## 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 12:28 am

This kind of thing takes G&T from awesome to mutherfuckingawesome.

September 14th, 2011 at 12:28 am

In a way, you can consider this "illusion" an artifact of the way our brains are made.

We innately guess with high accuracy problems of multivariable calculus (every day like when we figure out how much time we have to pull out in traffic for example, or when we catch something thrown in the air), and we identify patterns and apply heuristics in defiance of the best deterministic algorithms (figuring out that Google Maps' directions are not the best in the world, for example).

Our brains are made for those sorts of things. We have an innate sense of them. Our neural networks can fuzz them out with minimal effort, even if the precise numbers are more the domain of computers.

However, we didn't seem to evolve with the same knack for grasping probabilities. To a computer or some alien whose brain was naturally selected through a past of life-or-death probabilistic scenarios, the Monty Hall probably does seem intuitively correct. To us, it doesn't. I tend to think this is just because we lack the circuits to internalize the solution readily.

I think this has broader implications on how we behave as groups and as individuals in times of fear. We have trouble learning from past mistakes (i.e., applying statistical history to current decisions). We panic and overreact to the market's swings, for example, worsening the conditions. We sometimes succumb to mob mentality or Ponzi schemes or superstitions or badly placed bets as a result of our innate inability to grasp probabilities as they really are.

Suffice it to say, the Monty Hall problem has kinda baffled me for years, as well.

September 14th, 2011 at 1:02 am

Neo: Whoa…

September 14th, 2011 at 1:14 am

The trick is that Monty won't open the car door. Otherwise, we'd have to add the case where door 1 is a goat, and Monty opens the car. Then switching is a loss, and we have a 50/50 shot at the car.

The situation isn't pure 50/50 because you have more information- Monty never opens the car door.

September 14th, 2011 at 1:30 am

I always explain it with a pack of cards.

I lay out 52 cards (face down) and tell the you that you'll win a car if he picks the ace of spades.

You point to a card, and I flip over all but one of the other cards, showing you that none of them was the ace of spades. Two cards remain face down – the one you picked, and the one that I had left over after showing you fifty non-aces. Now, should you take my offer to exchange cards with me, or is it just a 50/50 chance at this point?

Well, obviously you should switch, because your original pick had a 1/52 of being the ace of spades, and the remaining card has a 51/52 chance. Now, having established that the final state isn't a 50/50 chance, it's easier to explain that the Monty Haul three-door selection is also not 50/50, and you should switch.

I've made several stubborn undergrads see the light using this example, it really works.

September 14th, 2011 at 1:57 am

When you first started, the probability that you won the car was 1/3. That doesn't change with the new information.

However, you have gathered new information about the door you *didn't * pick, and which Monty Hall didn't pick. Because Monty will never reveal the car. And Monty is only revealing information about the doors you didn't pick (because he's not going to reveal your door).

September 14th, 2011 at 2:01 am

The probability that the car is behind the door the contestant picked is 1 in 3.

So, the probability that the car is behind the two unpicked doors, taken together, is 2 in 3. When Monty opens one of those two unpicked doors to reveal a goat, the probability that the car is behind the opened door drops to 0. The probability that the car is behind the two unpicked doors, taken together, is still 2 in 3, all of which is now borne by the unopened unpicked door.

Monty's opening a door to reveal a goat doesn't change the 2 in 3 probability; it "concentrates" or "distills" that probability on to the door Monty didn't open.

September 14th, 2011 at 2:36 am

Neal and FMGuru's explanations are really good.

September 14th, 2011 at 3:17 am

I agree, lovegtm, that Neal's and FMGuru's explanations are great, but Will's note points to an important ambiguity in the initial formulation. Ed's pictures work under the assumption that Monty /always/ opens a /non-car-door/. If Monty were to always open /any/ of the remaining doors (thus sometimes showing the car), or were to open a goat-door whenever the contestant is right and half the time when the contestant is wrong, the odds would become 50/50 again.

Less clear but shorter: In the original problem, Monty has extra information that the contestant does not have, and reveals it in opening the door. In the first variation (Will's), Monty has no extra information that he can reveal. In the second variation, Monty has extra information but systematically hides it from the contestant.

Needless to say, I have spent way too much time thinking about this.

September 14th, 2011 at 6:20 am

You could test this with a simple computer program that gives you the Monte Hall choices over and over again. As long as your second choice is to switch, your success rate should approach 66% and not 50%.

September 14th, 2011 at 6:29 am

Continuing that thought, if you always stick, then your success rate should approach 33%.

And if you alternate switching and sticking, your success rate should approach 50%! That may be another contributing factor to the counterintuitiveness of the solution. However, if you look at the success rate of sticking vs. switching even in this run, they should demonstrate success rates of 33% and 66%, respectively.

