I initially had the wrong answer as well because it seems to reduce so nicely to 1:2. What’s counter-intuitive is that if Monty were to pull someone off the street after one of the unchosen doors had been open, and offered the same deal, it would reduce to 1:2. It just seems wrong that what seems to be the same choice would have two different odds but it does indeed.
Probability reflects ignorance. In a deterministic universe, the Omniscient assigns everything a probability of 0 or 1.
You place $100 on the Pass line at a Craps table, and the come-out roll is … 1. Another guy is blocking your view of the other die. How do you like your bet? 
Meanwhile I, also with $100 on the Pass line, see only the other die. I see 5 and I’m happy enough. 
It’s not clear what you mean here. In the usual game, your initial choice of door specifies the other two doors that Monte will give you information about. Monte compares the other two doors, and shows you the worse of the other two doors, if there is a difference between them.
In other words: suppose you choose door 1. Monte now has only doors 2 & 3 available. If 2 & 3 contain a car and a goat, Monte deliberately shows you the goat - i.e., the worse of doors 2 & 3. If 2 & 3 are both goats, he justs picks one.
The fact that he is comparing the other two doors and showing you the worse of them is the key to the information that you get. This is why it’s right to switch - you are getting the better of two doors by switching.
Monte, notwithstanding Stranger’s protestations, cannot just pick randomly. If Monte picks randomly and just happens to show you a goat by chance, but could have shown you a car, then the game has different posterior probabilities.
What do you envision as the rules of a game where Monte “pulls someone off the street”? If you have not specified which two doors Monte is to compare and give you information about, then it’s certainly quite different.
Not at the state at which the door has been opened. It doesn’t matter whether the host had foreknowledge or not; the contestant now has additional information about the group of doors he didn’t select which changes the probability of the remaining unopened door concealing the prize. The previous state, in which the prize might have been behind the open door, is now eliminated but the potential for the prize to be behind one of the unselected doors (open and closed) remains the same, 2/3.
Stranger
The problem also rests on the assumption that Monty always opens a door. On the original show, that wasn’t the case. Suppose, instead, that Monty’s policy is “If the contestant picks a goat, open their door and tell them that they lose. If they pick the car, then show them a goat and give them the option to switch.”. In that case, always switching is the worst strategy possible.
In the actual show, it wasn’t either of these strategies, but rather a mix of them: Monty would give you a choice if and only if he thought that doing so would trick you into making the wrong choice. It’s effectively like playing paper-rock-scissors with Monty, and he was in practice very good at it.
Sorry Stranger, but you are wrong. All I can ask is that you take the time to think it through again carefully.
So, technically, the “data” is the door that Monte shows you.
Now think through the prior probabilities for each version of the game:
Standard game where Monte chooses goat deliberately.
p=1/3 that I choose car, then Monte shows goat (p=1)
p=2/3 that I choose goat, then Monte shows goat (p=1 because deliberate)
Different game where Monte chooses randomly.
p=1/3 that I choose car, then Monte shows goat (p=1)
p=2/3 that I choose goat, then Monte shows goat (p=0.5) OR Monte shows car (p=0.5)
In the random game only, there is a prior probability of Monte showing a car. The posterior probability, conditioned on the data that Monte has by chance on this particular occasion shown a goat, must take the prior probability into account.
Honestly - this is fairly elementary conditional probability.
In the deliberate came, you should switch for a 2/3 win, the classic Monte Hall result.
In the random game, it’s 50/50.
Alternatively, go up to the simulation I drew up post #40. If you don’t believe me, actually do the simulation.
Mea culpa. On further reflection I have concluded that Chronos and others are right and I was wrong. Here’s my latest take on the question. The point is that when Monte chooses a door at random, there is one chance in three that he will find a car. This reduces by 1 chance in 3, your chance of winning and that just takes away the edge you had in the original question where you go from 1/3 to 2/3 by always switching. Another way of looking at this is to imagine what happens if your first choice was a goat (when you always win the original game if you always switch). That happens 2/3 of the time. But now Monte will choose a car half the time, so your chance of winning if you switch is now only 1/2 of 2/3 or 1/3. You can still play always switch; it won’t hurt you.
Maybe someone with programming skills can run a Monte Carlo simulation game. Brian Hayes, writing in American Scientist ran such a game on the original problem before he was convinced that you win 2/3 of the time if you always switch.
