## THERE IS NO SPOON

This isn't really political, but I did this in class on Tuesday and I had too much fun watching everyone tie their brains in knots not to share it here.

We are beginning to talk about probability in a course on research design / methodology, and I introduced the topic with a few basic examples. One of them is the infamous Monty Hall Problem. The MHP is like an optical illusion – no matter how many times I tell you that these lines are of identical length, your brain will keep telling you that the bottom line is longer. It is the kind of paradox with a solution that appears to be absurd even after it has been demonstrated to be correct. You simply cannot wrap your mind around it.

The MHP is a probability game named after the host of the 1970s/early 1980s game show *Let's Make a Deal*. The most popular part of the show was a game that has since become the subject of dozens of academic papers in math, statistics, and logic. Here is the basic setup:

The host confronts the contestant with three doors. Behind two doors there are goats (which the contestant did not actually win, but were intended as a gag "prize") while the third door hides a brand new car. The three prizes are distrubuted behind the doors randomly and Monty Hall knows what is behind each door. The contestant chooses a door, which remains closed. Hall must then open one of the other two doors to reveal a goat. He cannot open the door hiding the car (if the contestant has not chosen it). He then asks the contestant if they would like to stick with their original choice or switch to the other available mystery door.

An example is clearer; You choose Door 1. Hall opens Door 2 to reveal a goat. He then offers you the option of sticking with Door 1 or changing your choice to Door 3.

To 99.9% of humanity, this problem has an easy solution. Since Hall will open one of the two goat doors, the two unopened doors contain one goat and one car. There is no advantage to switching, since the odds of the chosen door containing the car appear to be 50-50. Going back to the example, if Door 2 is opened to reveal a goat, there's a 50% chance that Door 1 contains the car and a 50% chance that Door 3 contains it, right?

Switching, in fact, is always in the player's interest. When *Parade* magazine ran this problem in 1990 they received more than 10,000 critical letters, including several hundred from people with PhDs in math and science. All patiently explained that the odds are 50-50 and switching is not beneficial. The well educated letter writers were all wrong.

The probability of winning by staying with the original choice is 1/3, not 1/2. And the probability of winning by switching is 2/3. Most people, no matter how long they stare at this and work through the scenarios, cannot accept that. It took me days to absorb it when I first encountered the problem. It just does not make sense. Yet it's true. Since there are only two different objects behind the three doors – two goats and one car – there are only three possible combinations: car-goat-goat, goat-car-goat, and goat-goat-car. Perhaps the only way to understand why this creates a 2/3 probability of success by switching is to see the three scenarios spelled out. The yellow represents closed doors, the white door is the one opened by Hall to reveal a goat, and the arrow represents the door originally chosen by the player (which is always Door 1, for simplicity's sake).

And now that I've explained it and you've seen the irrefutable evidence, I bet your brain *still* doesn't comprehend *why* switching increases your odds. Mine doesn't. I know what the solution is but it will take me a few more years to understand why.

To paraphrase 50 Cent, I love this problem like a fat kid loves cake.

September 14th, 2011 at 4:52 pm

On the other hand, if through bribery or whatever, you know in advance that the car is behind either Door 1 or Door 2 but not behind Door 3 . . . then Monte's disclosure of Door 3 doesn't tell you a thing and doesn't change your odds.

Lesson: Information changes probability, see, e.g., five card stud. What's amazing about that?

September 14th, 2011 at 4:54 pm

I usually help people with this one by walking through it thus:

"Here's three doors. Pick one. Now….what are your chances of having picked the correct door?"

Unless they're having a really bad day, they answer "one in three".

"Right. So there's a two in three chance it's one of the others. Right?"

Yup.

"So if I were to give you your choice of either your one door (1/3) or *both* of the other doors (2/3), would you switch?"

Again, unless they're having a really bad day, they answer "sure".

"Well you can have both of the other doors if you like – but I'm just going to open one of them before you make the decision".

