Home/BlogA diversion from global warming topics.
The simple little probability problem below has apparently been debated for many years. It came to prominence when Marilyn vos Savant answered a reader’s question about it. Her answer was believed to be wrong by some of the greatest statistical minds in the world, and eventually most of them admitted she was correct after all.
A story about that debate is here.
But I maintain that the answer depends upon an unstated assumption, and so there is no correct answer. Of course, I could be wrong. Disagreeing with a person having the highest IQ in the world is, statistically speaking, not a smart thing to do.
The Monty Hall Problem
There are three doors, and behind one of them is a new car, and behind the other two doors are goats. You want the new car. You choose door #1, knowing you have a 1 in 3 chance of winning.
Monty Hall then opens door #3 and shows you a goat there. Should you change your pick from door #1 to door #2? Most people said no, that you still don’t know whether the car is behind the first or second door, and all that has happened is your chance of winning has simply improved from 1/3 to 1/2.
But Marilyn vos Savant said “yes”, that you should switch. Experts disagreed.
From what I can tell, through, it entirely depends upon why Monty Hall showed you what was behind door #3.
If there is a goat behind door #3, then clearly the new car is behind either door #1 or door #2. If Monty Hall was going to show you door #3 no matter what was behind it, then your chances are still 50/50… you might as well stay with door #1.
BUT…if Monty Hall was only going to show you a remaining door that had a goat behind it, then you should switch to door #2. The reason is you would have new information you didn’t have before…that if he knew that the new car was behind one of the remaining doors, he was going to in effect tell you that by not opening that door.
In that case, you actually have a 2 in 3 chance of winning by switching doors.
But, as far as I can tell, which of these two assumptions is in effect was never stated, and so there is no correct answer to the problem.
(RIP, Monty Hall).
I think the issue is confused by the way it was originally presented. If the host were to open either door #2 or door #3 to reveal a goat, then the contestant will increase his chances of winning by switching from his choice of door #1 to the other door that wasn’t opened, whether #2 or #3.
Here is an easy way to think about a problem like this.
Suppose you had to choose one out of 100 trillion doors.
Once you make your choice, you know you didn’t pick the correct one. You could bet your life you aren’t going to win with 100 trillion to one odds. The one you picked just isn’t correct. Now suppose I open every door except your pick or one other door. Do you keep your original one in a 100 trillion door or switch? You could make the claim that there are only two doors and it’s a 50/50 chance, but you can’t neglect that your original choice was made with much worse odds.
Thanks for that bit of logic – easiest way I’ve seen so far to wrap your mind around the Monty Hall problem.
Yes, but why would you open for me 999.999.999.998 doors if not to make me change my lucky choice and lose my price? The reasoning works only if you assume that the presenter doesn’t care if you win or lose. If he prefers you to lose, then he will try every trick to make you change your lucky choice, including apparently improving your odds.
The odds do not matter what his intentions are. If the first door you picked was one out of a hundred trillion, then those were your odds – 1/100-trillion. Those odds will not change no matter what the game operator’s motives are. Now if good old Monty opens up every other door except yours and one other door and then asks you to switch, the odds are now 1 in 2 and that won’t change if Monty hates you or wants to marry you. It is the same for 100 million, 1000, 100, 50, 10, or 3 doors. The odds are lower for your first pick and if all the doors but yours and one other door are then opened, what is left is a 1 out of 2 pick no matter what Monty’s motives are.
Agree except when you say odds are 1 in 2.
Mike O. What youre saying is incorrect.
The odds are indeed 50/50.
Everytime a new scenario is presented odds reset.
That’s how card counting is done.
Everytime you remove a door my odds of being correct increased, until it wittled down to 50/50.(assuming randomness)
The explanation you should be giving is that randomness is taken out of the probability.
If someone selected 1 of a trillion AND the other player owns the other 999.99 billion AND he chooses to show you everything except the correct door. Then it’s clear you switch- ONLY BECAUSE HE OWNED THE REST.
IF he had chosen at random from the remaining trillion its obvious he had it and its obvious it wouldn’t take until the last door to reveal it- If he did however, in a random draw, reach the last door- odds are 50/50
That’s why prize money goes up in who wants to be a millionaire, with every low amount taken out.
A better demonstration would be with 52 cards.
If I play Monty in a game to find the Ace of spades
Monty pulls at random and I pull at random, we both have a 1/52 chance of having the ace of spades.
The deck has the remaining 50/52
If he pulled all the other 50 cards revealing no ace of spades. it didn’t change his 1/52 or my 1/52 except to say its now 1/2 or 50/50.
However the game isn’t random.
If Monty could take 51 cards and leaves me 1 his odds were 51/52.
He then examines the 51 cards and chooses to show me 50 cards (leaving one face down) which would be anything except an ace of spades.
In this scenario the odds (51/52) of having the card never changed regardless of how many wrong cards he showed me. And i would prefer to have his 51/51 than my 1/52
correction: And i would prefer to have his 51/52 than my 1/52
“If Monty Hall was going to show you door #3 no matter what was behind it, then your chances are still 50/50… you might as well stay with door #1.”
As I understand it the problem derives from a US game show, presented by a Monty Hall, who on choosing the first door always revealed a goat rather than a car. The game would be daft if he revealed the prize at this point, since it would depend on Monty Hall’s whim rather than the contestant’s choice.
Indeed
Monty knew #3 did not have the car, and was thus free to open it without being accused of robbing the contestant. The question then comes down to his motive for jumping in ahead of the contestant. Assuming impartiality on his part, there is no point in changing to #2.
You double your chances by switching, based on the events as described. It’s a probability puzzle and there is a correct answer.
Why do I always agree with barry?
It’s easy to let someone you trust do the thinking.
It helped a bit I had been through it before.
You double your chances if Monty Hall always opens a door (with a goat behind it). That this problem is so hard compared to sleeping beauty problems, always strikes me with a surprise. Sleeping beauty is another problem or set of problems which trigger(s) wildly wrong, or erroneously stated analyses.
Motls explained Sleeping Beauty with lots of temper and sap some time earlier and even him failed to fully grasp the apparent problem on stating the question, i.e. defining the distinction between probability of initial variable value and probability of actually detecting that value.
Most interesting phenomena, intelligent people bashing each other over question they think they understand right and well as others do not.
What you are saying is correct. But Marilyn vos Savant was not the first person to pose the problem. It was originally posed by Steve Selvin in 1975. Selvin pointed out that there are “critical assumptions about Monty Hall’s behavior that are necessary to solve the problem”. Any proper description of the problem should include this criterion … alas most don’t.
Assuming Monty will always reveal a goat, the simplest explanation (I believe) is:
If you play this game over and over, and use the strategy of switching every time you play, you will win the car PRECISELY when your initial choice is a goat. That is … two out of every three games.
An interesting variation:
If when the contestant initially chooses a car, so that Monty has a choice of two doors to reveal, Monty ALWAYS opens the leftmost door, the probability of winning by switching reverts to 1/2. The 2/3 chance of winning is dependent on Monty choosing randomly when he has a choice of two doors to reveal.
This problem is related to another common paradox … the two daughters problem. I still see this problem taught incorrectly in online lectures:
Assume you know that your friend has two children, but don’t know their genders. If he tells you that at least one of his children is a girl, the probability that both are girls is 1/3. But if he tells you that his OLDEST child is a girl, then the probability that both are girls is 1/2.
These are examples of conditional probability. A sad case of conditional probability applied wrongly was that of Sally Clark who was convicted of killing her two children based solely on incorrect mathematical reasoning. I’ve forgotten the actual numbers, but let’s say that the chance of a particular child dying of SIDS is 1 in 5000. Clark had two children die without apparent reason. The prosecutor argued that the probability of two children dying of SIDS is 1 in 5000 squared or 1 in 25000000, so therefore it was almost certain that Clark killed her children. But he was asking the question the wrong way around. Instead of asking “What is the probability that two specific children will die from SIDS”, he should have asked “GIVEN that two children died unexplainably, what is the probably they DIDN’T die of SIDS”. When you compare SIDS rates to child murder rates, that probability is VERY low. Clark was convicted without physical evidence or motive.
If when the contestant initially chooses a car, so that Monty has a choice of two doors to reveal, Monty ALWAYS opens the leftmost door, the probability of winning by switching reverts to 1/2. The 2/3 chance of winning is dependent on Monty choosing randomly when he has a choice of two doors to reveal.
I think the 1/2 chance here comes from us *knowing* that the contestant has picked the car, not from which door Monte chooses.
If we don’t know if the contestant has picked the winning door, the odds remain 2/3 by switching no matter what door Monte tends to open (assuming he’ll never open the door with the car).
The contestant doesn’t know he has picked the car.
This iteration is mentioned in wikipedia:
https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors
You’d still switch, as the probability is still better than 1/2.
The probability given is 1/(1+q). In my scenario q=1, so the probability is 1/2.
The odds would change, I think, if the contestant knows Monte’s preference.
I think most people would assume the conditions of the problem as they were intended – that the door picks were random (and the host will never open the door with a car behind it).
That is always the point of the Monty Hall problem – that the contestant is aware of Monty’s strategy.
In the original and second iteration (vos Savant), seems clear to me that the only inferred strategy is that the host will open (any) door with a goat behind, not that the host has a bias to pick the leftmost (or rightmost) door with a goat.
Not that there’s anything wrong with exploring other iterations.
That’s all I’m doing – exploring other scenarios (I don’t think “iteration” is the right word).
I get it. But I’m a little confused by:
That is always the point of the Monty Hall problem that the contestant is aware of Montys strategy.
The only strategy I can evince from the wording is that Monty deliberately chooses a door with a goat behind it – no other bias in which particular door.
I was thinking about your scenario a bit today. I’ll spell it pout here for fun, see if I can see it clearly. Monty has a 100% bias for choosing the rightmost door-with-a-goat and the contestant knows this (I’m 82.4% confident this is the scenario you’ve described – do correct me otherwise).
Contestant picks door 1:
If Monty picks door 2, the car is behind door 3. You will be 100% correct switching.
If Monty picks door 2, even odds the car is behind 1 or 2.
Contestant picks door 2:
If Monty picks door 1, the car is behind door 3.
If Monty picks door 3, even odds the car is behind 1 or 2.
Contestant picks door number 3:
If Monty picks door 1, the car is behind door 2.
If Monty picks door 2, even odds the car is behind 1 or 3.
Still hard to let go of the idea that your first pick was a 1/3 chance, and that the other 2 doors are 2/3.
I need to think about this some more. Feel free to help me out.
