Marilyn vos Savant and the Monty Hall Problem
priceonomics.com
priceonomics.com
1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it.
But there's a 2/3 chance it's behind one of the other doors.
So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose).
You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B.
So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance
Make the switch every time.
Imagine a related Monty Hall problem, where you select a door, and then Monty immediately asks (without revealing anything), "Do you want to keep that door, or would you like to pick the other two doors?" Clearly you'd pick two doors instead of one.
When monty opens a door and gives you the "choice to switch" he is making noise designed to make picking two doors look like picking one door.
All I can say is that it's an incredibly effective verbal/logical obfuscation that relies on synergistic choice of number of doors and the (in my view) totally bogus part about Monty opening one of the other doors.
In fact, some people (in this very thread!) are still analyzing the problem as if the opened door represents new, 'very valuable' information! It is truly just slight of hand to make it seem like you're only picking one door when you are picking two doors.
In the "real world" problem there are additional variables to consider, such as whether or not the host would open a door containing a car, whether that would result in a win or a loss, whether the host would offer a switch in all cases, whether if so, there is any additional incentive for switching (which wouldn't adjust the probability of winning, so is a red herring, but "intuitively" it seems that if the host is trying to make you lose by trying to make you switch, probability should go down).
When you try to conflate the real world problem, which actually has multiple different probabilities depending on exact scenario, into an answer to what's generally accepted as the "Monty Hall Problem", some of this gets in the way.
In addition, there are two probabilities in the "Monty Hall Problem" - the first is whether your switch results in a win. The second is the probability that, if you switch, it was the correct choice.
I think this is what the article was trying to get across with the example of the doors but I felt that just confused the matter by talking about "shifting" the probability.
The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.
I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.
To be clear, if the host picks randomly (whatever happens to the door he picks) your odds are the same whether you switch or not.
Will try and run a simulation tonight or this weekend.
Specifically, if you have a 50% chance of winning on swap, you have a 50% chance of losing by not swapping. So, yeah, by the time you get to the swap, no matter what, you are at a 50% chance of winning. Swap or not.
I usually find this problem annoying, not because it's all that difficult, in fact it's quite intuitive - when you're told the exact parameters defining the Monty Hall Problem and systematically work through them.
In my experience though, it's used more often as an exercise in diminution, a sick wet dream of probability teachers, where the learning party isn't aware of the problem, and usually either hasn't been explained, or doesn't quite grasp, the exact circumstances around whether the host's choice is random or decided.
There are lots of "it depends" moments that can be applied to incomplete descriptions of the problem, including (amazingly) whether the host offers a choice at all - this is the one that seems to trip up most people, as they might start to question the "motives" of the host (which are irrelevant in the actual statistical problem).
I think to enumerate the possibilities you'd have to see that if you picked the winning door, there are two ways the host could leave doors for you to swap to and lose.
If you picked a losing door, there is only a single way for the host to reveal a losing door.
So, at the point you are looking at a losing door and making a swap, there are 4 ways you could have gotten there. You picked the winning door, and the host showed either of the two losers. Or you picked either of the two loser doors and the host showed you the other loser. Four possibilities, two of them you win if you swap.
It finally jives with me that your odds at swap time are only 50%. Seems kind of obvious when you think of it as random events and you are at the end with it definitely behind one of two doors. Either door is clearly as likely. Now, your odds of getting to this point are vanishingly slim, the more doors there are. Which makes sense.
I think my intuitive block comes in in that your odds of winning the game are not increased in this scenario at all. Which, I knew. I think I even stated it at some point. Still a hard block to get around.
That's really interesting and counter-intuitive.
https://gist.github.com/ecdavis/da8f67258860e9f35620
EDIT: After thinking about it for a while it seems obvious and I feel fairly stupid.
1/3rd of the time your initial choice was correct. The random host always reveals a goat, you take the opportunity to switch and lose as a result.
1/3rd of the time the random host reveals a car, the game ends without an opportunity to switch.
1/3rd of the time the random host reveals a goat, you take the opportunity to switch and win as a result.
First, let's make it clear exactly what variation of the game we are playing.
1. Prize is assigned to a random door with each door being equally likely. Neither you nor Monty know which door.
2. You pick a door. Because the prize was assigned randomly and you don't know where it is, it is irrelevant how you pick your door. Without loss of generality (WLOG) we can assume you always pick door #1.