Usually that helps it "click".

September 14th, 2011 at 5:19 pm

"See, that's what I always hated about this problem. I really want the goat, not the car. No one ever tells me what I need to do to get a goddamn goat."

I didn't know Mickey Kaus commented here.

September 14th, 2011 at 5:34 pm

There's a variation of this problem that we give to genetics students.

A couple has two children.

One is a boy.

What are the odds that the other child is a girl?

The answer is 2/3.

September 14th, 2011 at 6:10 pm

SoBe –

You got my goat, if that helps.

Bahhh!

JzB

September 14th, 2011 at 8:16 pm

"A couple has two children.

One is a boy.

What are the odds that the other child is a girl?

The answer is 2/3."

You ARE joking, right?

If not you fail conditional probability.

September 14th, 2011 at 8:19 pm

nevermind, Blieker's right, I read the question wrong.

September 14th, 2011 at 8:25 pm

"You ARE joking, right?

If not you fail conditional probability."

No (assuming 50% chance of either a boy or a girl). If a couple has two children the possible combinations are:

BB, BG, GB or GG. There's a 25% chance of each combination.

If one child is a boy, that eliminates the GG combination leaving BB, BG and GB.

Of the remaining combinations 2 out of three have a girl.

When I took the GRE there was similar problem involving drawing marbles from a bag. The genetics students complain that it's a trick question, simply based on how the problem is worded, but it's a common enough problem that it pops up in a lot of testing situations.

September 14th, 2011 at 8:26 pm

"nevermind, Blieker's right, I read the question wrong."

OK. You responded while I was responding. You do see why the students complain, it's how you read it.

September 15th, 2011 at 12:53 am

I've tried the ace of spades/deck of cards explanation with several people and they still think they have a 50/50 chance of holding the ace of spades. I think I'm going to start offering them wagers to supplement my income–perhaps losing money will help them come to a more sophisticated understanding of probability

September 15th, 2011 at 6:04 am

"A couple has two children.

One is a boy.

What are the odds that the other child is a girl?"

No answer can be given unless it is explained how you know that one is a boy.

In most real-life situations, the answer is 1/2.

September 15th, 2011 at 10:34 am

"No answer can be given unless it is explained how you know that one is a boy.

In most real-life situations, the answer is 1/2."

The fact that one is a boy is given in the question, how one knows that is irrelevant. The answer is not 1/2. If the question were worded differently, say a couple has a boy and are having a second child what are the odds that it will be a girl…then the answer is 1/2. But, that's not how the question is worded.

September 15th, 2011 at 10:56 am

The easiest way to think about this, to me at least is this:

You said correctly that the odds of choosing the car on your first choice is 1/3. That means that there is a 2/3 chance that the car is in one of the other two doors. Game show guys opens one of those two, which is never the car. That leaves one door left which has a 2/3 chance of being the car.

September 15th, 2011 at 3:25 pm

The first choice is a false choice. It doesn't matter, because the contestant can't lose. All it allows you to do is play funny math (because the contestant's chosen door can't be opened after the first guess).

The only choice that affects the outcome is the second choice, between two doors. 50-50

I never watched the show and have no idea what the outcomes were, but I have no doubt that if the data were compiled you would find the "switchers" and "non-switchers" to have shown an identical, statistically significant, 50% chance of winning.

September 15th, 2011 at 9:13 pm

The best thing that I appreciate is that you struggle with it as well. That makes you a good teacher fo' sho.' If the teacher still struggles with it, then the kids become that much more fascinated and their knoggins really start to kick into overdrive. I hope you expressed that in class, what you expressed in this essay.

September 16th, 2011 at 7:18 am

"The fact that one is a boy is given in the question, how one knows that is irrelevant. The answer is not 1/2."

Wrong.

The sentence "One is a boy" can be interpreted in more than one correct way, and the most natural interpretations wreck the punchline.

If you know that some guy has two children, and the first one you see getting out of the car is a boy, the chance the other is a girl is 1/2.