(I love this puzzle, by the way – it’s simple, nonintuitive, and elegant)
Based on the following:
1. That the host will always reveal a door with a goat behind it, leaving the contestant with an apparent 50% chance of winning.
2. The contestant will play by the the Vos Savant strategy, and switch doors when presented the choice.
…………………………………………………..
Note: There are 3 doors but only 2 outcomes. When the contestant first chooses, they have a 1/3 chance of selecting a car, and a 2/3 chance of selecting a goat.
If they happen to initially choose correctly (1/3), the Vos Savant method will always give them a failed outcome.
However, If they happen to choose incorrectly (2/3), the Vos Savant method will always give them a winning outcome.
Hence the Vos Savant method will give contestants a 2/3 chance of winning.
The two daughters problem is way more compolicated than you suggest. It depends entirely on how the information is revealed and in what context.
It is fiarly unusual for someone to say something along the lines of “At least one of my two children is a girl.” With no agenda or context. But if that was the case, then the possibilities are:
G-G, G-B and B-G. Making the odds the other one is a girl one in three.
But it is much more natural for someone to reveal that information a different way. In general, three-fourths of people with two children have at least one daughter. (G-G, G-B, B-G, but not B-B).
So the key is how and why the information is revealed. If the person first mentions having two children and later reveals the sex of one of them at random then the odss are back to fity-fifty.
It is inherent in the hypothetical that the host will only ever open a door with a goat behind it because the puzzle is presented as a game-show, and no game show reveals the correct pick to the contestant.
Odds demand you switch.
Even if the host does NOT know what is behind the doors, and opens one at random showing a goat behind, the odds are no different.
You switch.
If the host opens a door with the car behind it, then you, rather obviously, pick that door.
In all cases, you switch your choice.
Imagine the host opened door #3, revealing a goat, before you chose any door.
Now, should you choose door number 1 or door number 2?
It doesn’t matter. Equal chances of winning.
It only makes a difference if the game show host purposely avoids revealing the winning door.
The original question to Marilyn vos Savant never mentioned this, so it is ambiguous and cannot be answered:
“Suppose youre on a game show, and youre given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows whats behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?
Craig F. Whitaker
Columbia, Maryland”
Not ambiguous. We know how game shows operate. Monty revealing the car would be pointless and ruin the show.
If Monty eliminates a door at the start, then you have just one decision between two doors, so sure, just 50/50. (You could make two decisions if you get to change your mind once, but you have zero new information between the two decisions, so still just one decision.) But that doesn’t seem related to the actual problem. In the actual problem, you have two decisions with three doors and you have new information before the second decision. That new information is which of the two other doors WOULD have the car, if your initial guess was wrong. That was very valuable new info. (And, it seems to me you get that information upon seeing the goat, no matter why Monty chose that door?)
https://en.wikipedia.org/wiki/Bayes%27_theorem I have been informed that the answer she gave has to do with Baysian probability.
Dr. Spencer I believe the tip off is found in “and the host, WHO KNOWS what’s behind the doors, opens another door, say #3, which has a goat. In other words he will NEVER open a door with the prize behind it.
Imagine the host opened door #3, revealing a goat, before you chose any door.
You’ve changed the conditions of the problem, and so yes, the probability will change.
The wording of the challenge as quoted (it’s not the original – I’ve linked that below), strongly implies that the host’s choice of door is random – “the host, who knows whats behind the doors, opens another door, say #3,” – and the set-up of the gameshow tells us that the host will never open the door with the car.
I’ve seen various iterations of this problem that come up with different numbers, but they all change the conditions of the problem, or add a variable that was not there originally – such as the host having a preference for opening the left-most door, and the contestant knowing that.
Marilyn vos Savant said:
“Virtually all of my critics understood the intended scenario. I personally read nearly three thousand letters (out of the many additional thousands that arrived) and found nearly every one insisting simply that because two options remained (or an equivalent error), the chances were even. Very few raised questions about ambiguity, and the letters actually published in the column were not among those few.”
…to add, for some reason my comment appears as a direct reply to your comment when in fact I intended it for another comment of yours….
barry, the contestant can only switch to the remaining door, not to the one that is revealed. If the host does not know where the car is and the door he opens happens to reveal a goat it is then 50/50. The contestants that can continue after the host reveals a goat are the 1/3 of them that picked cars and one half of the 2/3 of them that picked goats.
Chances to win by switching in the MHP is calculated by the probabilities of both the contestant and the host picking goats, 2/3×1=2/3. If the game was done at ‘random’ then it’s 2/3×1/2=1/3, same as in staying which is 1/3×1=1/3.
“Disagreeing with a person having the highest IQ in the world is, statistically speaking, not a smart thing to do.” – R.S.
That’s hyperbole, right?
In a discussion about IQ with a friend from Smith College years ago, she told me that her brother’s IQ wasn’t measurable, i.e., he never got any questions wrong. My guess is there are probably others who can do this, as well, though obviously not very many.
My point is, how do we know how smart the smartest people are? Ask them? And if they have to pass a test, who writes it for them?
RIP – Monty Hall
I would switch in a flash but not for the reasons any of the “smart” people said. Since the host and the producer of the show understand that ratings depend on guests winning so viewers can gain satisfaction from watching. They also know that income depends on air time for the sponsors who supply cars. The odds of winning are modified in ways that have nothing to do with probability once the initial choice was wrong and everything to do with keeping the show on the air. If there have been a lot of winners on the show just before me, I could be tricked into changing to add some suspense for the next contestant, but since nothing is known about that from the way the problem is stated, I would still switch.
The question was originally posed as a probability problem. The correct answer is to switch, because you double your chances of winning by doing so.
There is a correct answer: to swich. So Roy is wrong.
Under the specific conditions he outlined, Dr. Spencer is correct.
The great thing about this puzzle is that it inspires thinking. Roy is not the only one to have come up with that iteration. This puzzle has had PhDs in maths and statistics disagreeing.
Yes, and some of them were clearly wrong, if you read the initial reactions to her column, and came off looking very foolish. I read it long ago when it first came out. It taught me one should never risk one’s reputation on a trivial matter before having thought things through.
It also taught me experts often aren’t, something you might ought to keep in mind in the climate debates.
How true is that.
Here’s the original statement of the Monte Hall problem.
https://www.jstor.org/stable/2683689?seq=1#page_scan_tab_contents
Thanks for that, Barry.
It’s always good to go to the source.
Here’s another source, later than yours, but that of the version Marilyn Vos Savant answered. Interestingly, it was not she who originally asked the question on her website.
http://marilynvossavant.com/game-show-problem/
In the question that Marilyn answered, it specifically states, “the host, who knows what’s behind the doors, opens another door, say #3, which has a goat.”
Yes, I read that as implying that the host picks a random goat-door, as opposed to having a bias when picking goat-doors (prefers to show the rightmost/leftmost goat door).
I always looked at this problem backwards. What are the odds of the contestant choosing incorrectly in the first place? 2/3 of course. When Monty opens a door, if it shows the prize then obviously the probability is zero that the contestant choose correctly, right. However, if Monty shows a goat, the odds still remain at 2/3 that the contestant choose incorrectly in the first place. Thus switching is prudent.
I remember very little of this show, did Monty ever show the prize after the contestant had selected ending the game? I seem to remember there were varied prize values with maybe one being the joke.
as I asked above, if Monty had shown you door #3 had a goat to begin with, should you then pick door #1 or door #2? It doesn’t matter. It only matters if Monty purposely avoids showing you the door with the car behind it.
If Monty were to show you the contents behind a door first, then I would recommend you think of a door first. If it matches Monty’s door then yes, you have a 50/50 chance from the remaining doors. However if you thought of a different number than Monty and he shows a goat, then by all means proceed to change your mind, because you had a 2/3 chance of incorrectly selecting in the first place.
Roy – you are right …. if the host always opens a door with a goat then you should always swap.
Of course it doesn’t make much sense in the game show context for the host to open a door and show the car …. that would be quite pointless.
The best way to imagine this is to have a 10 door version of the game. You pick one, then the host opens 8 with goats and asks you if you want to swap. Of course you should. Your original probability was 1/10 …. if you swap it becomes 9/10.
So in the 10 door game. You choose one door, the host only shows ONE goat, do you switch? Yes, but only just.
In the classic 3 door problem. You choose any one door. Probability is clearly 2/3 that it was wrong. That doesn’t change when he open any other door and shows a goat – but he has eliminated one of better choices, so choose the other one.
“It only matters if Monty purposely avoids showing you the door with the car behind it.
That is correct. The probability of winning if you stick with your original choice is 1/3. It does not matter what the host does after that. The probability of being right in your initial choice is 1/3.
However, if he picks a door randomly, his probability of success is the probability of success given that you didn’t pick the right door (1/2), times the probability that you didn’t pick the right door (2/3), plus the probability that you picked the right door (1/3) times the probability of success given that you picked the right door (0). So, his probability of success also is 1/3.
The remaining probability, if you switch, also is 1/3, so that all probabilities add to 1. So, your odds of success are the same whether you switch or not.
If Monty purposely avoids showing you the door with the car behind it, his probability of success is zero. Hence, your probability of success after switching rises to 2/3.
“The probability of winning if you stick with your original choice is 1/3. It does not matter what the host does after that.”
INCORRECT. If the host is choosing randomly and happens to choose a goat, the probability your initial choice is correct improves to 1/2. If the host instead deliberately reveals a goat then the probability your initial choice is correct remains at 1/3.
In the first case, the host is giving you information by revealing a goat, in the second case he is not.
You are talking about a conditional probability – the probability of getting the car given that the host did not choose it. But, there is no way in which you can game the system to be expected to win more than 1/3 of the time if the host is choosing randomly.
I think we can take it as read that the host is never choosing randomly. Monty will always show you a goat, never the car. The wording in the original 1975 problem and the 1990 Parade make this clear (to me at least).
If there’s a chance that Monty is going to reveal the car, then that should be part of the scenario.
“Do you want me to open one of the other doors? If it’s the car, you lose.”
Great minds have argued over the conditions as stated (host intent, host bias, etc), asserting they can be interpreted. That doesn’t seem so to me. But I do enjoy discussing it.
“If the host is choosing randomly and happens to choose a goat, the probability your initial choice is correct improves to 1/2.”
Nope. Still 1/3. One third of the time, the host will choose the car.
Basically, you’ve got a game with three players – you, the host, and the you that would switch. Each of these players have probability of 1/3.
If even 0.0000000000000000000001% of the time the “you would have new information” scenario occurs then wouldn’t that tip the scales given adequate sample size?
So as I see it the possibility of that scenario in this universe (or even a parallel one) makes it where you should switch.