3. The host picks a door and opens it. Because the prize was assigned randomly and Monty does not know where it is, it is irrelevant how Monty picks a door. WLOG we can assume he always opens door #2.
4. If Monty revealed the prize when he opened his door, the game ends and you lose.
5. If Monty did not reveal the prize, you are given the opportunity to switch to the remaining door (door #3).
6. Your door is opened. You win if the prize is behind it. Otherwise you lose.
There are three equally likely cases to consider.
1. The prize is behind door #1. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is bad.
2. The prize is behind door #2. This occurs 1/3 of the time. Monty opens #2. The prize is there and the game ends. Note that in this case, YOU ARE NOT GIVEN THE OPPORTUNITY TO SWITCH.
3. The prize is behind door #3. This occurs 1/3 of the time. Monty opens #2. You are given the opportunity to switch. In this case switching is good.
Note that in the cases where you are given the opportunity to switch (#1 and #3), switching wins in one of them and switching loses in the other. Each of these cases is equally likely (occurring in 1/3 of the games of played), and so in this version of the game switching makes no difference.
Here's another way to look at it. Since neither you nor Monty know where the prize is when you pick doors, we could change the game so that the prize is not placed until AFTER Monty opens a door, and this would not change any probabilities.
So, in this modified but equivalent game, we play like this:
1. You pick a door.
2. Monty picks a door and opens it. There is nothing behind it, because the prize has not yet been placed.
3. You are asked if you want to switch to the other unopened door.
4. The prize is placed randomly.
5. If the prize is placed behind the opened door, the game ends and you lose.
6. Otherwise, your door is opened and you win if the prize is behind it.
It should be clear that you have a 1/3 chance of winning the car in this game no matter how you pick your door or whether or not you switch. At the time the prize is placed, there is a door that is now your door, and you win if and only if the prize gets randomly placed behind that door.
That's part of the setup. If it's not, it's not being told correctly. It's usually in the form of "...and Monty opens a door that is always a goat..."
1: http://lesswrong.com/lw/ii/conservation_of_expected_evidence...
Imagine that after opening the door everyone goes home and a new participant who has no knowledge of the previous experiment is called to make a decision between the two remaining doors. Obviously in this case the probably is 1:2, as there is no information regarding the previous test. Actually, if we imagine a similar situation happened before the initial problem with a previous show with 4 doors, we may as well see that the initial probability of 1:3 is also wrong if we had access to the knowledge about the previous experiment.
I still agree with the switch suggestion, but the mechanism behind it is still not entirely clear.
A more intuitive example would also be to invert the problem: leave one of the non picked doors (C) out of the random opening and use the door the contestant picked (A) as a possible case in the random opening experiment. If we start by opening A, then we already know enough to see if the user won or lost, if we open B and there is the car behind it, we already know everything. If we open B and there is a goat behind the door, consider whether you would switch your choice from A to C. I guess you probably wouldn't and it would feel easier to justify your position, as this doesn't involve the "psychological pain" of changing a previous choice by reasons that you don't entirely understand (associated with possible loss-aversion feelings).
I have a friend who has two children, he has a son born on a Tuesday, what is the probability his other child is a girl?
Compared with this problem
I have a friend who has two children, one I know is a boy, what is the probability the other child is a girl?
And this one
I have a friend who has two children, the oldest is a boy, what is the probability the other child is a girl?
Why do these three probabilities differ (in particular, since the last two may be more obviously different, why is the first one equal to neither of the others)?
I spent the best part of a week in grad school pondering this until I came up with a deeper intuition. But the results are still surprising, I think.
The middle case is 2/3 likely to be a girl. Think of the possibilities of the sex of two children (sex given in birth order) M then M, M-F, F-M, F-F. I tell you one is a boy, so we have MM, MF, FM, in two of the three cases, the other is a girl.
In the first case the answer is 14/27 a girl, so almost 1/2, but not quite. To figure this out you can consider all 196 combinations of two children with birth-days-of-the-week, cross out all those that don't match the phrase "he has a son born on a Tuesday", and you're left with 27 possibilities: 14 of them are when the other is a girl (14 because she could be born first or second, and on any day of the week 2x7=14), and 13 when the other is a boy (6 when the other is an older boy born on a non Tuesday, 6 when it is a younger boy born on a non Tuesday, and 1 when both are both on a Tuesday).