No, Riemann is correct. It’s not just the additional information about what’s behind the door that’s important; it’s also the information about the circumstances under which you’re playing.
In the “classic” Monty Hall problem, Monty must pick a goat, and therefore there must be an endgame where you opt to switch or stay.
In the “random Monty” option, sometimes Monty picks a goat, and sometimes a car, so you don’t always get to the endgame. If you do make it to the endgame, you have an additional piece of information to factor in: the known probability that you got into the endgame in the first place.
I suspect this might be more intuitively obvious when contrasting similar games played with, say, 100 lottery tickets (i.e., the difference between being guaranteed a spot in the endgame v. landing there randomly with 2% probability).
A way to emphasize the point is to use 100 doors and Monte open 98 doors to reveal 98 goats.
If Monte chooses his 98 doors by intentionally dodging the car, the remaining door has a 99% chance of having a car in it (since the set of 99 doors that Monte had to choose from had a 99% chance of having a car somewhere).
If Monte chooses his 98 doors randomly, it is rather unlikely that he will manage to reveal 98 goats and 0 cars. But in those rare cases where Monte does manage to show 98 goats, the remaining door is on an equal footing with the player’s initially chosen door. It has a 50% chance of containing a car. This is more obvious by having Monte randomly choose which door not to open. This way, the player will have chosen one door randomly and Monte will have chosen one door randomly, and the other 98 random doors would be opened. If, by chance, 98 goats are revealed, the two special doors are still equivalent.
In an ensemble of many runs of the game, the opening of the 98 doors tells us the same amount of information either way. In each run of the Monte-knows-where case, we have a 100% chance of learning which door has a 99% chance of having a car. In the Monte-is-random case, we have a 99% chance of learning which door has a 100% chance of having a car. In the first case the information is useful. In the second case it is not since we aren’t allowed to switch to the revealed car. Since we are only interested in the cases where the car isn’t revealed, we’ve “thrown away” all the cases with useful information. But in the full ensemble games (including the ones we aren’t examining), the information is indeed there.
If you look up to my post #36, I commented on the two original statements of the problem that Stranger kindly posted. It was interesting, I had not seen them before.
The original problem, posted in a statistics journal, made it abundantly clear that Monte was choosing to a show a goat deliberately, he would never show a car.
In the later magazine version that popularized the problem, this was not made clear. Nor, remarkably, is it made clear in the current Wikipedia statement of the problem! They do say that Monte knows what is behind all the doors, but they do not say what he does with this information, how he decides which door to open. And honestly, when not otherwise stated, "randomly’ tends to be a default assumption in many problems.
I tend to think that the widespread confusion and the legendary “Nobel Prizes and PhDs getting it wrong” is at least in part attributable to the fact that the problem was framed ambiguously, and many people assumed incorrectly that Monte was choosing a door randomly.
The result is still a little counterintuitive either way, but it’s much easier to get a feel for it if you realize that Monte is deliberately comparing the remaining two doors and showing you the worse one.
Yes, the striking thing in this version is that under the Monte deliberately shows only goats rule you reach the endgame every time. And obviously you should switch to the one remaining door (out of 99) that Monte deliberately kept hidden.
In the Monte opens doors randomly version, of course 98% of the time you just don’t reach the endgame. Monte ends up showing you where the the car is by accident. In only 2% of the games is the car by fluke not among the 98 doors that Monte has been opening randomly, and it’s even money which of the two remaining closed doors has the car.
If I pick door #1, and Monty opens door #2 and reveals a car, I’m absolutely going to switch… to door #2.
If you want to get pedantic about it, there is no “last ball removed” as you always remove two at a time. If, when you remove two balls, you do so one ball at a time (that is, remove one ball, then remove the other)…
[SPOILER]There is a 50% chance that the “last ball removed” is blue, and 50% that it is red.
Notice that (a) the number of red balls either remains the same or is reduced by 2 after each draw (and replacement), and it starts as an odd number, so it will always be an odd number, and (b) the number of total balls is reduced by 1 after each draw (take 2 out, put 1 back in). Eventually, by (b) there will be 2 balls left, and by (a), one is red.
Note that, by (a), when you get down to one ball remaining, it is always red - but the question did not ask which ball would be remaining.[/SPOILER]
Moderators, please move this thread to the Monte Hall/Plane on a Treadmill subforum.