This comment suggests you are already living in a parallel universe.
Haha…How so? That was a joke (I dont see data to support a multiverse). I do think the statistics require a system approach. Thus the potential he will give away information at some point does create an advantage to switch in the system, even if the individual event is dependent on the situation.
The focus of the commentary, like the standard explanation of the Monty Hall problem, is misplaced. There is no “right” answer for a particular game. Individual events don’t have probabilities. Probabilities apply only to large numbers of events. The switching strategy is the correct strategy if you are allowed to play the game many times, and always use the switching strategy. But that’s all that can be said.
For a complete explanation, see this: https://politicsandprosperity.com/2014/05/30/the-compleat-monty-hall-problem/
There is a right answer to “are you more likely to win if you switch or stay?”
I don’t think anyone believes that there is a 100% winning formula, if that’s what you are trying to rebut (?).
Probabilities do NOT only apply to large numbers of events.
A large number of events is simply a requirement to test a probability experimentally.
Them MH problem appears to have it’s own website.
http://www.montyhallproblem.com/
There is an analysis of the problem, and links to other relevant sites.
The analysis there and elsewhere that I’ve seen convince me that you should always switch, though i’m still not intuitively comfortable with that. But then, that’s the point, isn’t it?
SIMPLE WAY TO CHECK
Watch reruns of the show.
Tabulate wins and losses separately for all who stick, and for and all who switch. If the correct solution is given by those who say we should switch, then two out of three of those who switch should win, while only one out of three of those who stick should.
The problem is that this is not actually how the show was run.
WHAT! We aren’t living in an ideal, and honest world? Who’d a thunk it?! //s//
But by tabulating the data, we might get some idea of how it was run, and how far from honest it was. Meant to write that. Probably should have. Thanks.
Honest?? Is honesty at stake here?
That’s what the //s// (sarc tag) was for, Des.
The game host doesn’t want to give away goats. There is no money in selling goats.
So basically host is giving you two choices and shown 1 of 3.
It better than 2/3 odds if switch doors.
The person can say I did pick the correct door, but if then picked goat, can say host changed my mind so I picked the other one, but I picked correct door as my first choice.
You also have what Marilyn vos Savant said, the host wasn’t going to show the car, as that is not a game and so picked the door where he knew there was a goat.
Or bias of game is to give away prizes, giving prizes allow advertising which ultimately what it’s all about.
Does switching increase the probability of winning?
I’ve tried to clarify that, in courses I’ve taught on probability, this way: Suppose the contestant is given, at the outset, two choices, not three: Choose door 1, or doors 2 and 3, as a bundle. He wins if he chooses 1 and the car is there, or if he chooses the bundle 2 and 3, and the car is in either 2 or 3. He should obviously choose the bundle 2 plus 3, with 2/3 probability of winning.
But there is No Difference, in terms of probabilities, between: (A) Choose, out the outset, the bundle 2 plus 3; or (B) Choose door 1 now and switch to door 2 if a goat is revealed behind 3, or to 3 if a goat is revealed behind 2. That a door in bundle 2 plus 3 is revealed to have a goat behind it is irrelevant. There is always at least one such door. Which door is irrelevant, as is the question whether or not the moderator knew that in advance. Thus, switching is the correct strategy, increasing the probability of winning from 1/3 to 2/3. With this little caveat:
Suppose the moderator does not know in advance where the car is. Then we must clarify what “switching” means. Because then the moderator will, at times, open a door behind which stands the car. Is the contestant allowed to switch to that door, or only to the door not opened by the moderator? In either case the game essentially ends, the contestant winning with probability 1, if he can switch to either door, or with probability 0, if he can switch only to the door not opened. “Switching” becomes totally uninteresting — the game is over. That would not be a viable “game show”, so we can plausibly exclude it from consideration. Or, if you want to include it, that scenario would change the “switching strategy” problem to one of interest only when a goat, not the car, stands behind the opened door. Limiting the problem to that, the only interesting scenario, renders totally irrelevant the question whether or not the moderator knew, in advance, where the car was.
Should be simple to verify using a Monte Carlo computer simulation.
Or maybe look at it with more doors. Imagine there are 10 doors. You have picked door 1. The host opens all the other doors bar 1. Would yo switch to the one he did not open? I would.
Exactly. I don’t understand why this is a decades long conundrum. If the host always opens a door to show a goat then you should definitely always switch. Its obvious.
And the way to see this is to look at a 5 door or 10 door version. In those situations, the host is basically telling you where the car is !
I dont understand why this is a decades long conundrum.
There are 2 reasons.
1) Most people do not get it at first. The probability here is counter-intuitive (to most people). Most people really do think the odds are 50/50 when the goat is revealed.*
2) There has been much argument about the specifics of the scenario, with disagreement about whether the host is choosing randomly (could reveal a car), or if they always chose the right-most door (door 3), or if they are always choosing goat-doors, or choosing goat-doors but with a bias towards the right-most goat door (happens to be door 3 in the scenario).
* I was first shown this problem in a pub with 3 coasters with $1,000,000″ written on the back of one of them. It took an hour of patient explaining for me to see why I was wrong that the chance was 1/2 – scenario given was that the host would always show a door with a goat, no other bias (2/3 correct answer for that scenario). I’ve loved this problem ever since.
I think the problem is asinine and has nothing to do with the purpose of statistical analysis.
Spoken like a true uneducated conservative.
AGW zealots would be the goat pickers. 🙂
I’m sure YOU believe there is humour somewhere in your comment. If only we could all see inside your head.
Heh. That ax must be worn down to the wood with all the grinding.
So by Hall knowingly eliminating the 3rd selection he is increasing your data points thereby statistically increasing the probability that #2 is correct. Kind of like when 98 out of 100 reasons why agw is true have been exposed as false?
IMO If you want a correct answer to a math problem, you should start by state it correctly and completely.
Open none just leave the show knowing that your inability to deal with disappointment has won again, 2 to 1 just cant do it
Regards
The missing bit of information is that in all previous shows, the host opened a door with a goat. The assumption is that the host opened that door because he always opens a door that has a goat behind it.
I explained it to some friends, but they didn’t believe me, so I demonstrated with 3 cards, an ace and two kings. They could choose one face down. After looking at the other two, I would turn over a king. By swapping, after 50 or so attempts they got an ace about 2 out of 3 times. I made it clear that I would always turn over a king, to eliminate the doubt you raised.
Regards,
Carl
I hate to reveal how much I watched Let’s Make a Deal as a kid, but if you pick door 1 and stay with it, you will almost certainly take home a goat. The ‘car’ was almost always behind door #2 or door #3. If door 3 is revealed to have the goat, then switching to door 2 would almost certainly win you the car.
The assumption of equal probability up front went out the window, or door as it were, on this game show. Use another situation.
If that’s what you think happened on the show then you didn’t actually watch it. The Monty Hall problem is based only loosely on the show – Monty didn’t actually do it that way.
From her website:
“Of the letters from the general public, 92% are against my answer, and of the letters from universities, 65% are against my answer.”
Kind of tells us something about “consensus”, huh?
It also tells you something about the knowledge of the general public vs the knowledge of educated people. Further, educated people are swayed by evidence, as they ultimately were in this example. The general public still provides opposition to this solution, despite the now almost universal agreement amongst professionals. So yes, this is almost identical to the climate debate.
Kind of tells us something about consensus, huh?
You simply repeat your comment?
Yes – it tell you that when opinion DIVERGES from a consensus, new evidence is suggesting the theory is wrong. In the case of the climate that hasn’t happened.
What it “tells” is that a “consensus” often has no basis in fact. Consequently , a consensus offers no guarantee of being correct.
What it tells is that when the consensus answer is wrong, it very quickly moves towards the correct answer as more information comes in. When the consensus answer is correct, as more information comes in the consensus solidifies.
After further thought, I do thing Roy is correct. It’s 50:50
Say the car is behind door #1 (for all cases below), and you pick #1.
Then Monty can show you the goat behind either door 2 or 3.
Here are the possible outcomes…
Monty shows you the goat behind door 3. Your choices are…
1=>1 WIN
1=>2 LOSE
He shows you the goat behind door 2, and you decide…
1=>1 WIN
1=>3 LOSE
Now suppose your 1st choice door conceals a goat. In that case Monty can’t show you the car, so has to show you the goat you didn’t pick. If you pick door 3, he can only show you door 2. Your choices are…
3=>3 LOSE
3=>1 WIN
And the same if you pick 2 instead of 3. After he shows you the goat behind door 3, your choices are.
2=>2 LOSE
2=>1 WIN
And the same if the car is behind any of the other doors. Stick to your first guess, and you’ll win half the time, or switch and win half the time. You only have a 50:50 chance of winning, either way.
The reason would seem to be in the degrees of freedom Monty has, or not, depending on whether you first choose the car, or not.
No, the chance of winning by switching is the inverse fraction of your chance of winning with your first guess.
Instead of 3 doors, do it with x doors. After the contestant guesses a door, x-2 doors are opened, revealing x-2 goats, leaving only 1 goat, 1 car and two doors unopened.
Substitute 100 for x, you have 1/100 chance of being right the first time, and 99/100 chance of being right by switching.
I saw that at one of the websites that discusses this, but bottom line is that after your initial guess, you are either (figuratively) holding the item you want, or you aren’t. And, after counting the number of ways you could win by exchanging it or win by keeping it, and then dividing the desired (or undesired) outcome by all the possible outcomes, I get that P(win) = P(lose) = 3/6 = 0.5…
It doesn’t matter what happens before you get to the point where you have to decide to keep or discard what is in your hand. At that point there are only 6 possible outcomes, and of those only 3 will win.
If you can show me where I miscounted, fine. But at the moment I don’t see anything wrong with it.
Correction, that should be 4/8, not 3/6. Had head still stuck in other “proof” there. Sorry.
OK, I think I see the problem.
If you choose the car first, then Monty has 2 goats to chose from. If you only consider the case of one, and neglect the other, you do indeed get 2/3 for switching. But that’s only because you haven’t correctly counted the number of possible outcomes.
There are two ways to lose by choosing the car first, not one as in the other cases. That’s the fly in this ointment. You have to count ALL possible outcomes, or you’ll get it wrong.
yonason,
I agree with you.
Cheers.
Thanks.
Forget the three door set up for a moment.
Try it as a 100 door problem. After you choose a door, Monty opens 98 other doors, revealing 98 goats. There are two doors left closed: your choice and one other door.
Do you really think you have a 50% chance of having already guessed the right door?
Of course not. You had a 1% chance of being right, and a 99% chance of being wrong. Hence, you should switch. Do it 100 times, and switching would be correct 99 times.
