I find it funny in these cases that people think it’s a choice to “believe” or not - as if this is something like believing in God or not. In one case, we have data, mathematical proof of the answer - how can you not believe this. It is not an issue of belief - it is an issue of whether you accept mathematics or not, and if you don’t accept the math… there’s a big problem. So what’s next Jeff? The airplane on a treadmill? Lol 
captcha: Mighty erect - WTF??!
@BA’s comment to @DavidW
If you flip 10 million coins and randomly distribute them, you can saw with almost certainty the sets of two results are evenly distributed. And, beyond that, all the coins in sets of 1, 3, or more don’t actually have any bearing on the sets of 2. In fact, all the other sets of two don’t have a bearing on the single set you’re looking at. It’s an isolated case. Which is explicitly stated in the problem.
What you’re neglecting to notice is that it is, essentially, the same as the Monty Hall problem. More specifically, it’s the issue associated with switching. Girls represent the door you chose, and Boys represent door left by the Host. If you wanted your guess to be ‘right’ (that is, guess the gender, or win the car), you’d swap to the opposite gender (door) that you’ve chosen. Well, 67% of the time, anyway.
@Allen: “Just like a bunch of programmers to go back to basic counting and complicated simulations rather than use an existing, well tested algorithm.”
No… because your the programmers here are arguing about how to apply your “well tested algorithm” (or else they don’t understand it). In this case it is easier to just flip a coin or count. 60 seconds later you see the error of your ways. It is the people who want to try and prove this through mathematical means and refuse to just test out their own hypothesis that cause this to go on forever.
Create a hypothesis… THEN TEST IT! Don’t sit here and tell us over and over again how something is supposed to work when you haven’t even taken the time to make sure it works. 
on this matter i guess i can consider myself to be lucky
Okay, for people who are still having trouble grasping the concept of ‘switching changes your probabilities’, try looking at it this way (which came to me while writing a simulator for the Monty Hall problem):
Don’t think like a contestant. Think like Monty.
Initially, there are two possibilities:
The sum of these probabilities of these two events is 100%. I think everybody can agree that the chance of #1 is 1/3, and the chance of #2 is 2/3.
Monty, then has to choose a door that has the following attributes:
A. The door is not the door you chose.
B. The door is not the door with the car.
Requirement B is what throws people off: they think that Monty is revealing a random door. The problem is – 2/3 of the time, he’s NOT.
In case #1, A and B are the same door, so Monty has two doors to choose from. But no matter which one he picks, if you switch, you are guaranteed to lose.
In case #2, however, you’ve cornered Monty. Requirements A and B each rule out one door. With only three doors, there’s only ONE door left for him to show. You picked the first losing door, and he picked the second losing door. Therefore, in this case, you are guaranteed to win if you switch to the third door.
Therefore, there are exactly two possible ways to win:
Once again, the chance of picking the winning door is 1/3. This means that 1/3 of the time, switching will lose.
Meanwhile, the chance of picking the losing door is 2/3 – so 2/3 of the time, switching will WIN.
Therefore, if you switch 100% of the time, you will win 2/3 of the time. If you never switch, you will win only 1/3 of the time.
@Jasmine:
There’s a serious problem in American culture where “what feels right” is believed over mathematical proof or scientific evidence. See any number of debates, from vaccines causing autism to evolution vs. intelligent design.
In the face of all of the evidence, some people replying to this post still believe that the odds are 50/50. Unless they’re trolling.
“Monty Fall is the same as Monty Hall once the chance of accidentally revealing the car is gone.”
“Intent” was bad choice of words on my part. Since Monty is not a random event generator, but has actual decision-making powers, his actions enhance instead of simply restrict the problem scope.
Lay out the possibilities end to end. It’s not hard, there are only nine of them. Here, I’ll start it for you:
1 0 0
0 1 0
0 0 1
That’s three possibilities, where ‘1’ is the door with the car, and the '0’s are the doors with the goats. Assume that you always pick the first one.