Thanks!
This is a good statement of the foundational source of (most of) the confusion. But this next point is also key:
Said another way, the actual TV show problem is not the simplified purified version that **Riemann **just posted about. Instead it’s the ambiguous version that wiki accurately describes.
Monty may do something with his foreknowledge or he may not. He may do a goat-reveal after the contestant’s first pick or he may not. And that decision is not strictly random, although it’s certainly not 100% rule-driven either.
THAT’s the game they were actually playing on TV. And it has very different posterior probabilities than the simpler case **Riemann **described above and in other good posts. Because we have lots more factors to consider. Right down to that Monty has a goal of his show staying entertaining from week to week and Monty being unpredictable added to that.
I don’t know if Monty’s repertoire included deliberately revealing the car after the contestant picked goat #1. With the goal of enticing the contestant to switch to goat #2 which is a funnier prize. I doubt he ever revealed a car on the show. If it did happen, I’d expect it to have been inadvertent. But maybe not.
That’s interesting. I’ve never seen the actual show, I did not know that it worked in an *ad hoc *way like that. As you say, if Monte is motivated to stop you winning, or even just to make the game arbitrarily “interesting”, and he has the option to not open a door at all, then it’s a completely different problem.
Was the game current on TV at the time Marilyn Vos Savant put out the problem in her newspaper?
There’s so much potential confusion about the unspecified rules that I think it’s a little lame to claim that even professional mathematicians and Nobel Prize winners screwed up.
Yes, but you didn’t do that. You looked for every single possible way the question could be even slightly ambiguous in an effort to avoid answering the question in the first place. You didn’t even ask for clarification–you just stated the problem was impossible.
You were told to add an extra blue ball, which means not from the original set. Furthermore, you had to add the extra balls, meaning they couldn’t be taken from the same source where you put them, or you’d be adding nothing. And taking the balls out was clearly stated as taking away, not adding, so it can’t mean adding them back to your hand.
Just because not everything was spelled out doesn’t mean the actual question was ambiguous. You had to reach for far fetched, nonsensical interpretations to make it not work.
I’m really not sure how to respond to this. Your post sounds very angry, as if you’re demanding an apology from me or something. And yet I’m completely at a loss to see what I should have done differently. You say I didn’t ask for clarification but that’s pretty much the bulk of post #2 – me asking for clarification. By my count, I asked four specific questions in that post. I tried to be thorough. But then you seem to be angry about the fact that I was too thorough. Then you claim I was making “an effort to avoid answering the question”. This is a very strange thing to say, considering the fact that I actually did answer the question, in post #12. My answer is under a spoiler tag.
When I first saw the problem, several aspects of it didn’t make sense to me. I considered the possibility that I had misunderstood what was being asked, so I tried to imagine various other ways it could have been phrased more clearly. I considered trying to answer it by following my own assumptions about what the rules were trying to say, but then I noted that the OP had specifically asked for a factual answer, after having been disappointed with a “large array of differing answers”. I decided it would be better to ask for clarification before trying to answer the puzzle. So that’s what I did.
This meta-criticism of sbunny8 all seems a little OTT to me. The problem was very poorly worded, and asking for clarification was reasonable. Quite honestly, when I first saw the problem on another thread it was worded the same way, and I didn’t bother with it, because it was ambiguous, did not come from a trusted source, and I couldn’t be bothered to try to parse out whether it interesting or not. It was only when I saw the solution that I realized it actually was an interesting problem.
So, was sbunny8 a little verbose in the way he asked for clarification? Maybe, but really, what’s the big deal? It’s ok to state problems carelessly, but there’s only one acceptable way to phrase a request for clarification?
And, it was a little ironic, to say the least, that one of the critics of sbunny8 then made an error on Monte Hall because of an ambiguity in the framing of the problem that he mistakenly thought was unimportant.
No. Thats like moving a question about the sun to a question about fusion.
Monte Hall is mentioned as serving as an example of
“well its absurd to use ambiguity down path B, so stick to path A.”
The OP didnt even contain the ambiguity,because the question made it clear the final ball was allowed to be removed as one. AND anway whether it was removed or you just had to look inside the bag/box,etc , the question about what colour it is wasn’t going to change, its no schrodingers cat.
(BTW schrodingers cat is no cat, schrodingers cat is a tiny particle allegorically named as schrodingers cat.)