Same deal for 3 doors, but less intuitive.
Joel
I get it now. All about opening the doors randomly versus intentionally opening just the ones with goats.
OK, Joel. Thought about again last night, and realized you were correct. I was counting incorrectly.
The heck with 100 doors. Take it to a trillion doors. Pick one. You do NOT have that car. But when all goats are put to pasture, and you’re left with one door to switch to, it’s now yours when you do. So, yes, switching is always the best, and with a 2/3 probability of winning, in the case of 3 doors.
Thanks.
I began having second thoughts since last night after thinking more about what you bring up, but am still not convinced I’m wrong. It all comes down to whether, when Monty has a choice of which of two goats to remove, you count that as two possible events or just one event. I’m “sticking” to my original analysis, but am open to arguments to the contrary.
And, yes, I see what you are saying. In the first round there are 2 ways to select the goat, and only 1 to select the car, so the contestant has most likely selected the goat. It certainly LOOKS like he should switch to optimize his chances.
Finally, since it doesn’t matter to me if he sticks or switches, and if I’m wrong switching is best, then by all means he should switch, just to be on the safe side.
Roy does NOT say it is 50:50. He says that if (as the problem is meant to be) the host always chooses the goat, then the chance of winning by switching is 2/3. But is the showing of the goat came about through a random choice then the answer is 1/2.
look up mythvbusters episode 177 for the experimental proof that switching is correct
If you choose door no. 1 and then door no. 3 has been revealed to have a goat behind it, there is no advantage to change. Your guess is that a car is behind door 1. Your odds have just got better but it doesn’t mean your odds will get even better by changing. You can apply all the laws of probability but the fact remains there is a car behind either door 1 or door 2. Your choice was door one (and you may be correct). Nothing has happened that should prompt you to change your guess.
Do you agree that, if you switch, you will get the car precisely when your initial pick was a goat?
Very well put.
OK, I read that link and I think I get the point. Your original guess is stuck with its original, one-in-three probability, which changing (assuming an original three choices) means that the door you switch to has inherited the probability of the opened door. I’m not sure why it should, but it does.
I just wonder if this win a prize competition could be translated, say in war. Suppose you’ve got a spy who has mixed feelings about his treason. He won’t tell you what his sides army will do. He’ll only give you limited information, much like this show host.
You have limited troops, so you must place almost all of them in the south, middle or north of the battle front. You tell this cagey spy, and he says, “Well, I’m not going to tell you anything else, but I will tell you the attack is not coming in the north.”
Does that mean that a wise general would move his troops to the middle, given that the MH reasoning said that the chance of the attack being there has become 2 out of 3. That strikes me as odd.
And what if your situation is different. Suppose instead of a spy on the other side who knows, you have a spy of your own, but the time is limited, so you choose to send him out in the North. He returns and tells you, “No attack in the North.” Should you switch from defending in the south to defending in the middle based on that? In one sense, the information you have obtained is the sameyou know there’ll be no attack in the north. But in another sense it is different. The source of your information doesn’t know if your choice to defend the south or north is correct or not.
All of y’all are wrong (it was Carol Merrill who opened the door… ☺)
I don’t get the 1/3 vs 2/3 logic. Seems stupid to me. I’m going with 50/50
If Monty only opens a door and asks if you want to switch if you have picked the door with the car, then you should keep your pick, because odds are 100% you have the correct door.
This is the scenario that is encountered more in real life and why people want to keep their original pick.
Why would someone offer the switch if the other option is better?
Faulty logic: Initially, there is a 2/3 chance that the car is behind either door 1or 3. Door 3 is revealed to have a goat, so therefore door 1 now has the 2/3 chance.
Hang on … just 6 minutes earlier you said “I dont get the 1/3 vs 2/3 logic. Seems stupid to me. Im going with 50/50”. Now you accept the correct answer?
Sorry Snape: faulty reading comprehension. Try read my first IF statement and understand I changed the problem.
Dan
Sorry for the confusion. I was not replying to what you had written. By “faulty logic” I was talking about the strange way the 2/3 chance is apparently arrived at.
The premise of this question is that Monty ALWAYS offers the switch.
There are no motives here – it is a straight maths question.
No, read the original question. It says nothing about what always happens. It indicates what is happening in a single situation. Your premise is assumed. If you go to Marilyn vos Savant’s site, she later explicitly states that it always happens. But the original problem does not state it. This is Dr Spencer’s point.
I don’t care how the original problem was written. I care how it was INTENDED, and how it WOULD be written if she ever presented it again, and how ALL university lecturers present it in lectures. Without that assumption it is not an interesting problem.
Although I guess “stay” could be the same as “makes no difference.”
I think it is very clear from the wording that Monty picks any door with a goat behind it. The clues are in the word, “say,” and the words, “the host, who knows what’s behind the doors…” Here’s the full quote as it appeared first on vos Savant’s column.
“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?”
Stating that the host knows what is behind the doors is unnecessary if the host picks purely at random. This statement strongly implies that he is being selective (he will show you a goat always).
“Say” means, “for example.” Which door numbers are chosen is not the point of the exercise. It could be any 2 of the numbered doors selected (contestant) and then exposed (Monty).
The question posed is about whether it is better to switch or stay. This by itself suggests (but does not necessitate) that 1 of 2 answers is correct. With the rest of the wording, though, I take it pretty clearly to mean that there is a correct answer of the 2 stated, and that a third, unstated option (makes no difference) is not on the table.
Although I guess stay could be the same as makes no difference.
I’ve run a simulation on a computer. Switching wins and the computer had no knowledge of Monty’s intentions. It confirms the statistical treatment above. Counterintuitive, yes.
As far as I can tell, the same flawed logic is used for claiming door 2 has a 2/3 chance.
What flawed logic are you referring to?
Des
I must be missing something. You can use the same logic to conclude that door 1 and door 2 both have a 2/3 chance of winning the car.
Assuming your initial pick was door 1, and the host opened door 3, the probabilities of winning associated with each door are:
Door 1: 1/3
Door 2: 2/3
Door 3: zero
Only one door has a probability of 2/3 of having the car.
A. Initially, there is 2/3 chance the car is behind either door 2 or 3. Door 3 is revealed to have a goat, so that means door 2 must have a 2/3 chance of having the car.
B. Initially, there is a 2/3 chance the car is behind either door 1 or 3. Door 3 is revealed to have a goat, so that means door 1 must have a 2/3 chance of having the car.
By the logic above, doors 1 and 2 BOTH have a 2/3 chance of having a car. This can’t be true, so the logic is flawed.
We are talking about probabilities AFTER THE GOAT IS REVEALED.
Before the goat is revealed the probabilities are all 1/3.
After it is revealed they switch to 1/3, 2/3, 0.
At no time do the probabilities add up to more than 1.
Dr S, i don’t think that intent matters… In either case, there are just 4 options. Stick with 1 win a car, stick with 1 win a goat, switch to 2 win a car & switch to 2 win a goat. So, odds ARE 50/50.
i think what folks are confusing here is a broader array of options: Let’s say there is a car behind 1. You pick 1, a goat is revealed in 3, stick with 1, win a car. Pick 2, goat revealed in 3, switch to 1 and win a car. You pick 3, goat is revealed in 2, switch to 1 and win a car. Only in a broader sense do your odds increase to 2/3 by switching…
(i think marilyn is wrong about this specific case)…
Nobody who says switch is confused. Those 4 scenarios you gave are NOT equally likely outcomes.
Any kiddo with modicum coding skills can model this puzzle on a computer.
Problem is … people who are getting it wrong due to false assumptions will also be making those false assumptions when coding.
Des
I explained the flawed logic for concluding that door 2 would have a 2/3 chance of having the car. So what’s the correct logic?
Stick to the thread where we were talking instead of chasing me around. The topic of this thread has nothing to do with what we were discussing.
One things for sure. I think. Monty would always have a goat to show you.
I heard that the two goats were named Mike and Gordon.
I worked it out this way:
Whatever door you initially choose will have a 1/3 chance of having the car. Then, the host will ALWAYS show a door with a goat. Thus, your odds of having guessed right wouldn’t change with this new information.
When then presented with two choices, and knowing your initial guess still has 1/3 odds, then the other choice must necessarily have a 2/3 chance.
So now you’ve flip-flopped are are no longer claiming that this is “flawed logic”. AND you’re pretending you worked it out yourself when all you are doing is presenting the same argument that has been given many times before in this discussion.
Des
Of course I flip flopped. That’s what’s so fun about this brain teaser. You start out sure it has to be 50/50, then it’s humbling to find out you’re wrong. I didn’t “get it” until I went back and read Joel’s comment.
Des
I didn’t work it out myself and wasn’t trying to pretend that. Here’s what I had previously said to Joel:
Sir Isaac Snapelton
October 1, 2017
Joel
I get it now. All about opening the doors randomly versus intentionally opening just the ones with goats.
You’ve got it now, Snape.
Sir Isaac,
If you wander in after doors have been opened, and all the other diversions have taken place, then you are faced with one favourable outcome, one unfavourable.
Equally likely. 50/50.
The diagrams in the link are deceptive, and lump more than one occurrence together. Classic misdirection.
Any computer simulation showing otherwise has been written by a foolish Warmist, in all likelihood.
Maybe you could look at it this way. Initially, two out of three outcomes are unfavourable. Throw one outcome away. Now, of two possible outcomes, one is favourable, one not.
Switch, dont switch? Toss a coin?
Cheers.
Mike
Lol! People are so sure its 50/50 they make an ass of themselves. Myself included.
Sir Isaac,
You pick 1. Car happens to be behind 1.
If door 2 is opened, showing a goat, you switch. You lose.
If door 3 is opened, showing a goat, you switch. You lose.
Might as well stick with your first choice, eh?
Work through the others if you wish. It comes out the same. Picking car, or not picking car. Two doors, fifty fifty.
Cheers.
Mike Flynn:
Do you agree that if you switch, you will win PRECISELY if your initial choice was a goat?
You pick 1. Car happens to be behind 1.
If door 2 is opened, showing a goat, you switch. You lose.
If door 3 is opened, showing a goat, you switch. You lose.
Might as well stick with your first choice, eh?
Run the iterations with a car behind each of the doors in turn. Any door could be picked, but let’s stick with door number 1 for easy conceptualizing. Host always opens alternate door with a goat (never the car).
Car behind door 1 – switch you lose.
Car behind door 2 – switch you win.
Car behind door 3 – switch you win.
Switching gives you a 2 out 3 chance of guessing correctly.