That’s three possibilities from your perspective. Now, cross-tab that with the three possibilities from Monty’s accident perspective, and you have nine scenarios at the end. For the sake of discussion, daw this out as columns going across, where the first column is where Monty hits the first door, the second is where he hits the second door, and the third where he hits the third door. Because all three variables are independent, this is a valid and complete depiction of the problem space (with one of the three variables fixed; technically you could multiply the space by three, with identical setups where you chose the second door each time and then the third door each time but, again, being independent variables fixing one is legal).
100 – 100 – 100
010 – 010 – 010
001 – 001 – 001
Now, we know because the problem says it is so that the end state is such that Monty did not reveal your door nor the door with the car. That means that the entire first column gets eliminated from consideration, as well as the possibility at (2,2) and (3,3) (in both cases Monty would be revealing the car.
X00 – 1-0 – 10-
X10 – 0X0 – 01-
X01 – 0-1 – 00X
This leaves four outcomes:
50% of the time sticking your guns wins; 50% of the time switching wins. Thus, because the problem space has been reduced, we have a straightforward probabilities calculation.
Another way of thinking about this is this:
In the first starting condition, there’s a 33% chance that Monty will mess up the whole scenario. In each of the other two starting conditions, there’s a 66% chance that Monty will mess up the whole scenario (either by revealing the door you have already chosen or by revealing the car behind the other doors). This is the crux of it: since Monty has no guidance here, he is more likely to eliminate LOSING scenarios than winning scenarios.
Now, if Monty has INTENT, he is not just randomly reducing possibilities. He will always have two to choose from, and so what we need to look at is the pre-choice solution graph, not the post-choice one (because his choice is not an independent variable). Here, instead of a crosstab matrix, you have a tree.
The pre-choice graph looks, again, like:
100
010
001
The post-choice outcomes map from those starting points as:
100 – 1-0 OR 10-
010 – 01-
001 – 0-1
In the first of the scenarios, sticking to your guns wins (no matter which Monty chooses to reveal). In two of the scenarios, switching to the other closed door wins.
Again, Monty isn’t a random dummy here, and has exactly 0% chance of eliminating you in any scenario. Thus, the chance of your original choice selection being “right” remains at 33%, while the new information makes it twice as likely that the non-revealed other door is the right one.
Does that make more sense to you?
The fact that Monty can choose either of two outcomes in the first case does not make that case twice as likely to happen. Imagine giving Monty additional choices like opening the door with his left hand or right hand but only when you’ve picked the wrong door to start; these also would not make those possible outcomes any more likely.
Again, really: these problems are well-known and documented. If you think everyone has the Monty Hall or Monty Fall problems wrong, and you’re not a professional statistician, it’s like telling the world you know quantum mechanics better than Stephen Hawking. That’s the realm of the crank.
Oh no! It’s sadnessbowl! At least this time we agree.
What’s this about vaccines causing autism? I heard that it was watching television at a young age (with some research to back up the correlation, although it may not be causation).
I read your post in my blog reader, and this description of the problem is very misleading.
“”“It appears to be a pretty silly question. Two doors are available – open one and you win; open the other and you lose – so it seems self-evident that whether you change your choice or not, your chances of winning are 50/50. What could be simpler? The thing is, Marilyn said in her column that it is better to switch.”""
When I see the problem, it is the classical Monty Hall three doors, and you do a switch. Your description makes it sound like you have two doors to choose from, you choose, and then without any additional information, you switch your choice.
Like I said, your quick description is IMHo a very misleading description of the scenario.
BTW, this has made it into popular culture, being used in the movie “21” as a class exercise.
Since my previous explanation was very similar to a previous one. Here’s another attempt with the million door extreme example. Your first choice has a one in a million chance of being correct, so it is very more likely the winning/correct choice is not the room you picked. Now the host reveals all the incorrect choices until you have your original choice and the only remaining room that you didn’t choose. If confronted with the these two rooms without the previous knowledge then you would have a 50/50 choice. Since you know your guess was “really” was a guess/“random” choice you now know the remaining room has the odds of being correct of 999999 times out of 1000000 tries, because the host told you what 999998 of the choices were wrong!