So that's not the intuition, as much as the answer.
The intuition is that as you add more specific knowledge about which child you mean, that question contributes less to the probability of sex, and is more an irrelevant detail. Being more specific can be anything. Do it with birthday-of-the-year, (more specific than day of the week) and it is even closer to 1/2. Do it with something unique, like their phone number (or their birth order, or 'the taller one is a boy') and it is exactly 1/2. Be less specific, make it so they both toss a coin and it is 'one of the children tossed heads and is a boy' and it is 3/7.
The actual Tuesday problem requires a little more thought. Somehow that extra information makes you drop a case! I can't help but think that the ordering is the critical factor, and that it's not a coincidence that the "extra information" has to do with time. In other words, the dropped case happens because we drop the distinctive ordering when the two boys are born on a Tuesday.
To answer your question: we treat the MF and FM differently, just for ease of calculation.
What we're doing is taking a set of possible situations (different 2-child families), each with an attached probability (I'll come back to this), throwing away those that don't match some known information (i.e. those without a boy), taking those that do, grouping them into sets (the set of families with the other being a girl, and the set where the other is a boy), and calculating the probability the actual scenario was in each of those sets.
Now the 'attached probability' probably sounded odd in there, because that's not what we've done, right? We've just been counting situations. True, but that's because we carefully started with situations that are all equally likely. So rather than bugger about with those attached probabilities, we could just use raw counts, knowing that it would work.
So, if we start with four possible sexes of two children: MM, MF, FM, FF, those are all equally likely. I'm using the birth order as a way of breaking the two MF cases apart, so I've got two cases that are equally likely, letting me do the counting trick. I can just count the 2 out of 3 remaining situations that match the information, so 2/3. Easy.
But I don't have to do it that way.
Let's say we follow your intuition and start with three possible sex combinations: MM, MF, and FF. Now, those aren't equally likely, right? There are twice as many families with MF children as with MM children. So I need to use the attached probabilities.
MM (1/4 of 2-child families), MF (1/2), FF (1/4)
We can remove the FF as inconsistent with the information (as before), then the probability the parent has an MF family (i.e. their other child is a girl) is
1/2 over (1/2 + 1/4) = 2/3
as before.
In fact, for any reasonable calculation based on data, we're almost certainly not going to be able to use the counting trick. In the case where I was using actual birth rate statistics for sex (rather than assuming as many girls as boys are born), I'd have to do it this way. The counting trick is really only useful for toy problems and teaching basic probability: you were right to call me out on it.
---
The Tuesday's child doesn't drop a case. There are certainly seven cases where, in an MM family, the eldest is a Tue child, and seven where the youngest is, but one of those cases is shared: a case where both are Tue children.
In terms of the intuition you can think of it this way, identifying by birth-day-of-the-week is identifying which child is which, so we're almost at 1/2, but there is one case (both Tue boys) where this information does nothing to distinguish them, so that tiny bit of ambiguity remains, and it can't quite reach the full 1/2, there's still a trace of the 1/3 result that came when the ambiguity was total.
"""Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3.
In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male."""
I advise students to always work this through from the combinations of possible scenarios.
The deeper truth of the vos Savant story is that verbal descriptions are ambiguous, and the even deeper truth is that probability calculations are highly volatile on information.
Eventually disagreements come down (among people who can competently do the calculations) to arguments about what such and such a phrase means, or what underlying probability distribution it implies.
Hence in the Monte Hall Problem, there is no advantage to switching if Monte selects the unchosen door to open at random, and it happens to be a goat. That situation is consistent with the description, but perhaps not the 'feel' of the game show setup.
In science, it is wise to be aware of this. If you have probability calculations that are so volatile based on interpretation, then you probably want to avoid trusting the results, no matter how careful your hermeneutic.
This isn't quite right. Obviously if there's a chance Monty might open the car, it screws up the gameshow, but it doesn't actually change the odds in the situation that he opened a door with a goat.
The knowledge in Monte's head doesn't change the odds.
To simulate this properly, you'll need to change the problem statement somewhat... you'll either have to throw out cases where he opened the door with the car, or state the problem such that we're only look at cases where he randomly opened a door with the goat. But you'll still get 2/3.
