the prize is there you win. If not, you lose. You will only win 1 time in 3. That is, your intial (and in this case only) pick has a 1/3 chance of winning.
Now, add the twist. You pick a door. You know from the case above that you only win one time in three. After you pick, Monty tells you that he will open one door (let’s make it door 3) that doesn’t have the prize behind it. (Because he knows where the prize is, and he will always open a door without the prize). After that, you can stay with your pick, or switch to the other door. Now, your initial pick had a 1/3 chance of being right. The odds of that door being right don’t change after he opens door 3. So, you know that door 1 had a 1/3 chance of being right, and door 3 has a 0 percent chance. This means that door 2 must have a 2/3 chance of being right, so you should switch (since the probablilities must add to 1).
If this still seems complex, try making a table, with all the possible results. You pick door one, prize is behind door one, and Monty opens door two. You pick door one, prize is behind door one, and Monty opens door three. And so on. Then for each case figure out whether you should have stayed or switched. You’ll see that switching wins 2 times out of 3. And I’ll suggest again trying it for yourself. It can really help to show what’s going on.
There are indeed two remaining possibilities, but the odds are not split evenly between them. Although you now know that the probability the eliminated door has the prize is zero (because it’s been opened and revealed to be empty), the probability that your chosen door was the winner is still only 1/3. That is, if you played the game 3 million times, and always stuck with your initial choice (so you never switch), you will only win about 1 million prizes. If instead you always switch, you’ll win the other 2 million.
This scenario just isn’t analogous.
Suppose you played the Daily Pick 3 game over and over, say about 10,000 times. To simply things, let’s say you always choose number 123 (because it’s your favorite; maybe your first dog was named “123”). In total you should expect to lose 9,990 times, and win 10 times.
Of those trials when the first ball didn’t come up 1 — which is 9/10ths of the time, or 9,000 times out of the 10,000 — you will have won zero times. Of those trials when the first ball did come up 1 — which is 1/10th of the time, or 1,000 times out of the 10,000 — you will have won all of your 10 victories, which is indeed 1/100th of the trials that begin with a 1.
Your probability of winning a Pick 3 game is always 1/1000, assuming no shenanigans. Your probability of winning a game, given that the first digit is a 1, is indeed 1/100, but that condition only holds for 1/10th of all possible trials.
In the case where the fat woman blocks the door that Monty reveals (and you can’t see whether it’s blank or not), then the question becomes, “How much do you trust Monty?” If he told you that he’s only going to reject a blank door, and you believe he’s honest, then we’re back to the original Monty Hall problem; your best strategy is to switch.
Of course you’re trusting him that there’s a prize somewhere anyway, otherwise the whole game is futile. So in my view you might as well believe him, you might as well play along, when he tells you he’s rejected one of the “zonk” doors. For a one-shot game, with no other information available, it’s your best bet.
In an iterated series of games, you could tell whether Monty is being honest or not by comparing your running score with 2/3. If it’s significantly less than that (significant in the statistical sense), then you’ll have a pretty good idea he’s revealing the prize door sometimes.
I’m confused as to why you would ever stand pat if a second door gets revealed, regardless of Monty’s motivation.
Your initial choice has a mere 1/3 chance to win. Period. If another door gets revealed, and you are allowed to switch, then you double your odds by switching, regardless of Monty’s motivation.
Just because your odds are doubled, doesn’t mean you’re guaranteed to win. But I’ll switch every time no matter what. If I’m always allowed to switch, I’ll win 2/3 of the time.
Just read my post #15 above. The answer is as plain as the nose on your face. Always switch.
Again, I don’t agree that’s the case here. If we are given that (1) Monty has opened one of the two doors you didn’t pick, and (2) has in the process revealed that door to be a zonk, then a strategy of always switching will win you the prize 2 times out of 3. That is, you’ll win 2/3 of the times when these conditions hold. It doesn’t matter how Monty arrived at his choice: whether it’s a zonk by sheer luck, or a zonk by deliberate arrangement. The observation that it’s a zonk is enough.
In the usual descriptions I’ve seen of the Monty Hall problem, it’s generally implied (though admittedly not stated outright very often) that Monty’s choice of door can’t be the prize door. I suppose it’s possible he could reveal the prize, though I’d have to ask: what happens next in that case? The prize is now out there for all to see. Surely the player has just lost? Or has he just won? In any case, the player wouldn’t seem to have any more decisions to make. The game ought to be over at that point.
I don’t think the probability can ever reach 0%, unless Monty is lying about the existence of the prize in the first place, or he’s surreptitiously moving the prize around after you’ve picked your final door.
I would break down the possibilities like this…
If Monty can decline to offer you a switch, then in those trials when he doesn’t do so (however many those are), you obviously have no further decisions to make. You’re stuck, I assume, with the door you first picked, for better or for worse. Therefore your probability of winning trials of this class is 1/3, and you have no worthwhile strategy. (This version of the game would seem to me to make bad television.)
If Monty offers you the chance to switch, but doesn’t open any door beforehand, then you have no particular motive to switch. You have no particular motive to stay either. So switch if you find it entertaining to do so, but it won’t make any difference. Whichever door you settle on, you’ll win about 1/3 of the trials in this class (however many those are). Again, you have no real strategy.
