# Monty Hall, Monty Fall, Monty Crawl

Remember The Problem of the Unfinished Game? And the almost 2,500 comments those two posts generated? I know, I like to pretend it didn't happen, either. Some objected to the way I asked the question, but it was a simple question asked in simple language. I think what they're really objecting to is how unintuitive the answer is.

This is a companion discussion topic for the original blog entry at: http://www.codinghorror.com/blog/2009/06/monty-hall-monty-fall-monty-crawl.html

Hah. Jeff just wants to boost his comments count.

I recommend doing 0.999999… == 1.0 next. That always generates lots of comments.

Ridiculous BS. Mathematicians can bite me. Once the goat door is revealed there is a 50/50 chance of getting the car, regardless of anything else. And the earth is round (mostly) and time is relative, etc. etc.

If you stick with your initial door, you basically selectd a single door, so your chance to win is 1 in 3, so 1/3. However, if you switched, you basically selected two doors and the moderator tells you the wrong door of these two doors. It is just confusing that the moderator first opens the door (thus telling you the information about the wrong door) and THEN you decide to take the other door (which in fact, is the sequence of selecting the TWO other doors and then trivially selecting the LAST door (because the other door has a goat, by moderator)), so overall, this is 2 in 3, 2/3.

@Kenneth:
I lol’d.

I wanted to toss two other statical quirk problem in humans.

You conduct two experiments of tossing a coin repeatedly. In the first experiment you toss the coin until you see HTT as a pattern. In the 2nd experiment you toss the coin until you see HTH as a pattern. You repeat both experiments a large number of times (to make the results accurate).

Will the two experiments have the same average # of tosses to see the pattern. If not which one occurs sooner and why?

A test for a disease is 99% accurate, 1/10,000 people have the disease. You are chosen randomly to be tested for the disease and you test positive. What’s the probability that you have the disease?

See the problems and solutions here there’s also a lot of other good ones.
http://www.ted.com/talks/lang/eng/peter_donnelly_shows_how_stats_fool_juries.html

How many goats behind each door? What kind?

Jeff, you missed the point of Paul Bucheit’s critique of your boy/girl question.

Just because it’s “a simple question asked in simple language” doesn’t mean it’s well-formed and unambiguous. In fact, if anything, “simplicity” probably makes ambiguity MORE likely. This is especially true when calculating probabilities, where your assumptions make all the difference.

Let’s translate your question into less ambiguous terms. There are 2 possibilities:

1. Ted has 2 kids. You ask him to tell you the sex of one of them. (either one, his random choice). He says “Girl”. Given this, what’s the odds the other child is the opposite sex? Answer: 50%

2. Ted has 2 kids. You ask him if one or more of his kids is a girl. If so, what’s the odds the other child is a boy? Answer: 66%

Do you see the difference:

In the first question, the “Girl” reported to you is a random draw (the question would be the same even if “Boy” had been reported) and is thus independent of the second child’s sex.

In the second question, the “at least one Girl” result is screened for, so the two-Boys possibility is being filtered out.

Your question is ambiguous because it doesn’t say how the first reported result (the one he tells you) is chosen.

You may think this is trivial, but it’s not: What if you wanted to write a simulation to test your calculation? Which algorithm do you choose? Do you throw out the Boy-Boy results or not? If you don’t have enough information to write a simulation, you don’t have enough information to give a definitive probability.

The problem with the Unfinished Game is that it does NOT distinguish properly between what Jeff Atwood intended it to be (hence ending with a 2/3 probability) and the “nosy neighbor” scenario.

It comes to the same downfall as Monty Hall/Fall: intent is KEY because it shifts the probabilities.

In Monty Fall, the probabilities shift to 50/50 because Monty is not adding information (more information is coming in, but that information is as likely to drop us out of the scenario a (he accidentally opens the door we chose, or he accidentally opens the door with the car behind it) as it is to guide us. Thus, the problem domain has contracted but no new information has been imparted on the existing choices.

In the Unfinished Game scenario, we are supposed to realize that the information offered is NOT random, but as a result of a pointed question (“Is one of your children a girl”) but not TOO pointed question (“Is your oldest child a girl”). If it were random, you end up with the “nosy neighbor” scenario, wherein the extra information contracts the problem set (you are as likely to have seen a boy walking across as a girl) instead of ellucidating the remaining domain.

Which brings me back to the first paragraph of Atwood’s piece, where he scolds us for telling him his formulation of the Unfinished Game problem was incredibly poorly stated. Sorry, but it was. He does not say if this person pointedly told us he has a girl at random or with some surrounding information, and supposes that we must infer that there was some surrounding context to that information.

Personally, as a parent, I answered “~1”. If you are a parent and tell someone that “one of my children is a girl” you are saying that the other one is not. Simple human nature. If I have two girls, I’m going to say I have two girls, not one. The only possibilities then are:

1. The other child is a boy
2. The other child is a hermaphrodite of some sort
3. The person is being deliberately coy about their children’s genders

Since 2 and 3 are generally very small probabilities, that leaves just about a 100% chance that the other child is a boy.

Of course, Atwood’s scenario posits that (3) is the dominant probability, which then overwhelms the odds of the situation, and further that there is not a REASON that they are being deliberately coy (for instance, some odd societal stature loss from having two girls).

@Lint I found that explanation a little confusing so after working it out myself I’m going to restate my reasoning through it to hopefully help anyone else out as well.