This python gist appears to simulate the situation correctly, and shows that it becomes 50/50.
https://gist.github.com/ecdavis/da8f67258860e9f35620
That said, I'm not sure I completely agree with this solution just throwing out cases where he opens the winning door, without replacement. I feel like at this point we're calculating a different problem where we haven't precisely defined what it is we're measuring, so you could tweak things to get either answer.
The distribution in order of toss:
HH 1/4
HT 1/4
TH 1/4
TT 1/4
Looking at the first three rows of the distribution, weight of rows where there's one tails is 2/4 while there's only one row with both heads, whose weight is 1/4. So (2/4)/(3/4) = 2/3 probability of tails.
Problem three: Two coins were tossed, first was head. So we look at the first two rows of the distribution. HT and HH each have equal weight so it's (1/4)/(1/4+1/4)=1/2.
First problem: Two coins were tossed, one was tossed on tuesday and is a heads. What is the probability that the other is tails?
Let's assume all coins are always tossed on random days as well. We are now interested in a sub-population where one was tossed on a tuesday and was heads.
If we tabulate all combinations in the population that we are interested, x axis is first coin toss week day and y axis is second coin toss weekday:
tuesday.txt
mon tue wed thu fri sat sun
mon HH HT
tue HH TH HH HT TH HH TH HH TH HH TH HH TH HH TH
wed HH HT
thu HH HT
fri HH HT
sat HH HT
sun HH HT
$ egrep -o "HH|TH|HT|TT" tuesday.txt | sort | uniq -c
13 HH
7 HT
7 TH
Edit: to put it in words, the slight asymmetry is that if the other child was also born on tuesday, from the population where both children were born on a tuesday there are more boy-girl + girl-boy samples than boy-boy samples.
Winning by switching doors requires your first choice to be incorrect.
You have a 2/3 chance of guessing incorrectly when all three doors are closed.
Thus, you have a 2/3 chance of putting yourself in the position from which switching will win.
If there are three doors, the odds that the car is behind any of them is 1 in 3, or 1/3. Once you select a door, the odds are 1/3 for your door, 1/3 for each of the other, unselected doors. Put more simply, 1/3 for your door, or 2/3 that it is not in your door. Once another door is opened that does not have the car, then the odds of it being your door are 1/3, and the odds of it not being your door are still 2/3, so it makes sense to switch.
At the time of selection you freeze the probability of the sets (as this is the macroscopic world and the implications of the state in the other rooms must have collapsed due to interactions with all sorts of radiation from the rest of the universe).
Thus what you say is exactly correct.
* T0 - {1/3} {1/3} {1/3} * T1 - {1/3 Your pick} {2/3 = {1/3} + {1/3}} * T2 - {1/3 Your pick} {2/3 = {0/3 Host Pick} {2/3 Remaining door}}
> The short answer is that your initial odds of winning with door #1 (⅓) don’t change simply because the host reveals a goat behind door #3; instead, Hall’s action increases the odds to ⅔ that you’ll win by switching.
Well, why don't they change simply because the host reveals a goat? If you show me a three sided coin, I'd say there's a 1/3 chance of getting a particular side. If you then magically change gravity such that one of the other sides will never ever be landed on, would the two remaining sides not now have 50/50? I guess the article asserts that the "probability" doesn't get redistributed amongst the remaining options evenly, and I want to know why it doesn't and why it instead all "goes" to the one I didn't pick.
I don't mean to be confrontational in asking, and I don't doubt the answer, I just don't get why. Could anyone explain it better, or point me somewhere that does?
Your mistake is somehow thinking they are the same state.
Instead, Monty does the statistical equivalent: He allows you to know for sure which of the other doors definitely has a goat.
To answer the question about the 3-sided die: the question is flawed for 2 reasons.
First, the way you phrased it, you know the information a priori. It's like me learning that the car isn't behind door #1 before I choose. In that case, I have a 50/50 chance on the other doors.
But what if you didn't tell me before? This introduces our second problem. Let's say I roll the die and you hide it under a cup. I tell you I chose side A and you tell me A is physically impossible and ask me if I want to switch. Well of course I want to switch ... but I didn't learn any more information about sides B and C. It's a 50/50 toss-up. In terms of the game show, you basically just opened the constant's door.
A critical element of the original problem is when the information is learned and knowing that the contestant's door will not be opened. It is what allows the phrasing from zaksoup above.