The odds are the same no matter which door you choose first, as long as the host always opens a different door revealing a goat.
Des,
Yes. Of course, the host has only one choice of door (unless he wants to show the car).
If you have picked the car, he has two choices.
So, car picked, switch loses twice – two goat choices. Host shows one, you switch to the other!
Either goat picked by you, switch wins – but host only has one choice. Has to show remaining goat.
Two switches win, two switches lose. Fifty fifty.
Doesnt matter which item is behind which door.
Cheers.
Two switches win, two switches lose.
Nope.
Two switches win, one switch loses. Only one switching is the contestant.
If the contestant chooses the door with the car and switches, he loses.
If he picks a door with a goat, he switches and wins.
Thing is, there are two doors with goats, and one with a car. So he’s twice as likely to win by switching.
Or look at it this way….
http://www.drroyspencer.com/2017/10/the-monty-hall-problem-there-is-no-correct-answer/#comment-266083
Think of it this way, Mike.
What are the odds you picked a door with a goat in the first place?
2/3.
That odds on your first selection don’t change one bit when he opens a door with a goat. If you stick, there’s still a 2/3 chance you have chosen a goat.
And still a 1/3 chance you picked the car.
The door Monty opened has a 0/3 chance of hosting a car.
That means there is a 2/3 chance the car is behind the other door.
If you wander in after doors have been opened
You’ve changed the conditions of the problem and are now answering a different question.
I was convinced for a whole hour it was 50/50. It took a lot of patience on the part of the guy who introduced the puzzle to me to sort me out.
The key is Monty gives you information by opening a door with a goat behind it.
Your first guess was a 1 in 3 chance. The other 2 doors are a 2 in 3 chance of having the car. It doesn’t matter if Monty paints one the 2 doors aubergine, hangs a sign on it, or opens it to show you a goat. The odds of the car being behind one of those two doors is always 2/3. Those odds don’t change just because Monty did something. But by opening one of the two doors to reveal a goat, the chance that this door has a car is now 0/3, and the remaining door takes on the odds for both – 2/3.
Extended version!
There are a million doors. You choose one. Monty, knowing where the car is opens every door but one (it takes him a few months), revealing a goat behind all of them.
Do you think your 1 in a million chance has now magically converted to 1 in 2?
No, the remaining door has a 999,999 out a million chance of having the car behind it.
barry,
Million doors makes no difference. If you have the chance to switch to another unopened door (he only opens one, remember), whats your advantage?
You had a one in a million chance of a favourable outcome, you are taking on another one in a million chance. Dont forget, you dont know whats behind the door you picked.
But no million doors, that is another foolish fantasy. Only three.
Monty gives you no useful information. He has a choice of two doors – one of which at least has a goat behind it. Whether you have picked the car or not, he still shows a goat!
You pick a goat – he shows a goat!
You pick a car – he shows a goat!
If you believe you are better off, good for you! You might even believe there is a testable GHE hypothesis – who knows?
Cheers.
The testable hypothesis is that switching increases the chance of winning a car.
see:
https://www.youtube.com/watch?v=mxZNpqEFNac
where the mythbusters demonstrate this.
Thus, your logic is faulty (as usual).
Mike Flynn,
“Monty gives you no useful information.”
YES!! He has given you no useful information about your initial choice, and PRECISELY why the probability of winning with your initial choice does not change from 1/3.
Des,
Except that you have only two choices now, not three. One door is out of contention. Two doors. Which item is behind which door?
Even money.
Cheers.
Million doors makes no difference. If you have the chance to switch to another unopened door (he only opens one, remember), whats your advantage?
No, he leaves 1 unopened door and gives you a choice between them.
Extend that out to a million doors – he opens all but one, still leaving you with 2 choices.
This extension is meant to make it damned clear that the odds you have picked the correct door is one in a million, and the odds the door remaining is the inverse probability – virtually certain.
It’s a more intuitive leap to figure out why you should switch, when the odds you picked correctly to begin with are so abysmally low. With only 3 doors, the intuitive leap is harder for almost everyone (me included – I didn’t get it for a long time).
No, no!
Obviously Mike and Gavin!
Mann and Schmidt, that is.
You dont need to thank me. My pleasure, as always.
Cheers.
Mike !
What are you doing here?
Get behind the door!
If the geomagnetic activity falls Russian high-pressure system will develop in Europe. This means winter conditions in October.
https://www.facebook.com/Sunclimate-719393721599910/
Stick to a CLIMATE page.
Agung boils.
https://magma.vsi.esdm.go.id/live/seismogram/
Not sure why you think so much of volcanoes. If it’s VEI4 or less it won’t effect climate. VEI5 will have an effect measurable by instruments but not noticeable otherwise. VEI6 will give us a slightly noticeable dip for two or three years. VEI7 will have serious consequences, but the effects will again be gone in a few years. You need VEI8 to have a hope of sending us into an ice age. But the last two of them were 26000 and 75000 years ago, and that is more frequent than typical because they come on average once in 100000 years.
Stick to a CLIMATE page.
Very nasty skin condition.
The best way I’ve seen this explained, so as to relieve your intuitive thinking, is to think of the same game with a million doors. Your odds of choosing correctly are one in a million to begin. Now open all doors but two to find no cars. Think it through. Given your odds were one in a million to start, should you switch?
There are plenty of people who still don’t get it after this. They assume the odds switch to 1/2 because now there are only 2 doors.
I’ve just come up with the Russian roulette version. 3 guns, one has a full chamber. You choose one, Monty fires another. Click! Maybe crisis mode would produce critical thinking…
barry,
Why not stick with the original problem?
Cheers.
I’m having fun.
Another way of looking at it:
First imagine a slightly different version of the game where, after you make your initial choice, Monty doesn’t open a door. Instead he offers you the opportunity to “swap” and open BOTH of the doors you didn’t originally choose. You will win the GREATER of the two prizes behind those doors. Of course the probability that you will win the car by swapping is 2/3.
In the real version of the game, you are basically playing EXACTLY the game just described. The only difference is that Monty has first opened one of those two doors on your behalf, and it is guaranteed that the door he didn’t open will contain the greater prize. He is basically saying “So you don’t want the 2 door option where you win the greater prize?? Then let me start you off – I will now reveal the lesser prize of the two …. hint hint wink wink ….”
Des,
Let us stick with the original. Your imaginings are not relevant. Why do you want to shift the goal posts?
Fifty fifty.
Cheers.
So you have no interest in learning? Figures.
I myself have had to go through many different ways of explaining it before light dawned. Part of the fun of this puzzle is the to and fro, and the creative thinking about how to explain it so that the listener eventually sees it for themselves.
One I’ve used to try and get them to make the leap to the subtlety of the probabilities is…
A velvet sack contains a million marbles. One is white, the rest black. The host directs a friend of yours from the audience to blindly choose one marble and hold onto it. You don’t see the pick, you keep your back to your friend. Host pulls out all marbles but one from the bag. He shows you 999,998 black marbles.
What are the odds your friend picked out a while marble?
What are the odds the white marble was in the other bag after your friend picked?
Quite a few people get it after this explanation. But not everyone. They see 2 marbles left, and immediately revert to 50/50.
Perhaps the clearest way to see it is by laying out the basic permutations, with sticking and switching. Remember, Monty opens a door with a goat after you select, so you only have 2 doors left to choose from
Car behind door 1:
Contestant picks 1 –> sticks –> win
Contestant picks 2 –> sticks –> lose
Contestant picks 3 –> sticks –> lose
Car behind door 2:
Contestant picks 1 –> sticks –> lose
Contestant picks 2 –> sticks –> win
Contestant picks 3 –> sticks –> lose
Car behind door 3:
Contestant picks 1 –> sticks –> lose
Contestant picks 2 –> sticks –> lose
Contestant picks 3 –> sticks –> win
In each case, contestant has a 2/3 chance of losing by sticking.
Does this change if contestant switches? Let’s see.
Car behind door 1:
Contestant picks 1 –> switch –> lose
Contestant picks 2 –> switch –> win
Contestant picks 3 –> switch –> win
Car behind door 2:
Contestant picks 1 –> switch –> win
Contestant picks 2 –> switch –> lose
Contestant picks 3 –> switch –> win
Car behind door 3:
Contestant picks 1 –> switch –> win
Contestant picks 2 –> switch –> win
Contestant picks 3 –> switch –> lose
Switching gives you a win 2 times out of 3.
Remember, Monty is always showing you a goat when he opens one of the other 2 doors. If you selected the door with a car to begin with, you will lose by switching. If you selected a door with a goat first, you will win by switching.
Your odds of first picking the winning door are 1/3, and your odds of first picking a door-with-a-goat are 2/3.
It’s more likely you picked a goat door to begin with. That won’t change no matter what Monty does.
Roy, you are absolutely right that the answer depends on us knowing the game host’s strategy. As the problem is stated we do not know whether he planned to open another door. What if he had a secret strategy to only open another door if the contestant had picked the car? I.e. his strategy is to sway you away from your correct choice?
Jay
For dramatic effect, Monty Hall wants the contestant to have to make a choice, “stay or switch”, which means he always has to show a goat. That’s the basic premise of the puzzle.
If Monty Hall randomly opened one of the two doors, and a goat appeared, you might as well not switch.
He would not do this because one out of three times a car would appear and the game would be over.
You are correct. However, if you read the original problem, you will see that he starts by offering money for the contestant to switch before revealing the goat as if he is trying to trick you. If he doesn’t always present the goat to the contestants then it is more like a poker game. Hence, the assumption here is that he always gives you the option to switch.
That’s an interesting take, but from what vos Savant said, nearly everybody assumed that was Monty’s usual pattern – always open a goat door. The wording strongly implies this, and that was how it was intended.
Very interesting problem but poorly defined so causes confusion as multiple answers could be generated.
Under the strict condition that Monty MUST open a goat door from the doors not chosen by the player then this offers no additional information about the player’s first choice. Hence its probability cannot change.
The problem should be stated in terms of the conditional probability:
Pr(Acar | Bgoat OR Cgoat)
Now expanding this in the usual way the denominator which is the term :
Pr( Bgoat OR Cgoat) = 1
The numerator is Pr(Acar) x Pr(Bgoat OR Cgoat | Acar)
The second term is also unity as it is not influenced by the condition that A has the car (i.e. Monty still opens a goat door irrespective of what is behind A)
Hence the original probabilty Pr(Acar) does not change (Pr = 1/3). Hence the switch to the other door is optimal (Pr=2/3).