G-G, G-B, B-G, B-B are the possible combinations.

If I ask the sex of one of the child I’m not in 3 of 4 states as you might assume. I’m putting myself in one of 8 states. The way I’ve asked the question makes ordering relevant for the G-G and B-B situation. So now I have the following states…

G-G expands to…
G1-G2
G2-G1

G-B expands to
G1-B1
B1-G1

B-G expands to
B2-G2
G2-B2

B-B expands to
B1-B2
B2-B1

Using those states 4 of them (G1-G2, G2-G1, G1-B1, G2-B2) are descriptive of my current situation of knowing the first kid I’m told about is a girl. Of those 4 states 50% of them are Boys 50% are Girls.

Now when you ask “At least one Kid is a Girl” you are wording the question such that it is order independent. G1-G2 is the same as G2-G1. In this case you have the following possible states you can be in

G-G, G-B, B-G.

Of the three states 2/3rds of them has a Boy for the other child.

I think the key to understanding stuff like this is to identify if ordering matters, break down the possible states, then count the probabilities.

For the record, a better statement of the “Unfinished game” problem would be:

In a random gathering of people, we asked everyone with two children to stand up and everyone else to sit down. We then asked those standing to only remain standing if one or more of their children was a girl. One woman was standing at this point. What are the odds that she had a girl and a boy?

The problem, though, is that making the problem clearer starts sending warning signals to the quizee. Not sure how to best pose this scenario in a “natural” manner without either causing critical ambiguity or tipping the quizee off that something unintuitive is afoot.

Hmm… trying to explain the Monty Hall problem may be almost as bad as trying to convince your drunk friend not to bet \$100 on the Superbowl coin flip.

“Well the last three Superbowls were heads, so this year it’s definitely going to be tails!”

I won’t name names.

Please, if you want to explain the answer, create your own blog that nobody reads, because that’s about how interesting this problem is. The topic has been beaten to death. RIP

Oh dear I’m going to get dragged into this again. Here goes:

@Tom Dibble

It comes to the same downfall as Monty Hall/Fall: intent is KEY because it shifts the probabilities.

I don’t agree with this statement. I don’t think the intent matters, just the actual outcome:

Monty Fall is the same as Monty Hall once the chance of accidentally revealing the car is gone.

However, intent is a good indicator of predicting what you should do before the event has passed.

It looks like the only way to finish the game is to close the threads.

@eshepherd OMG! THANK YOU! You’ve struck what, for me, is the best possible, most intuitive explanation of why the probabilities work out the way they do.

So people don’t have to search, eshepherd said: “You’re not dropping one, you’re getting both doors that you didn’t pick.”

Perfect! If you switch it’s probabilistically the same as getting to open two doors! 2/3! All the extra info about the host knowing and choosing is just there to make sure no situations arise where he accidentally opens the door containing the prize.

A better interview question would be:

“Given that the answer to the Monty Hall problem is to switch your answer, can you explain why that’s true?”

I build robots. Check out my blog at:
http://forums.trossenrobotics.com/blog.php?u=3182

@Logo - I wasn’t trying to detail the alternate solutions so much as make the point that the question phrasing critically matters in what result you arrive at. (Which Jeff seems to have dismissed as sour grapes)

Probably clearer than my case 1 question would be to phrase it like the “Nosy neighbor” scenario that Tom Dibble refers to, or “Nebulous Neighbors” in the pdf Jeff links. Tom didn’t actually elaborate on it, so I’ll paraphrase from the pdf:

Nebulous Neighbours: Your new neighbours have two children of unknown gender. One day you catch a glimpse of a child through their window, and you see that it is a girl. What is the probability that their other child is also a girl?"

To me, this makes it totally unambiguous what is being asked.

It really is a fantastic problem though, if for no other reason than the fact that it shows how violently even the most thoroughly educated people can react when something runs counter to their intuition.

Anyways, for anyone’s made it alllll the way down here while still in the 50/50 camp, I’m going to take my own (hopefully non-redundant) crack at explaining it.

The question isn’t about the probability of a car (prize) being behind a door, it’s about the probability of the door you PICKED having a car behind it.

At the outset:
There are twice as many ways to pick a losing door in the first phase as there are to pick a winning door.

So you have a 2/3 chance of picking a loser, and a 1/3 chance of picking a winner.

Now we take one of the losing doors away… what’s the probability of the car being behind the door you picked? The same as before! 1/3.

Why? Because there are two different ways (2/3) to pick a loser at the outset. This means the probability of having a loser after a door is taken away is 2/3 (there’s two losers to chose from), which means the probability of having the winner is still only 1/3.

Another way to look at it is: How can taking one door away after you make your first choice change whether or not the door you picked is a winner?

The intuitive paradox on this problem comes from looking at the end game without considering the prior state. If you were to walk into a half completed Monty Hall game without prior knowledge of the original guess, the odds would truly be 50/50.

In reality, the prior state plays a huge part of the final odds.

Consider doors x, y and z. We’ll define x as always being the winner without specifying the mapping between x, y, z and 1, 2, 3. Therefore when you pick a number, you don’t know what letter you picked.

If you picked x, Monty will pick y or z and your best move is to stay.
If you picked y, Monty will pick z and your best move is to change.
If you picked z, Monty will pick y and your best move is to change.

You do not know what letter your initial selection was but two possibilities favor changing while only one favors staying. You’re best bet is to act on the more likely probability that you initially chose y or z and not x.