When you choose door 1, you have a 2/3 chance of goat. That means that 2/3 of the time, Monty doesn't get to choose a door (You're on one goat, he can only reveal the other goat). So he only has one choice, the other goat. Since 2/3 of the time he's forced to avoid the car, that means that 2/3 of the time you will switch to the car.
when you "switch", you're also switching the representation of a door from 1/3 to 1/2. and the odds are better when there are only 2 doors remaining.
remember, you don't always win by switching - it's not magic. over time, it just wins more often than not.
The host who reveals the goat door has more information than you, he didn't randomly open a door that just happen to have a goat behind it.
If that assumption isn't true (e.g. maybe the host only shows you a goat when you pick the car), switching isn't necessarily beneficial.
The same logic applies to 3 doors.
If you still don't have an intuition after reading the 2 explanation in the article (especially the one with 100 doors), I'm not good enough right now to explain it any better. But the one way that works for me is to play around with more Bayesian statistics -- one day it just suddenly doesn't feel weird anymore!
In Bayesian interpretation at least, knowledge does changes what the probability would be, as the number just correspond to your belief of an event happening. Just tangentially related, but this might have some interest to you: http://lesswrong.com/lw/oj/probability_is_in_the_mind/
Are you going to change your choice? No matter what you selected initially, the host was still going to open 98 doors and give you the chance to swap your choice. Would you? It's obviously not a 1/2 selection at that point, is it? There were still 100 options, but now 98 of them are eliminated. Your original door was (and is) 1/100, making the other option 99/100 for a total of 100/100.
If you extend that out to 1,000,000 doors the game is even easier to play. What are the odds that you just happened to select that one special door out of 1,000,000 options?
Then he says, "Okay, the car is either behind the door you selected or the one I selected by skipping over it. There's only one car, so if you picked the right one, then I had to pick a random goat door to skip, but if the door you picked was the wrong one, the door I skipped was the one with the car. And, unlike you, I know which door has the car. Would you like to keep the door that you originally picked or switch to the one that I just picked?"
You don't know where the car is; he does. Do you want your pick or his?
Increasing the numbers might help. Suppose you had to pick one out of 1000 doors, then 998 were revealed to have no prize. That might make it more intuitive that your first pick had a lower chance of being right.
2 out of 3 times, you will not pick the door with the car. This means, 2 out of 3 times, your door has a goat.
Revealing a door does not change that fact- that your door 2 out of 3 times has a goat.
His constraints are: (1) He has to open exactly one door, (2) he can't open the door you chose, and (3) he can't open the door with the car behind it.
If you've picked the car then he can't do anything for you (2), so he opens a door at random.
If you've picked the wrong door, he can help you out by showing you the other wrong door, leaving you with only one remaining option - an easy win!
So assuming you understand Monty's intentions, you can see that there's a 1/3 chance you've picked the right door and should do nothing, and a 2/3 you've picked the wrong door and should listen to what Monty's telling you.
(Monty's intentions are of course irrelevant because his behaviour is completely described by the constraints, but he doesn't know that and it's not polite to point it out.)
Let's say I have 10 doors, and put a goat behind 1 of them at random. You pick 1 door at random, leaving 9 for me. So, there is 9/10 chance that I still have the goat, and now 1/10 chance that you have it.
Now, I have 9 potential hiding places for this goat. BUT, I am willing to communicate to you which door I put it behind (if I have it, which 9/10 times I do). I tell you this by removing all but one (i.e. 8) of my doors.
So, there's 1/10 chance that you have the goat. There's 9/10 chance that I have the goat, and have just told you where it is. I then give you the option to choose my door instead of yours. Clearly, you should switch.
The key is that I'm removing not "one door" but "all but one" of my doors. In this way, I find the 3-door case to be the most confusing, because the minimum amount of doors is removed. With 3 doors, I only remove one door - which doesn't seem that spectacular. If there were 100 doors, I would be removing 98 of my 99 doors - which I think is easier to understand by intuition.
In fact, because of this your odds of winning-by-switching increase with the number of doors(see n-variant http://en.wikipedia.org/wiki/Monty_Hall_problem#N-doors).
This still gives each of your doors a better chance than the door I initially picked, and I should still switch to one of them.