If the statement as originally presented didn’t specify whether the host’s suggestion/reveal is dependent on where the car is then the answer should be conditional, since the correct answer branches dependent on that fact. Any answer asserting exactly one or the other (either that the contestant should switch or that it doesn’t matter whether the contestant switches or not) to be true is not a correct answer since it is not always correct. Any answer that assumes the host will only reveal a goat (perhaps because the person who provides the answer has decided that such a situation makes the most sense for the game show) needs to have said assumption explicitly indicated in order for the answer to be considered correct.
As it is presented as a game show, it would be daft to reveal the car. If revealing the car was part of the game show, then one assumes that would have been stated as part of the conditions.
“Do you want me to open another door? If it has the car, you lose.”
I suppose there is an element to the puzzle, for those who try to imagine permutations of the game show not expressly stated, of figuring out the likeliest permutation.
The point is not what would be daft or not daft to happen in a game show. It is entirely reasonable to suppose that the host knows the location of the car and will not reveal the car, but it must be explicitly noted that the problem is being modified if such a supposition is made. The important thing is what is daft or not daft in solving a problem. There is a statement of the problem. If additional assumptions are made when answering that statement those assumptions need to be made explicitly (this is the important issue). Tacit assumptions usually end up being daft. They cause confusion and incorrect answers. Dr. Spencer argued that there is no correct answer as the problem was originally posed. I argued that the correct answer is conditional, and, in any case, there are wrong answers regardless, and any non-conditional answer that doesn’t make additional assumptions is wrong.
I didn’t read the debate concerning Marilyn vos Savant’s solution to the problem, but if the question posed to her was as Dr. Spencer quoted it in these comments then I can easily imagine that the criticism levied against her had to do with her answering a different question from what was posed to her without explicitly indicating that she was doing so. It is a serious error in the mathematical sciences. It’s not an entirely academic matter, either. Massive and costly mistakes can be made because of such “obvious” assumptions.
Yes, and the statement implies clearly that the host will not simply stop the game by showing where the car is. If you allow that, then of course the game changes; and the whole question stops being interesting.
barry,
You wrote –
If the contestant chooses the door with the car and switches, he loses.
Indeed. However, the misdirection occurs by not pointing out that there are two individual goats, behind two separate doors, thus two different ways of losing – not just one.
The outcomes need to include the possibility of the host choosing goatAs door, or goatBs door. The host may choose either one, if the contestant has chosen the car door.
To the viewer, each goat appears the same, but they are not.
To cover all outcomes, you should include the cases where goatA is revealed, and goatB is revealed, as separate outcomes. If this reality is considered, then the contestant can lose in one of two ways, depending on which goat is revealed. Switching after seeing either one, will result in switching to the other goat, and losing. Two losing switches. Of course, regardless of which goat he might choose, the revealing of another goat, followed by switching, results in a win. Pick either goat and switch – two wins.
Two loses, or two wins – fifty fifty.
Pointless, but interesting.
Cheers.
Mike
When the contestant makes his initial guess, he has a 1/3 chance of being right. There is thus a 2/3 chance that one of the two remaining doors will have car behind it. Do you follow so far?
If the car IS behind one of those two doors, how will the contestant know which one? The host ALWAYS reveals the door with a goat, that’s how. The car is always behind the other door.
If the contestant switches, the only way he loses is if the car is behind his initial guess. He wins regardless of which of the other two doors has the car.
Again
If the contestant switches, the ONLY way he loses is if the car is behind his initial guess, and his original guess only had a 1/3 chance of behind correct.
Sir Isaac,
The car is not always behind the other door.
If the contestant picks the car, there is one of two goats behind each other door. It doesn’t matter which individual goat is revealed, switching loses, both times.
For example, if the car is behind door 2, and that is the player’s pick, then the host can either reveal the goat behind door 1, or the other goat behind door 3.
The player loses by switching. Two ways, depending on whether the host reveals door 1, or door 3. His other 1/3 guesses ensure a win by switching – but there is only goat or the other to for the host to reveal. The player has picked one, and the car is behind another door. In that case, the player switches and wins.
However, he doesn’t know whether he has picked the car, with two certainties of losing by switching, or one of the goats, with one certainty of winning by switching, for each goat revealed.
For any door picked, the host may reveal goatA, or goatB, but only providing that the player has picked the car (host can reveal either), or the goat that has not been picked by the player.
Still fifty fifty. Two wins, or two losses. Depending on the player’s pick, and depending on the hosts choice of goat to reveal in the case where the player has picked the car.
Luck. Just stick with your original pick. Why waste a good worry?
Cheers.
Mike
One more time. This is the key:
If the contestant goes for the switch, the ONLY WAY he loses is if his original guess was correct. That only happens 1/3 of the time.
Sir Isaac,
No. 2/4. There are four outcomes, not three. He has two ways to lose if he picks the car. Two different goats. Two different outcomes, each a loser. on two different turns, of course.
Each games, he has 1/3 chance of picking car, goatA, or goatB. Irrelevant to the outcome.
After he makes his choice, then the game begins. Each turn for one pick has four outcomes.
Two positive, two negative.
If he picks the car, he can lose two ways – switching to the non revealed door will lose. The host can pick either one door or the other to reveal. There is a goat behind each.
If player picks a goat, initially, host has only one choice. On each turn player switches and wins. The host has no choice – can only reveal the goat that player has not picked.
Maybe if you write the outcomes down for one door, and not worry about the initial 1/3 chance too much, it might help. Just assume that all outcomes happen one after the other. There are four, of course. Two good, two not good. The key might be to remember that there are three objects – goatA, goatB, car. Pretty well everyone thinks only two items are involved – car and goat.
I notice that some foolish Warmists stick to the “consensus”, without actually being able to challenge anything factually. Deny, divert, and confuse. And still without benefit of even a disprovable GHE hypothesis! Their science is obviously as lacking as their logic!
Oh well.
Cheers.
Mike, you are in effect double counting by not allowing for the probability of choice in the case of both goats being available to Monty.
There are only 3 cases for Monty’s doors; 1. goat/goat 2. goat/car 3. car/goat
Each of these has an identical probability (1/3). In the event of the goat/goat case it has a 1/3 chance and only a 1/6 chance of Monty disclosing any specific door with a goat ( x 1/3). Together, however Monty decides, they still end up comprising a 1/3 a priori chance of a goat coming from this goat/goat event.
If Monty must disclose a goat (from his two doors) then this event does not give any information about the player’s choice because a goat door will be disclosed in all three cases with certainty (by definition). Monty may as well not disclose anything if the player does not wish to switch.
From a comment above that Monty was offering money incentives to switch before opening any door I suspect there is more complexity to the game than is apparent from this simpler situation if only to enhance viewer appeal.
hehe I was robbed the 1/2 in computerized form did not display here and should have read:
.. and only a 1/6 chance of Monty disclosing any specific door with a goat (1/2 x 1/3).
TonyM,
3 cases and four goat choices to reveal.
1. Either GoatA or goatB is revealed
2. Player picks goatA – goatB is revealed.
3. Player picks goatB – goatA is revealed.
For either of the first (and it is up to the host – either one is the same to him) – switching loses.
For either of 2 or 3 – switching wins.
Host has four choices – two optional, two “compulsory”.
If Im wrong, Im wrong. So far, people just say “youre wrong”, without showing why.
An analogy might be people asserting that there are only three outcomes from tossing two coins – two heads, two tails, or one of each. Absolutely correct in one sense, totally irrelevant as far as probability goes.
Cheers.
Cheers.
Mike,
The two coin analogy is totally different as the coins are two independent events.
A better analogy, and a correct one, is the pea and thimble trick. Once the pea has been allocated it does not matter what colour or variable is under the other two; they are both losers and dependent only on where the pea is. You are trying to differentiate the losers.
The goats just represent losers and it matters not whether they are roosters or chooks or just nothing as they are wholly dependent on where the winner is placed (ie. car); not so for coins where the second coin has no dependent relationship to the first coin outcome.
I gave a more formal treatment using conditional probabilities above. If the event space cannot reduce from 1.0 then there is no chance of improving the chance of the players original choice. This is the case here as the chance of Monty displaying a loser is 1.0 which is obviously the maximum possible.
So formally the question is what is the probability of A having the car given Monty shows a goat in either B or C?:
Pr(Acar | Bgoat OR Cgoat)
Expanding this the denominator probability is 1.0 as follows:
Pr( Bgoat OR Cgoat) = 1 (i.e. Monty must show a goat irrespective of where it is in B or C)
That is sufficient to show my point but I continue with the numerator as :
Pr(Acar) x Pr(Bgoat OR Cgoat | Acar)
The second term is also unity as it is not influenced by the condition that A has the car (i.e. Monty still opens a goat door irrespective of what is behind A)
So it reduces to
Pr(Acar | Bgoat OR Cgoat) = Pr(Acar)
meaning player’s first choice probability is independent of what Monty shows as long as he shows a goat.
This is a standard treatment of conditional probabilities. You might like to consider it in Venn diagram form.
Cheers
Mike,
To cover all outcomes, you should include the cases where goatA is revealed, and goatB is revealed, as separate outcomes.
Oh, I get how you’re thinking now.
To run the odds with goatA and goatB being revealed would require the contestant to pick the same door twice. The permutations would look like this.
C = contestant: M = Monty
Car behind door 1:
C picks door 1 > M shows door 2 > C sticks –> win
C picks door 1 > M shows door 3 > C sticks –> win
C picks door 2 > M shows door 3 > C sticks –> lose
C picks door 3 > M shows door 2 > C sticks –> lose
The contestant only gets one choice – Door 1, 2, or 3.
She cannot pick door 1 twice. And Monty can only open door 2 or 3 after that choice, not 2 and 3.
Here’s how the permutation looks if Monty can only open 1 door per contestant choice:
Car behind door 1:
C picks door 1 > M shows door 2 or 3 > C sticks –> win
C picks door 2 > M shows door 3 > C sticks –> lose
C picks door 3 > M shows door 2 > C sticks –> lose
Amazing how this thread mirrors so many other threads.
1) One vocal skeptic, upon hearing an novel scenario that highlights interesting theoretical aspects of a topic (probability), decides that learning from something new is asinine.
2) Another vocal skeptic insists on a wrong answer, even when the correct answer is repeatedly explained clearly and completely.
Yes, it’s the usual pattern.
So don’t waste too much time on these.
I find this entertaining. There are three doors and he correctly describes three cases. And then he talks about four possibilities as they’d were independent 1/4 each. Stunning mistake.
The error is that the probability is set by selecting one door out of three without any prior knowledge, i.e. randomly. That the host has two options to continue in one case does not reveal new information and thus does not change the probability.
Amazingly similar to deniers’ responses in threads about climate change.