September 16th, 2011 at 11:00 am

"If you know that some guy has two children, and the first one you see getting out of the car is a boy, the chance the other is a girl is 1/2."

Wrong.

September 16th, 2011 at 11:43 am

Wrong.

Work it out very carefully.

September 16th, 2011 at 2:56 pm

"Work it out very carefully."

It's basically the same problem and solution I gave above. As soon as you state that a "guy has two children" you set up a certain probability. That is the child combinations are either: BB, BG, GB, or GG. Each of those has a 25% chance of occurring. As soon as you know one child is a boy, that eliminates the GG combination leaving just the BB, BG and GB, each having a probability of 1/3. Of those 2 have girls, giving you a 2/3 chance that the second child is a girl.

September 16th, 2011 at 6:00 pm

The answer always seemed pretty simple to me, once explained. The odds are that you picked wrong the first time (2 to 3 that you picked a goat). Monty gives you a shot to switch your vote, AND takes one bad option off the table.

THE POWER OF PRIOR INFORMATION

September 17th, 2011 at 12:36 am

You have BB, BG, GB, or GG, with equal probability.

Then, either the elder child gets out of the car first, or the younger, with equal probability.

(B)B, (B)G, (G)B, (G)G

B(B), B(G), G(B), G(G)

Cross out the possibilities in which a girl gets out first.

(B)B, (B)G

B(B), G(B)

Tada! A real-world condition is imposed and the paradox has disappeared.

September 17th, 2011 at 9:39 pm

But you'll notice that only THREE of the original possibilities are present, BB, BG, and GB. Just because you've counted one of them twice doesn't change anything. There's still a two-thirds chance that the other is a girl.

You may be thinking of this: if I say a couple has two kids and the OLDEST is a boy, then you're right, it's 50-50 what the other's gender is. That's because the only possibilities are BB and BG but not GB or GG.

I'm not sure what you mean by "the paradox has disappeared" because there is no paradox; there's just a careful, right way of analyzing it, and a less-careful, immediate, but ultimately wrong way.

September 18th, 2011 at 1:56 am

I'm with asw on the "two children" question — the odds are 50%.

The first condition is simply: "A couple has two children." Since there's nothing in this condition or the next one ("One is a boy") or the question ("What are the odds that the OTHER is a girl?") about the order of birth being a factor, the BG and GB options are identical.

So after the first condition is imposed, there are three possibilities: that they have two boys, two girls or one of each.

But adding the second condition eliminates "two girls," so you have two remaining possibilities: "two boys" (other child is a boy) or "one of each" (other child is a girl).

September 18th, 2011 at 2:37 am

…Nope, never mind, I take it back; bliekker's right. His statement that this was a question posed to genetics students flagged the problem with my argument:

Of those three possibilities for two kids — two boys, two girls and one of each — the first two each have a 25% chance, and the last one is twice as likely, for the same reason that flipping a coin TWICE gives only a 25% chance that it's heads both times, 25% for tails both times, and 50% for mixed result (possibilities: HH, HT, TH, TT).

So once "two girls" is scratched as a possibility, "two boys"= 1-in-3 chance; "one of each," 2-in-3 chance. My bad.

September 18th, 2011 at 3:24 am

"But you'll notice that only THREE of the original possibilities are present, BB, BG, and GB."

But that's irrelevant. Each of (B)B, (B)G, B(B), G(B) has probability 1/8.

September 18th, 2011 at 2:04 pm

Putting parentheses around some of the letters doesn't make them a different case. There's only one BB. You can survey all the families in your town if you want. For all that have 2 kids and either 1 is a boy, 2/3 will have the other as a girl Give it a try.

September 18th, 2011 at 7:20 pm

I Love this problem too. Thanks Gin or Tacos, whichever one of you two came up with it.