Tangent: I wonder if this is related to how light demonstrates the properties of waves and particles both.

September 14th, 2011 at 7:22 am

I ran into this in "The Drunkard's Walk: How Randomness Rules Our Lives," by Leonard Mlodinow. He spends a few pages on this problem. This book can be a life-changer for those smart enough to ingest it in full (I failed).

It was Marilyn vos Savant who was asked for the solution of this conundrum in her column, "Ask Marilyn," I believe in Parade Magazine. Her answer was correct (she has, by the way, the world's highest recorded IQ), but she brought calumny down on herself from irate math professors all over the country. They embarrassed themselves.

September 14th, 2011 at 7:27 am

The odds are 2/3 that the original choice was a goat. Monty revealing an "unchosen goat" behind one of the 2 remaining doors eliminates the chance of trading one goat for the other. That means you would either be trading a car for a goat or a goat for a car. And going back to the probabilities of the original choice, it is 2-1 in favor of it being the latter.

September 14th, 2011 at 7:31 am

BTW, the most maddening optical illusion I've ever encountered is Roger Shepard's. As artists, we train ourselves to adjust to these illusions, but this still amazes me. I guess perspectival distortion is the most stubborn:

http://www.michaelbach.de/ot/sze_shepardTables/index.html

September 14th, 2011 at 8:02 am

The probability that I am now late for work for reading this post and the comments has increased from 1:2 to 9:10.

September 14th, 2011 at 8:33 am

Building on Neal's excellent answer, you can help yourself see that it is true by imagining the problem with 100 doors. You pick one; Monty opens 98 to show goats; do you switch or not? You surely don't think that somehow your odds of having picked the correct door at the outset just jumped from 1/100 to 1/2.

September 14th, 2011 at 8:38 am

Funny. This is the second time in a week that I've seen this famous problem referenced. Two days ago I started reading "The Curious Case of the Dog in the Night" and it's explained by the autistic character in the book. Really good book, btw.

September 14th, 2011 at 9:13 am

There are three options: A, B, C.

A is correct, let's say.

Scenario 1: Contestant chooses A. Host discards B. Contestant changes.

Wrong.Scenario 2: Contestant chooses A. Host drops C. Contestant changes.

Wrong.Scenario 3: Contestant chooses B. Host drops C. (host can't drop A) Contestant changes.

Right.Scenario 4: Contestant chooses C. Host drops B. (host can't drop A) Contestant changes.

Right.(Did I forget a scenario?)

In the case of a one-off, what advisory role do probabilities play? Because it looks like there are never three choices, just two opposing chances at right/wrong within one problem. (pick right first, change, wrong. Pick wrong first, change, right.)

September 14th, 2011 at 9:31 am

I have done the simulation in an Excel spreadsheet and it does indeed give you a 66.7% chance of winning over the long term.

What others have said is correct: the key assumption is that Monty KNOWS where the car is, and will NOT open the door that has the car. It clicks for some people once they take that into account.

ADM: Your scenario 1 and 2 are essentially the same thing. They each have a 16.7% chance of happening, while 3 and 4 each have a 33.3% chance of happening.

September 14th, 2011 at 9:36 am

ADM, indeed if you pick correctly first and change, you lose. And if you pick wrongly first and change, you win. Undeniably, you have a 1/3 chance of picking the right door out of the original three doors. So changing gives you a 2/3 chance of success. (2/3 chance you initially picked the incorrect door). Did you mean your example as a refutation of this?

September 14th, 2011 at 10:20 am

It's some sort of Bayesian conspiracy!

September 14th, 2011 at 10:35 am

My professor gave us this problem in my very first stats course in grad school. I believe he used in for two purposes. One was to discuss the difference between conditional and unconditional probabilities. The second was to grind it into our heads that statistics is not intuitive in any way and that we needed to rely on math and computers to do it properly.

Now, after 5 or 6 grad level stats course in which I have been pummeled with formula after formula, I now realize that he was right…and that I suck at math and computers. I should have done qualitative methods. Too late now!

Ironically, I'm now teaching undergrad statistics. Fortunately, it's so preposterously basic that even someone who sucks at it as much as I do can probably pull it off.

September 14th, 2011 at 10:40 am

The Mensa savant in Parade magazine took weeks of abuse from readers for this problem and its correct analysis. Comments were amazingly hostile. She had an interesting argument to explain the solution, asking what we would do if there were 100 doors, and having chosen 1, Monte then reveals 98 goats behind doors and then leaves you to choose between your present choice or the 1 unrevealed door.

I find the key issue to be that Monte NEVER reveals a car when he chooses to reveal a goat — and that is because Monte KNOWS THINGS, such as where the car is.

Duplicate bridge players use a rule called 'restricted choice' that is rooted in this same problem.