The problem as stated is: "you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3". Why would you think the host has the option of not opening a door? And if you think they could do that, why couldn't the host also do other things, like offer the contestant cash to walk away, or swap what's behind the doors, or add more doors, or open all three doors, or sucker-punch them in the gut?
Because the actual TV show of Let's Make A Deal worked that way. Here is Monty's own take on it (click to page 2.) http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...
This is the source of all the confusion from vos Savant's original column. vos Savant got the math right for her version of the problem where the host always opens a door, but that isn't how the show worked. She presented a simplified "spherical frictionless cows" version of the problem.
The actual show did not compel Monty with any rules on opening doors. He had free rein to do whatever he wanted. He could reveal an incorrect guess right away with no opportunity to switch. He could and would offer the contestant cash to switch or not switch. Monty could try anything to trick or bluff or deceive the contestant, all for the sake of producing entertaining television.
Of course, there is no correct mathematical answer at all to a game of bluffing and outwitting. At best you might find a Nash equilibrium in game theory, but even that is hard to pin down. You need to know Monty's tendencies precisely, but even that is hard when his goal isn't to maximize expected value but to produce entertainment.
A bunch of PhDs aren't going to have a problem with a precisely specified version of the problem, which is what was asked of her in the original column. (which, incidentally, didn't use the name "Let's Make a Deal" at all).
And really, the thing is, none of that matters. Maybe in this version he could reveal an incorrect guess! Maybe in this version he could reveal the door that the contestant picked! That's not what happened in the situation you're asked to analyze. All that matters is that the host picked a door that (a) was not the door the contestant picked, and (b) contains a goat. Nothing beyond that is part of this problem, and nothing beyond that is necessary to solve it. You're given a very specific situation, and asked what the best decision to make is.
"The host has zero incentive to reveal what's behind the door you already picked (or the choice of switching would be a no-brainer) so the fact that they left it shut doesn't tell you anything."
"However, the fact that they didn't (or couldn't) open the remaining shut-door hints that it may contain the prize."
The key is that the host's choice is not random.
By selecting the other door your are now betting against your initial choice, which was a 1/3 chance. Thus your new odds are 2/3.
When someone gets declared as the smartest/best in the world it first off sets people's expectations so high that they expect perfection, and secondly it places the individual onto a plateau for which others would love to knock them down.
The article doesn't discuss this, but I'm sure that played a part in why her detractors jumped on her so fast and so fiercely. They got told that this person was smarter than them, and they were inferior, and now was their chance at proving that that wasn't the case.
She was of course right, and she likely didn't declare herself the "world's smartest woman" but unfortunately for her, that connotation was still attached to her character, and no matter how modest she was it would have been difficult for other people to take an ego knock by feeling like this person was "better" than them.
Essentially the title of "world's smartest woman" set her up for something like this to happen. Luckily for her her detractors jumped on something she was correct about making themselves sound like morons. However it could have happened when she made her first public mistake (it would have happened eventually).
Fortunately this episode may have helped release the pressure somewhat that was placed on her character. She saved face, but her detractors feel like they took her down a peg so didn't continuously stalk her looking for mistakes.
PS - Just to reiterate I am in no way trying to blame Marilyn vos Savant or deflect any of it onto her. Someone else likely declared her the "world's smartest woman," that is outside of her control, and she dealt with this as well as anyone could hope to.
If only they had stopped to see if she was even wrong first, they wouldn't have proven the claim.
Have standards for what makes something confusing and difficult to understand risen significantly over the decades? When I brought this problem up with my group of (admittedly geeky and mathy) friends no one was left unconvinced by the end of the afternoon, and most even had the intuition as well as the mathematical proof of the answer. Clearly we aren't any smarter than the public figures who couldn't comprehend this puzzle a half a century ago-- so it all in the framing of the problem?
"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 #1, and the host, who knows what’s 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"
(a) she didn't originally pose the problem, it was a question asked of her, and (b) the wording of the problem is not ambiguous.
The intuition is simple. Each time you initially pick a door with a goat, you win the car by switching. This clearly happens 2/3 of the time.
"Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999)."
Furthermore, at the time she wasn't listed in the current Guinness Book so anybody that looked it up would see the column as a fraud unless they checked a specific old edition. Claiming somebody "is listed" when they had been listed, for one year many years ago, is a lie.