It took me about a day to finally see how the odds are 2/3 when given the choice to switch. Until then, I kept thinking it was 50-50 when the choice is given. Since it took me so long to get it (even after reading a bunch of other explanations) I came up with a new explanation that would have cleared it up for me sooner.

The odds are SET IN STONE at the moment the contestant chooses the one door to start the game. The other two doors have a 66% chance of containing the car. It doesn't matter if Bob Barker throws some personal knowledge by displaying a wrong door. He has inside info. There's no change to the odds for that. Now, if someone walks in off the street after half the game has been played, and knows nothing about the choice of the contestant and is told there is a car behind one of the two remaining closed doors, the odds for that person are 50-50. The original 1st contestant did not know which one of the doors did not have a car behind it at the time of his/her choosing.

Using a deck of cards: it's as if you get one choice to pick the Ace of Spades. Without looking at the card yet, if Bob asks you if you want to switch to his 51 to see if that set contains the Ace, say yes! It matters not a bit if he shows you any cards or not. Once you put your hands on a choice the other LARGER set has a way better chance of containing the right answer.

September 19th, 2011 at 1:17 am

"Putting parentheses around some of the letters doesn't make them a different case."

Yes it does. The parentheses show who got out first. If you don't believe me, simulate it. Choose one of {GG,GB,BG,BB}, then choose either the first or the second (getting out of the car). If you chose B, record whether the other is B or G.

There is a fundamental difference between "I met one of the two children and it was a boy" and "It is not true that both children are girls". The first implies the second, but they are not equivalent. This is the whole point of the puzzle! The infuriating thing is that so many of the well-educated people who redistribute the puzzle don't really understand it, thus further confusing everybody else.

September 19th, 2011 at 4:11 pm

asw…again it comes down to how the question is worded. I see the point you're trying to make, though posed as I originally put it 2/3 is the right answer. For those interested, this problem is discussed at:

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

September 20th, 2011 at 7:08 pm

How would the MHP apply to Deal Or No Deal? You start with 26 cases, take one, and reveal the 24 cases. What is the probability that if you switch cases you get the case with the higher denomination of the two that are left?

April 9th, 2012 at 6:29 am

FMguru is right. I've used a deck of cards to convert a math grad student. The New York time did ask Monty his opinion in 1991, and he gave a more detail response according to how he played it on his TV show.

http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doors-puzzle-debate-and-answer.html

The stayorswitch.com is the best attempt to explain the concept that I have seen yet.

October 18th, 2012 at 10:42 pm

No.

The numbers can be tweak back to expectations by changing the focus…

Your error has to do with which door he opens.. Look at the white squares.. The missing fourth possibility is that you selected car and he opens door 3… In which case, switching loses the car.. And the odds remain 50/50

Redraw your pictures, but instead use (goat, car, furniture).. Cause let's be honest.. Sometimes it was just a crappy hutch.

When you are forced to separate the two goats into the real possibilities.. And show ALL possibilities….. It goes back to a 50/50 switch.

If you don't believe that.. Make all three different names and all possibilities.. Including where you select door 2 and the car is in 1 or 3….

When you do Every possibility.. It's back to understandable

December 4th, 2012 at 6:52 pm

But Monty isn't allowed to pick the car door…. Am I the only one thinking this is retarded?

June 10th, 2013 at 4:56 pm

Here's my idea on why this is so hard to "get":

When I pick the first door, I am picking one out of three, so my chances of picking the correct door are one in three. Once Monty opens one of the remaining doors, I have to "pick" a door again (stay or change). It now "feels" like a new situation, where I only have two choices, and therefore have a 50/50 chance of picking the correct one. And therefore my brain says I have as much chance if I stay as if I change.

November 5th, 2013 at 8:47 pm

Much easier to illustrate when you increase the numbers. 100 doors 99 goats 1 car. Host opens 98 doors revealing 98 goats. 2 doors left: 1 car 1 goat. When you chose initially you had a 1/100 chance of picking the car. Most likely you have a goat behind your door. Switch doors.