September 14th, 2011 at 10:48 am

My favorite part of this 'problem' is that the only way you can lose is to pick the car originally, because of course you will switch away from the car when given a choice because MATH(TM) proves that switching is best.

There's something profoundly wrong with giving someone a 1/3 chance at a car and then giving them a 100% chance of losing when they guess "right". It starts getting especially painful when Monty doesn't have to show a goat but can choose to show a goat — if he shows a goat you still switch, right? But what if he shows a goat 100% of the time when you guess the car initially and only 50% of the time when you guessed a goat initially? Or 100%/66%? Or 66%/66%

Stupid math — I think I'd try to win a goat — better odds and all.

September 14th, 2011 at 10:49 am

Ah, an oldie but a goodie. 2/3s is the correct answer, with the given assumptions.

But, in terms of choice strategy, not entirely the entire picture. The probabilities change if one of the assumptions change.

Big huge fucking hint: Monty Hall knows what is behind each door.

September 14th, 2011 at 10:51 am

Strangely, I never had a problem with this Problem. It only took me a minute or two to realize that if you always switch, the only way you can lose is if you initially pick the car (a 1/3 chance). Otherwise, Monty will reveal the car door by opening the remaining goat's. I'm more baffled by people who don't get it – like those math professors – but to each their own. I'm certainly no genius; there are many things in this world that still confound me (like the existence of the Republican party).

September 14th, 2011 at 10:53 am

Ah, I see Robert Trenary is already on to it. Fine. See "The Art of Probability" by Richard Hamming.

Short answer: Monty doesn't know. His door choice is random. Draw out a decision tree.

September 14th, 2011 at 10:59 am

Here's a cute site with an animated game of the problem to play and win some cars: http://www.stayorswitch.com/

September 14th, 2011 at 11:03 am

Wonder how this plays out when applied to Republican Presidential candidates..

September 14th, 2011 at 11:08 am

@Joseph Nobles There's a site up that keeps track of percentages and it does come out to 66% and 33% depending on if you switch or stay. http://www.nytimes.com/2008/04/08/science/08monty.html

September 14th, 2011 at 11:14 am

All good answers. When I teach stats, I use 1000 doors and say "Assume you pick Door 1 and Monty opens every other door except Door 724. Do you really think it's a 50-50 chance that the prize is behind Door 724, or is it more like 999/1000?" Using specific numbers helps them visualize the situation.

Another explanation with three doors is to assume at the outset that the prize is behind either Door 2 or Door 3. This assumption will be right 2/3 of the time. But if it is right, you have a sure-fire strategy for finding the prize: namely, choose Door 1 and switch.

And I strenuously object to the notion that probability is counter-intuitive. Most of prob and stats is actually intuitive, but bad teachers make it otherwise.

September 14th, 2011 at 11:18 am

The key assumption in this problem, which is usually NOT stated (although it is here), is that Monty always opens a door. On the actual show, he didn't always open a door. If Monty doesn't always open a door, then it's not at all clear that switching improves your odds.

On the other hand, if you have literally no information on Monty's door-opening habits besides the fact that he just opened a door for you, the implied rate is 75%. That's comfortably above the 66.7% rate that makes switching worthwhile no matter what procedure Monty uses to decide whether to open a door.

September 14th, 2011 at 11:22 am

The truly weird thing about this is that I was just thinking about that problem yesterday, and then an entry shows up on this blog about it. This is not the first time something like this has happened. Very suspicious if you ask me. Get out of my head, Gin and Tacos!!!

September 14th, 2011 at 11:28 am

It's the same logic used in card-counting in blackjack. Once a card is played, the probablity of *that* card coming up (in single-deck) is zero.

September 14th, 2011 at 11:34 am

Sorry to take this a bit off topic, but the optical illusion that anotherbozo posted gives me an excuse to post my favorite:

http://web.mit.edu/persci/people/adelson/checkershadow_illusion.html

I've yet to show it to anyone that doesn't say that it's BS. Then you print it and cut the two squares out and they go slack-jawed. It's great to illustrate how the brain isn't a neutral collector of data, but an active (and sometimes inaccurate) interpreter of it.

September 14th, 2011 at 11:56 am

The reason the probability isn't 50/50 is because the system, while appearing random is not. The non-random element is introduced because the HOST knows where the prize is and will never open that door. The system may appear random to the contestant, but because an informed agent has meddled with the system, it introduces information that can be used to improve the odds.

September 14th, 2011 at 12:01 pm

I think the key to understanding this problem is "independent probability." There are two problems:

Problem 1.) Three doors – choose one – probability 1/3

Because of the way Problem 2 is presented we are deceived into thinking that it is an independent probability question just like problem 1 and the probability is 1/2. Naaaaaaaah

The Goats and Tawny (G&T) boxes shows the dependence of problem 2 on problem 1 and why it is better to switch than fight.