So there were a lot of readers who thought of the column as a joke. If you open 99 doors that are full of mindless drivel, what are the odds the 100 is the one that has something interesting? This is why so many people were fooled by it.
So they thought "ha, this is foolishly wrong, I'll write in and tell them that—of course, without doing the 5 minute math check to make sure, first!"
You don't bring up Monty Hall first. It bends minds too much.
I give a parallel game, where basics are similar, but you can choose to reveal 3 doors, 2 doors, or just one door each time. Everyone (after making sure they aren't being tricked) chooses to reveal three doors. Winner every time. Then I change the rules: Can't open all three doors. Have to choose between 1 or two. Everyone picks two.
Only then do I give the Monty Hall problem, and show (as @bhc3 put it succinctly) that it's essentially the same game they just learned: your choice is between revealing two doors, or revealing one door.
The most interesting part of this, and why I do it this way, is that I specifically optimize for my students to be able to explain what they know. I can teach someone to win an argument with someone who is convinced it's a 50/50 chance this way. Other ways, people sorta, kinda get it, but can't explain it to others who are confused.
EDIT:
The general principle here IMHO is teach something simple first, and show how it can be applied to something more complicated. When people think something is complicated, "breaking it down" into simpler concepts or making easy analogies doesn't have that great a track record. Their brain seems to have already classified it as "complex" and that has to be overcome —they have to be convinced to drop the tag #complex— in addition to however you teach the right understanding.
An intuitive explanation is that if the contestant picks a goat (2 of 3 doors) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (1 of 3 doors) the contestant will not win the car by switching. The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.
Moreover the basic argument is flawed. The knowledge of Monty has no barring on the knowledge of the contestant. If Monty simply presented you with three doors that you were to pick from, but opened an incorrect door before you choose your first door, then what would your odds be? The fact that he opens one after you choose does not matter. Monty has total knowledge of the setup and you still have zero.
The odds of initially choosing correctly are 1 in 3. In that case if you switch you will loose, but if you stay you will win. The odds of choosing wrong initially are 2 in 3. In which case if you stay you will loose, but if you switch you will win. So it is actually that simple. If you always stay you will win 1/3rd of the time and if you always switch you will win 2/3rds of the time.
It only seems like it should be 50/50 at first blush b/c that is what the odds would be if you choose to stay or switch arbitrarily. That is also why if you write down all the possibilities and simply add them up it will appear to be 50/50 too (which is what I did).
The only thing I still can't quite figure out is exactly how the exposure of one of the renaming two doors increases one's knowledge. I thought I could take the list of possibilities and eliminate those that no longer apply and then add them up, but that doesn't seem to work.
Let's make the number even bigger, so we don't even need to deal in probabilities -- a sextillion (10^21) doors. That's more doors than there are grains of sand on the planet. Say you're given a sextillion doors, and you pick one lonesome door, call it door A. What are the odds that door A is the right one? Essentially zero, right? I'm going to say straight up zero because now is not the time for you to be a dreamer.
Ok so now that you've made your futile pick, poor Monty has to go and open every other door, all sextillion of them, except for yours and one other (let's call it door B) which he leaves closed. Every door he opens contains a goat.
So now out of this vast universe of doors, there are (10^21-2) doors opened with goats bleating away, and only two doors left unopened: door A, aka yours; and door B, the one that Monty mysteriously left unopened. One of them must contain the car.
(I'm guessing at this point it will be obvious for many people that door B contains the car, but I'll elaborate further in case it isn't clear yet.)
So we're left with your door A and Monty's door B. At this point there are only two scenarios that matter. Either you picked the right door originally, in which case staying put is the winning move. OR, you picked the wrong door originally, in which case switching is the right move. And we already know you picked the wrong door (remember, no dreamers allowed here), so you should for sure switch to this by-now-extremely-conspicuous-looking door B.
Door B isn't just some random door; the fact that it was chosen as the only door to not be opened tells you a tremendous amount of information about it. Think about the game from Monty's perspective. Either you pick the correct door initially and he gets to make door B a random goat door, or you picked the wrong door and he has no choice but to make door B the door with the car behind it. Since we know you picked the wrong door, we know what Monty's gonna have to do.
> He retained the authority to offer the contestant cash NOT to switch. "I wanted to con [them] into switching there."