A widely read blogger has run out of useful and relevant material (a long time ago) but he must maintain some readership so that his ad revenues keep flowing. He looks into his bag of tricks to select topics that generate lots of traffic.
What is the probability that he will choose one of the following topics:
Whenever I am reminded of the Monty Hall dilemma I always wonder (and can never find out) if Monty and the producers of Let’s Make A Deal were aware that they were giving the contestants a 2/3 chance to win the car. And if they were aware, who was the brain on that show who understood the counter-intuitive odds that stumps (apparently) 92% of the population. Is anyone aware of an interview with Monty where he is asked about the math puzzle named for him?
Paul Erdos was brilliant, but even he realized his own limits when presented with the highly unintuitive Monty Hall problem. For his epitaph, he suggested, in his native Hungarian, “Végre nem butulok tovább”. This translates into English as “I’ve finally stopped getting dumber.”
If only the rest of us could be so lucky.
So you are suggesting we would be lucky to be dead?
There’s another way to think of this problem that may make it easier to understand.
Imagine that there are 3 doors to pick from and you initially pick 1 of them like in the original problem.
But then BEFORE Monty Hall exposes which of the 2 doors you didn’t pick has the goat, you get to decide whether to switch or not. You can have what’s behind 2 of the doors or just the 1 you initially picked. (BTW, you don’t have to take home any of the goats if you don’t want to).
You don’t have to be a math genius to know that your odds are better if you get to pick 2 doors as opposed to 1.
"Or perhaps you’re objecting to how unintuitive it is that the answer depends on what seem to be absolutely trivial pieces of information about how the information is obtained?"
Agreeing with Jon Skeet here. I think you are being a bit condescending to the smart people who simply disagree with you here. They do not find the math problem unintuitive but your wording made it unclear and ambiguous. I was surprised that you never made a correction/update to that post.
Thanks for this explanation!
There are two DIFFERENT questions being argued.
1/ If you have a unbiased choice of two unopened doors, you have a 50% chance of winning the car. (The instinctive probability…)
BUT
2/ Monty WON’T open the door with the car! Therefore Monty is giving you better odds. Your odds are improved if you CHANGE your choice (as you demonstrate) to a 66% chance of winning.
(Still can’t work out the baby thing though…)
Guy
Probability is much easier than all this. See The Daily Show’s coverage of how most things are a 50/50 shot. (http://www.thedailyshow.com/video/index.jhtml?videoId=225921&title=large-hadron-collider)
Marilyn is wrong, and her logic is flawed.
Remember, Monty is ALWAYS going to choose a door with a goat! When Marilyn says that 2 out of three choices result in winning if you switch, she’s assuming that one of the choices is the door that Monty opened…which is a goat.
In other words, you can’t include the option where Monty opens door #2 and a car is there…because that can never happen! If you eliminate that option, the odds are 1:2…exactly what you would intuitively expect.
It’s disturbing that eggheads at MIT wouldn’t see the flaw in Marilyn’s logic.
@BA:
“By concluding that all combinations of of boy and girl are equally likely you are making a statistical fallacy in assuming that a randomly selected subset has the same properties as the superset.”
Umm, no. By concluding that the PROBABILITY of an outcome in a small sample is equal to the outcome in a large sample you are assuming that there is no small-sample bias. This is not a statistical fallacy, but instead the underlying assumption in most statistic-based calculations.
The question was not, at all, “What is the gender of the other child?”. It was: what is the probability that the other child is a boy.
To run into your issue, we’d have had to state that there are four families each with two kids. Family 1 has two girls; family 2 has a boy and a girl; family 3 has two boys; family 4 has what? And then answer "a boy and a girl"with no margin of error. THAT would be a subset-superset fallacy, or more precisely, a failure to keep account of statistical error.
More interestingly, though, you do have a point on family size. Consider the social impacts of a male-dominant society crossed with a small-family-size inclination. If the main goal of having children was to have a boy to “carry on the family name”, a lot of families would stop after their first boy was born. Thus, a large portion (we’ll say all just to deal in round numbers) of the families with 2 children had a girl first, which means that the likelihood of having two girls is 50%. This type of information completely changes the problem space by reducing the possible outcomes.