You can actually play the game online here:
https://betterexplained.com/articles/understanding-the-monty-hall-problem/
Scroll down and click the doors.
Have 20 goes at it (don’t hit reset to replay, just click on a door).
Play with sticking for 20, and then switching for 20.
Compare the 2 results that appear as a rolling score in the top right hand corner of the box in which the game is housed.
If you think switching makes no difference, you’d best be prepared to play numerous times after resetting to get a result (by chance) that matches what you believe.
tonyM,
The original problem with the cash incentive is here:
https://www.jstor.org/stable/2683689?seq=1#page_scan_tab_contents
Thanks Barry.
It covers the problem well.
The first offers were just opening gambits and even the $1000 offer after opening a door would likely not entice even a risk averse player as just looking up the price of a Lincoln in 1975 shows a price tag of about $10,000 (very expensive car for those days).
I note that Monty did not offer a switch; the contestant expressed the desire to switch and may never have been given that option.
Given that Monty offered such a lousy deal the contestant had more incentive to switch if he worked on the assumption that there was a greater chance that Monty wanted him not to take the cash but stay with his box (and optimize the costs of the show). This is additional to the increased chance given by a switch if he could be granted that option.
Dr Spencer is right in that the game depends very much on the intent of the show and even each game which would be a balance between making the show exciting and how much to give away over time. I can’t see such a car deal could keep viewer interest for too many shows but there are many creative minds capable of finding ways to keep up the stimulus.
In the original, the point of the exercise is the same as intended in the later iteration in Parade magazine.
In the link, the original author answers his own scenario. The object of the exercise is to show that the odds increase by switching. This thesis is passed on in the later iteration.
I know that the problem (Parade 1990) can be interpreted differently by cracking open possible ambiguities in the scenario wording. I’m a little puzzled by that. The original wording seems clear to me (interpreting written text is one of my jobs). Various interpretive permutations I’ve seen seem forced.
The host opens “another” door – it cannot be that the host would ever open the contestant’s door.
The word “say”, means “for example” – the host is not constrained to favour a rightmost or leftmost door.
The host will always open another door that is a goat door, because he “knows what’s behind the doors,” inferring strongly to me that he will never intend to show the car.
Here’s the 1990 version in full:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
The question posed at the end requires that you must have a choice. The car cannot be revealed. Another door must always be opened, leaving 2 unopened doors. In terms of the probability lesson, this version matches the intentions of the original, which was made clear by the original author giving the answer.
Vos Savant, who published the 1990 version in her column, later remarked that very few among the thousands of letters she received misunderstood the wording – but that a huge number of those who did get the scenario as intended got the answer wrong.
I have a theory (unverifiable) that some who came up with the *wrong* answer attempted after the fact to ameliorate their error by squeezing the wording for other interpretations. And no doubt some gave the other possibilities thought in a genuine attempt to poke at the puzzle.
Barry:
We are in total agreement about the problem posed in the new reference i.e. the 2015 article covering the 1990 vos Savant column. I have just realized that this 2015 article was also referenced by Dr Spencer in his piece. Sounds silly but I had not read it until you referred to it.
Re the 1975 “American Statistician” article you referenced earlier I also agree with your comment:
“In terms of the probability lesson, this version matches the intentions of the original, which was made clear by the original author giving the answer.”
But all of this has really been covered by your solution and my solution which concur because the conditions assumed are the same – the key one being Monty must reveal a goat door out of the two boxes not selected.
But the actual show seems quite different as shown in the 1975 article. This is confirmed by descriptors in the 2015 article such as:
“variation of the Monty Hall Problem”
” Loosely based on the famous television game show Lets Make a Deal”
The same solution need not apply if the conditions change such as was shown in not being offered the switch or monetary offers made by Monty for the contestant’s box or variations, which then start to bring in subjective probabilities associated with Monty’s motives which modify the “textbook” probabilities as well as utility for the contestant.
Probabilities are always conditional on the knowledge available and Monty’s behaviour will certainly form part of those conditions. This, I think, is the type of variation that Dr Spencer was highlighting
OK, I haven’t read through all 180 something responses so I don’t know if this has been said already but here is an easy way to think of the problem. Instead of having the contestant select a door, the first action will be to have one of the doors selected at random (by roll of a die or whatever). Then you separate the doors into two groups. The first group contains the randomly selected door and the second group contains the other two doors. Now you ask the contestant to choose one of the groups. If it is the group that contains the car, contestant wins. The obvious choice for the contestant is to select the second group every time (assuming the original selection was indeed random) because the odds are 2/3 as opposed to 1/3 if selecting the first group.
This, of course, does not change Roy’s assertion that it depends on whether Monty was deliberately choosing a goat door but it makes it easier to see why the strategy of switching doors is advantageous.
Actually, I take that back. It doesn’t matter why Monty chooses a certain door. The odds are still better to switch, unless of course the door that Monty was going to select is the same one that the contestant chose. In that case, if Monty reveals a goat, the contestant has to choose one of the other two, at 50/50 odds. Um, I think :>
Kevin
Dealing with just 3 doors, I have a hard time understanding why random/deliberate is important.
The contestant makes his guess. Then, if the other two doors were opened at random, 1/3 times a car would appear. The constestants choice would then be between two goat doors.
By always showing a goat, the host eliminates the possibility of this losing choice!
I think I’m still missing something.
Oh.. ..when the host opens the goat door, we know that door had a 1/3 chance of having the car, which means It’s that much more likely to be behind the unopened door.
unless of course the door that Monty was going to select is the same one that the contestant chose
Yes, that changes things. But that action doesn’t fit with the original wording of the challenge.
“…You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door…”
What’s bugging me is why the two doors opening at random would ever produce a 50/5O chance.
Random: a car appears. The other doors have a zero chance of winning
Random: a goat appears. Wouldn’t the other door still have 2/3 odds of having the car?
Where is the 50/50?
It comes from Monty opening the door that the contestant chose having a goat behind it. Kevin wrote:
The odds are still better to switch, unless of course the door that Monty was going to select is the same one that the contestant chose. In that case, if Monty reveals a goat, the contestant has to choose one of the other two, at 50/50 odds.
That’s correct reasoning, but the premise doesn’t fit the game show as described.
Thanks Barry, that makes sense.
I’ve just read a great variant which might make it easier to get.
There are a thousand glasses on a table, 999 with poison that will kill you, 1 that is safe.
You pick one up.
Your captor, who knows which glass is safe, picks up one of the remaining glasses and tells you that one of the two glasses being held is safe. He never lies.
Which glass would you choose?
—————————————————————
It’s an inversion of the original problem, but the essential likelihood solution is the same (with larger odds)
—————————————————————
You can re-invert this to the Monty Hall problem (with larger odds):
After you’ve picked up your glass, the captor injects a solution into the remaining glasses but one. The solution will turn any glass with poison black, but will not change the colour of the safe glass.
You see all the other glasses are black.
He holds up a glass that he hasn’t injected.
What are the odds that you are holding a glass that is safe?
(1/1000)
Barry
Nice. That makes the choice, “crystal clear”…….Lol.
I had a “duh” moment yesterday after reading Kevin’s explanation:
If you’re given the choice between one or two unopened doors, the two door choice is obviously better. Equally obvious? One of the two doors you chose will definitely NOT have a car behind it. That’s the one Monty shows.
I should clarify. If all three doors are unopened, would you rather get one guess or two? Obviously two. Now point to the two doors you want opened. At least one of them will not have a car, right? That’s the one Monty opens first.
This is brilliant!
Vos Savant is correct. I can show you the proof. Is that her real name?
Yes
I was agreeing with her.
It is clear to me that the person should switch (2/3 odds vs. 1/3), but I do not see why the host’s motivation matters at all. The new information the player gains is which of the other two doors (those two having 2/3 odds of having the car) would be the one of those two with the car. That info is the kicker, not why he showed you that door. Sure, if Monty just picked at random, then sometimes he would reveal a car and game over before you ever get to decide about a switch, but when he reveals a goat, then you still have learned the same new information about the remaining two doors.
CM
I get what your saying, but consider this: 1000 people buy raffle tickets. Only one is a winner. 998 have been turned in and verified as losers. You own one of the remaining two.
I think your chances are 50/50.
This has been driving me nuts!
Correct. The difference to Monty Hall is that the 998 have been randomly “opened” and found non-winning, where Monty Hall would open 998 and carefully not touch the winning ticket, if any of the 999 as usually the case.
Your chances are 50/50 because your setup is such that you have been lucky already. In Monty Hall, your initial choice is not usually the lucky one, but we’re not told that.
I think the initial probability of 1/3 is a red herring. You end up with two choices 50/50.
You could construct some rule ,known to you,used by Monty to reveal a goat behind #3 which ruled out #1 or #2 as having a goat. In this case, you would then be able to choose the remaining door with a 1/1 probability.
Oh Christ! All this explaining how you could easily be fooled to think it is 1/2, people turn up and explain it is 1/2.
The only good thing is you’re humble; you say you think it is a red herring. Well, actually it is not. The probability your first choice is right is 1/3, and Monty does not provide information on if it is right, so it remains 1/3. That it is one of the others, is 2/3. If you can rule out either of the others, you have two options, one with 1/3 and one with 2/3.
Conditional probability is difficult.
Wert
You explained the situation with the 1000 raffle tickets correctly. Opening the tickets at random versus intent MATTERS. What bugs me, is when I try to apply the same logic to three doors, it fails. Intent versus random doesn’t seem to make any difference.
Try 100 tickets, 10 tickets, 5 tickets, four and finally three. The intent does not matter, only that if Monty was absentmindedly showing the winning door/ticket, he’ll switch and just draw one non-winning ticket away. He’s random only if you chose the winning door originally.
Conditional probability is something a poker-player must master. But it comes handy in many situations.
Let’s assume 1/10000 have a cancer and there is a test which gives 1 false positive in 1000 for it. Let’s test one random person with the test.
In 9 cases out of 10, the test will give a false positive.
However, if there is another indication that 10% of the population has including all the cancer patients, your chances improve a lot. How much?
100% correct. There was an unstated assumption in the problem as set out so many years ago. I came to this conclusion at the time.
Strewth, Appell gets it right for once!
What if the car was the cheapest, least attractive bag of new nails on the market while the goat was a prize-winning Billy stud-muffin worth hundreds of thousands? Eh?
If given the choice of picking TWO doors, instead of just one, in a modified version of Dr. Spencers statement of the Monty Hall problem, then anyone would select that choice because of the 2/3 chance of winning. That is precisely the opportunity offered by being allowed to switch doors in the original game. Thus one ought to switch.