//bb

September 14th, 2011 at 12:41 pm

If you liked this post, I recommend this site, which is a goldmine for these:

http://mindyourdecisions.com/

Maybe you know about it. It's one of the few besides G&T that I check every morning.

September 14th, 2011 at 12:51 pm

To bring this crassly to the world of brain-dead game shows, wouldn't this imply that when offered to swap briefcases if you reach the last two in Deal or No Deal, you should always make the switch? From the wiki (the last sentence is telling):

"Should the player refuse the final Banker's offer, with the selected case and one other case left in play, the player is given the opportunity to swap cases, values unseen, and win whatever case they end up keeping at that point. This offer, and its mention when delivering the final offer, is often unaired on television; only three contestants have ever swapped cases, and none in active play. All of them would swap for one of the two highest amounts."

September 14th, 2011 at 1:32 pm

Has anyone ever sat down and watched every episode of Let's Make A Deal and verified this empirically?

September 14th, 2011 at 1:34 pm

As far as I can tell the only reason that this "problem" has ever been controversial at all is because people who share it insist on proving it through the most complicated means available. What confuses people isn't the original problem, but the 99 goats and decks of cards and flowcharts of every possible solution.

Stick with the simplest explanation: You choose a door, and without opening any doors at all, Monty gives you the option of sticking with the door you chose, or picking BOTH of the other doors. It’s the same thing.

September 14th, 2011 at 1:34 pm

While 2/3 is the correct probability for the problem, what I don't understand is why it is called a Monty Hall problem, since the situation described does not match the way things work on Let's Make a Deal, thus making any inferences to most situations problematic. The hosts behavior is rarely as constrained as Ed makes it in the problem, which is far too representative of bad economics

September 14th, 2011 at 2:17 pm

I agree with you Bill, but I don't think it's the fault of the person posing the question. The game-show setting is obviously just a framework for the math, but for some reason people very quickly jump to psychoanalyzing Monty, or trying to apply the rules of the show to the problem. I doubt that you’d get this kind of violent resistance if it were presented as a simple probability question, but it mystifies me that people always seem inclined to make the problem more complicated than it is.

September 14th, 2011 at 2:28 pm

I never knew that Monty would never open the big prize door. Makes a difference, then, as someone said up above.

But I'd be happy with a goat. I don't think that they depreciate after you take them off the lot or that you have to pay prize tax on the thing. Win-win.

September 14th, 2011 at 3:29 pm

Look, it actually is simple. The problem is, like with Republicans, people make it simplistic when it is not.

This problem, as phrased, is a conditional. GIVEN that the contestant has chosen a door, and GIVEN that Monty knows where the car is, what is the probability of getting the car if you switch? Duh. 2/3rds. It really is simple.

What if Monty don't know? What if Monty knows and wants to fuck up the premise of the show, which is… suspense! Then he immediately shows you the car: "Oh, you fucked up! Ha ha ha. Give 'em a hand, ladies and gents".

Or… Monty doesn't know, in which case, his choice is random (50/50).

Or… Monty knows, and he wants you to switch or make the deal or "play it safe" and take what's in the box! So, he's going to show you the goat. Oh, piss in your pants! What should you do??!!

The beautiful thing here is, this is a classic Bayesian heuristic, the exact same type of thing that lets Google operate robot cars without drivers. I could go on, but what would be the point?

The general solution is (from Hamming's "Art of Probability"):

Big P = Probability of staying (not switching, and getting the car) is: 1/(1+2p)

where little p is action state of Monty (this can be either knowledge state or choice of action), with the values of p=0, 1/2, or 1. Plug 'em in. Figure it out.

September 14th, 2011 at 3:40 pm

Silas — Again, I think the issue is that the problem is often presented without the statement that Monty always opens a door. Without that statement, it's genuinely not a straight math problem.

FYI, I remember reading an article in TV Guide as a kid that talked about the "gonk prizes" on the 80s Let's Make a Deal revival. It turned out that the contestants did have the rights to the animals presented, and that they were often more valuable than the actual prize. The producers would offer hundreds of dollars to "losing" contestants to get the steer or whatever back.

September 14th, 2011 at 4:03 pm

There is now a 2/3 probability that I now understand this problem.

But still only a 1/3 probability that I will still understand it tomorrow.

Cheers!

JzB

September 14th, 2011 at 4:19 pm

Chris beat me to it – extend the problem to n doors where n is a large number. Then the solution becomes easier to see.

September 14th, 2011 at 4:50 pm

But what if you want the goat, not the car?

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.

September 14th, 2011 at 4:50 pm

Also, I was told there would not be math.

(someone had to say it …)

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.