If Monty opens one of the two doors that isn’t your initial door, and his chosen door has the prize behind it, then we’re in the scenario I was puzzled over a minute ago. What happens next? It seems sadistic to force the player to choose between the door he first picked (which he now knows is a loser), and the other door that he didn’t pick and Monty didn’t open (which he also now knows is a loser). If Monty offers you this choice, your strategy should be to punch him in the nose. Not a useful strategy from a strictly mathematical viewpoint, but rewarding in its own way. (And it makes for some good television.)
Finally, there’s the Monty Hall problem as it’s usually described — or at least as I interpret it — and which to me is the only interesting scenario. After your initial pick of door, Monty opens one of the other two doors, revealing that it does not hide the prize. He now offers you the choice of staying with your door, or switching to the other one. Given that this particular sequence of events has already happened, you should always switch. For the reasons discussed in this thread, by me and other posters, this strategy wins you the prize 2/3 of the time.
That 2/3 is of all trials from the class of trials described in the last paragraph. Maybe all the trials will go that way, or maybe only one in a million will go that way. That could depend on many things, including Monty’s motives and policies, the flips of coins and the rolls of dice, the time on the clock, or votes from the audience. Nonetheless, we don’t have to care about any of that once we’ve observed that Monty eliminated a door that is not the one we picked, and not the prize door. Given that observation by itself, and regardless of what Monty is up to, we already know enough to switch to the other door.
One thing to remember: Not all zonks are created equal. Some zonks will actually result in the player earning some money. Let’s say you zonk and win his and her pigs. You will be given the option to take the cash value of two pigs at the end of the show. No one in their right mind will take the pigs, but they are worth a certain amount of money.
Here’s a different version of the Monty Hall puzzle that might give you some insight:
Monty gives you the choice of three doors. Only one of them has a prize behind it - the other two are empty. You choose one - let’s call your choice door A. He then gives you the option of keeping door A or trading it for the two doors you didn’t choose - let’s call them doors B and C. When he makes the offer, he promises that after you’ve made your choice, he’ll open one of doors B and C that happens to be empty (we know that at least one is empty, and maybe both are). Once you’ve seen that door, he’ll open the remaining doors and you’ll know whether you’ve won the prize.
Should you switch door A for doors B and C? Of course you should - at the time you have the opportunity to make the switch, you don’t know anything about where the prize is, so choosing two doors doubles your odds of winning the prize. The fact that Monty will open one of doors B and C before opening all the doors doesn’t change anything, because you’ve already made your choice.
The original problem is sort of like the one I’ve just described. When you pick a door, the probability that the prize is behind the door you picked is half the probablity that the prize is behind one of the other two doors. Now Monty opens one of the other doors with nothing behind it and offers to let you switch. You know in advance that Monty wouldn’t open the door with the prize at this point in the game, so the probability that the prize is behind one of the two unchosen doors is still the same, even though one of them is open now. You do know, however, that one of those doors is empty, so the remaining unopened door must have twice the probability of having the prize as the door you picked.
When I first thought about this I thought about it the same way. It’s tempting to think that after Monty has eliminated one door, you have two doors left and the car is behind one, so that’s a 50/50.
The analogy that got it clear in my mind why this is wrong is this:
You have a tank which contains one fish. Monty pours the content of the tank into three compartments A, B and C, so that the fish falls into one. You try to pick which compartment, and choose A. Monty then runs a net through the content of compartments B and C, and dumps the catch (if any) into compartment B. You then choose from either A or B. Straight away you can see that despite the fact that you have two choices, the odds aren’t 50/50, because one of the compartments contains 1/3 of the content of the original water, and the other contains the content of 2 of the original compartments, condensed into one. So you’re clearly better picking the latter.
Going back to the original “pick a door” game, when Monty eliminates one of the two remaining doors, he is in effect “filtering” two of the original three choices (each of which originally had a one third chance of containing the car) and giving you the chance to pick the filtrate, thereby effectively giving you the chance to choose a filtrate that contains 2/3 of the possible original choices.
Here’s what bothers me about the “correct” analysis.
Suppose Monty allowed 2 contestants to choose doors. He then reveals the
third door that wasn’t the winner.
How is it that one contestant has a 1/3 chance of winning and the other 2/3?
Suppose Monty tells the 2 that they can swap their choices. Should they swap?
According to the prevailing analysis, contestant 2 should stay pat, but 1 would beneft from switching.
Help please.
That’s not the same situation at all. The unchosen door might be the winner. In the original, Monty always one choice, sometimes two.
OK, using the original problem, after Monty reveals the “loser” door, he invites a second contestant over and say to him, you can pick one of the two remaining doors, I’ll even let you force contestant 1 to give up door 1 if that’ the one you want. What should contestant 2 do?
Now in my eyes, contestant 2 has a 50-50 chance of picking the right door because there are 2 doors and only 1 winner. But this contradicts what you all have been saying, that taking door 2 has a 2/3 chance of winning.
How is this reconciled?