The fact that Mr. Halls motives, rules, attitudes, etc., are unknown is not relevant. Switching cannot reduce the chance of winning, and MAY increase the odds to 2/3. Thus one ought to switch regardless. Not being able to understand Mr. Hall prevents calculation of the exact odds of winning, but does not prevent making the right decision.
The solution is “simple”. Everything is “set” in the first choice: 2/3 of choosing booby prize leaves you with 2/3 probability of winning when switching your choice, and 2/3 probability of loosing when keeping your original choice.
Sorry, folks. Logic dictates that if the third door were opened randomly, and found to have a goat, you might as well stick with your original choice – 50/50.
If Monty intentionally opens the door with the goat, you should switch.
My understanding is that Monty ALWAYS opens a door with the Goat behind. This Was part of the game protocol – the game would be rather pointless if he picked a random door, because then he would occasionally expose the car and ruin the game. Note the Wiki explanation of the game:
…. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat…..
I don’t know whether this was advertised to the contestants or not – but it would rapidly become fairly obvious to anyone who followed the show. Dr Spencer, I think, has erected rather a strawman – it is trivially correct to point out that a random opening choice would leave the probabilities at 1/3, while it is bordering on incorrect to say that there is an ‘unstated assumption’ that the goat is ALWAYS exposed. That rule may not be pasted up on the wall, but it is obviously there.
The point, of course, is that it is no immediately obvious that the rule has any bearing on the probability of winning, so it can be left exposed as an obvious game feature which need never be discussed. A good example of hiding something in plain sight…
Dodgy
Yes, Monty always opened the door with a goat, and so the contestant should always switch. But the really strange, head scratching thing, is what would happen if Monty were to open the door based on a coin toss, and a goat appeared. In this case the odds that each door has the car is 50/50. So you might as well keep your original choice.
People think it shouldn’t matter why Monty opens the goat door, that your odds if you switch would be 2/3 regardless. Not true.
Monty can’t open the goat door randomly. He’d end up showing the car. If he’s allowed to show the car sometimes, the game is totally different.
The point of the game is to make the right decision, not to calculate odds. Switching doors does not lessen ones chance of winning, and may increase that chance. Why Mr. Hall selected the goat door is irrelevant to decision making.
Jerry
We’re not playing a game. It’s a brain teaser. I think it’s very strange that the odds change based on whether the goat door is opened at random versus with intent.
50/50 chance when down to 2 choices.It doesnt really matter if you switch or not.Its still 50/50 chance.
That is true only when there are two EQUALLY LIKELY choices.
Im saying I think the mensa person is wrong.
For example, if you flip a coin 1000 times and 999 come up heads ,its irrelevant to the next flip.Next flip is still 50/50(but I wouldnt bet on it knowing how irrational hot streaks are a reality)
Is there some aspect where there’s expert poker player analysis here?
If its purely a number quiz ,mensa is wrong.
Mojo
I was positive the odds would be 50/50……had to read a bunch of comments before I changed my mind.
Suppose you had two guesses instead of one. That would double your chances of winning the car, right? Even then, one of the two doors you choose will always have a goat behind it. That’s the door Monty opens first.
This particular problem is entertaining exactly because it will put many people in the state of aggressive denial. The question is simple yet misleads many who have quite a solid arithmetic skill.
Wert
I think it should be a two part question.
1. Should the contestant stay or switch?
2. Why are the odds 50/50 if, in theory, someone were to open the 3rd door at random and revealed a goat?
Number 2 is the mindbending situation that gets overlooked.
With global warming, clearly you should prefer the goat to the car.
It is about how the 1/3 probability for door #3 is distributed across the door #1 and #2. It’s a probability so it should sum up to 1.
If there would be only two doors, a host picking one would cause 1/2 of the prob to “flow” to the other door.
The situation with 3 doors is not much different. If door #1 is picked by the guest, it is out of the equation. The host now picks e.g. door #3 and the probability flows to the only other door still part of the equation. Door #2 now has probability 2/3.
If you have 10 doors, guest picks door #1 and the host opens from #2 – #8 all the probability flows to door #9, not to door #1. The guest has now probability 1/10 of winning when staying, and 9/10 of winning when moving.
In the three door case, if the host would pick door #3 no matter what. The probability is still 2/3 for combined door #2 and #3. If the host picks the door with the car, the guest definitely has to switch! The probability of winning is then 1. 🙂
Anne
But if doors 2 thru 9 were opened by someone who didn’t know where the car was, and by chance they all had a goat, then doors 1 and 10 would each have 50/50
Anne
In the three door case, if the host always opened door #3, and assuming placement of goats is still random, then there would be two possibilities:
1. door 3 has a goat, in which case door 2 has a 50/50 chance of having the car.
2. door 3 has the car, in which case the contestant is screwed, because both unopened doors have goats.
There is a free Resampling/MonteCarlo computer program called ‘Statistics101’ that has the Monte Hall problem already set up as an ‘example program’.
It can downloaded from http://www.statistics101.net/
The code might seem a little unclear at first sight but if a line said SAMPLE 1 LotteryNumbers Winner , it means take one sample from the collection of ‘LotteryNumbers’ and call it ‘Winner’.
Hope that helps!
Chas
Maybe you could check this: if a door is opened at random, and a goat appears, the original odds of that door having a car would get divided between the remaining unopened doors.
For example: in the case of 3 doors, if the host opens one of the remaining two at random, revealing a goat, the original odds of that door having a car was 1/3. This would get divided (1/6) and added to the other two unopened doors. 1/3 + 1/6 = 1/2
Correction, now I think the randomly opened goat door would have it’s CURRENT odds divided rather than it’s original. The math gets a little complicated.
4 doors. The guest chooses 1. The host opens another at random revealing a goat. 1/4 is divided by 3 = 1/12
1/12 gets added to the remaining 3 door’s odds: 1/4 + 1/12 = 1/3
Now we have the same situation as described above (3 unopened doors, each with 1/3 chance of having the car) where the odds end up 50/50 when a goat door is revealed by chance.
So you pick door A. In doing that you know there is a goat behind B or C at least anyway. You also know that there is a 2/3 chance of the car being behind B or C and 1/3 behind A. Without revealing a goat behind B or C you are offered a chance to change your choice to B & C. I’d change because there is a 2/3 chance of a car behind B or C, even though I know there is a goat behind one of them. The only difference is that Monty Hall knew where the car was but in practical terms that should make no difference to the probability of A vs B&C.
What If Monty were to forget where the goat was, but instead of consulting with someone backstage, decided to just guess?
A 1/3 chance a car would appear. A 1/3 chance a goat would appear. If a goat appeares, onky 50/50 chance a car is behind door B
Whereas when Monty remembers where the goat is and opens the door intentionally (which is the show’s premise), then door B has a 2/3 chance of having the car.
If Monty forgot and opened a door with a goat behind it,so what? There was and still is a 2/3 chance the goat was behind B or C. Just because in a one-off random choice by Monty he got a goat doesn’t affect the initial balance of probability in a one off go.
Consider further, what if Monty doesn’t tell you that he found a goat by chance? Or even further what if he lied to you that he found the goat by chance or not?
I summary 2/3 times the goat is behind B or C, so other than Monty telling you the car is behind A there is no reason to stick with A when offered a change.
Son if mulder
“Just because in a one-off random choice by Monty he got a goat doesnt affect the initial balance of probability in a one off go.”
It actually does, which is why I love this brain teaser.
Let’s say there are 5 raffle tickets. One is a winner. You buy one ticket.
Three tickets get turned in by other players, each with a 1/5 chance of winning, but are found to be losers. The two remaining tickets now have equal chances.
The opposite scenario is this: 5 raffle tickets, you buy one. Someone, who knows, points to 3 losers, leaving one unverified. In this case, your ticket only has a 1/5 chance. The other has a 4/5 chance.
Darn it, I forgot. In my first example, the first ticket found to be a loser had a 1/5 chance. The second had a better chance, the third better still. That leaves a 50/50 for the remaining two.
Moving back to 3 doors:
A door, opened randomly, has a1/3 chance of the car. If a goat appears, these odds are split amoung the other two (+ 1/6 each).
OTOH, a door opened BECAUSE it has a goat, had ZERO chance of showing the car, meaning the other door must have had all of the 2/3 chance.
Sir Isaac, I agree your comment about 1/6 being allocated to each remaining door but I was focussing on there never being a justification not to change.
I wrote a short program to prove it. A contestant wins 1 out of three times if they don’t change their guess andv2 out of 3 times if they change their guess. I also wrote a mathematical proof. But the verbal or intuitive grasp of it is not so easy.
The part that confuses people is that Monty Hall’s revealing a door that doesn’t contain the car is a misdirect.
Assuming an honest game, if Monty Hall had framed it initially as “you can choose either door 1 or get both doors 2 and 3 together”, then everyone would realize that it was a 1/3 chance vs a 2/3 chance.
But this IS what he offers you; he just does it in words that confuse people, and does so only after they made their initial choice.
The first choice had a 1 in 3 chance, i.e., a 2/3 chance of being wrong.
The moment he reveals that door that a) you didn’t pick is b) not a winner, what he is REALLY saying is:
“Your door has a 1 in 3 chance of being correct; the other 2 combined have a 2 in 3 chance of being correct. Now, it MAY be that you got it right with your 1 in 3 chance, but there’s a 2/3 chance you’re wrong; and just let me tell you that *if* the 2 in 3 chance pans out (and it might not, even though it’s twice as likely to than the 1 in 3 chance), then it will be behind *this* door, not *that* door”.
There was a 1 in 3 chance that your first choice was correct and a 2/3 chance it was wrong; but which door Monty Hall chooses to open *after* that first choice did NOT have a 1 in 3 choice; some of the time, Hall is going to have a 1 in 2 choice of which door he reveals (e.g., if you picked door 1 and it IS door 1, he can choose either door 2 or door 3), and some of the time he will deterministically have NO choice, but be compelled to open one particular door (e.g., you pick door 1, and door 2 is correct, he HAS TO open door 3).
What Hall offers is the chance to switch from your 1 in 3 choice to getting both of the other 2 put together, i.e., a 2 in 3 choice. But he words it in a way that constitutes a misdirect.
The fact that so many people can’t grasp it means it is a very good misdirect indeed.
It has nothing to do with ‘framing the question’ or assumptions. Mathematicians routinely ask hypothetical questions It is a simple question of listing the possible outcomes. Example: Car in Door A, Player picks Door A, Monty opens Door B. If the player stays with Door A- win. Just list the possible outcomes. Six wins from not switching, six wins from switching- as you expect. No Genius required.