So he wanted them to win the car. That seems interesting. I suppose it makes sense as this was an entertainment program. Viewers like a happy ending. Or if the guest accepts the cash and stays with the lower probability door, it's okay that they get the goat because they have cash. So win-win just for appearing. Also prevents unhappy guests from saying bad things about the show.
I'm even more appalled that no one tried, you know, actually running the game. Have someone take a quarter and two pennies, mix them up under cups, let you choose one, show you a penny, then switch or not. Repeat several times. Switch places. When you're playing the role of Monty, the truth will reveal itself...
Bizarre. But I suppose this kind of thing that goes against instincts can really upset people.
The randomized variant, by the way, tends to garner roughly the same degree of "you're completely wrong" responses when given to people who know just enough about the normal Monty-hall problem not to be fooled by it.
IIRC, you get less information in this case, since there was a 1/3 chance that the host would accidentally pick the "car" door, but you still have a 1/2 chance of winning by switching.
It seems that you are referring to the other variant, in which case this is false, because that one has a 2/3 chance of winning by switching.
So, your overall odds aren't as different as it seems. Just you have a much better odds of winning by switching, if you are still in the game.
Best quote in the entire article.
And if I had to explain the problem in one sentence: Opening the first door obviously gives you a 1/3 chance, which remains unaffected by the host opening another door.
You had a 1/3 chance if you opened the second door too. Why does the second door's chance change, but the first door's chance is somehow locked by you having chosen it?
The key here is to understand that no matter what you do, the game organizer will ALWAYS open one of the two others doors.
Now, what if the organizer was to open a door only in some cases? He could open the door only when you choose the correct door. In that case, you have 0 chance of winning when you switch doors when the organizer opens another one.
What makes the 1/2-1/2 probability - that so may math PhDs defended - acceptable is this: the organizer is selecting a door at random and opening it. Then if I choose a door and the organizer opens another one, I know that I have a 50% chance of having selected the correct one. Of course, he could also have opened my door in which case the game ends.
This is a good example of how an ill-posed problem have different solutions depending on its implicit interpretation.
I actually don't agree that "the original problem clearly states it." Fully cognizent of the issues at hand, and with space for reflection entirely outside of the context of trying to solve the problem, I agree that it is clear that this is what the statement of the problem was trying to get at (particularly with the note that Monty knows which door has what) but it strictly does not say that that informs his choice (or how).
I don't have an IQ of 200+ nor do I have a PhD in math, but I'm pretty certain that the probabilities in this problem are unknown unless you know the behavior of the organizer in EVERY situation, not in one particular case.
It's not implicitly assumed, it's stated in the original question: "You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat." You pick a door, the hosts opens another door which contains a goat. He can't open the door you chose, he can't open a door with the prize, he can't not open a door. Introducing a possibility which is not a part of the specified problem turns it into a different problem.
> I have shown that by assuming a different knowledge - in this case, the organizer only opens the door that contains a goat when you choose the door that contains the car - you can end up with different probabilities.
For every "the organizer only the opens a door that contains a goat when you choose the door that contains a car", there's an "the organizer only opens a door that contains a goat when you open a door that doesn't contain a car". Not knowing the host's behavior means that it isn't part of your analysis. If you don't know why he does what he does, all you can use are the facts in front of you: there are three doors, you pick one, you are shown a second door which doesn't contain the prize, and are offered a chance to choose.
> I'm pretty certain that the probabilities in this problem are unknown unless you know the behavior of the organizer in EVERY situation, not in one particular case
Not so. If you knew the host's behavior (which the original problem does specify precisely), you can use it in your analysis. But not knowing it, means the best decision has to be independent of it. And if you knew the door that contained the prize, you could use that in your analysis. But not knowing it, means the best decision has to be made independent of it.
you have 1 black ball and a 2 white balls in 3 different urns each containing a ball. You select one urn A. Some external agent randomly overturns an urn. That ball it contains is white. What's the probability that the ball in urn A is black? The answer is 1/2. Not 1/3. It's 1/2 because you the assumption you make on the external agent's behavior is that it's random. If you don't make an assumption on its behavior, you can't calculate probabilities. If you make the assumption that the external agent picked up the white ball on purpose and will repeat that behavior when faced with similar circumstances, the probability is 2/3.
My point is that if you don't make an assumption on the host's behavior, you can't calculate probabilities.