She does specify that Monty always reveals a goat, at 1:06. It's a bit quick and you might miss it, but it is pointed out.

Not really. She says she opens the door to reveal a goat; it's not exactly made clear that he _has_ to reveal a goat. It's obvious that he would - the show would be retarded if he revealed the car - but when you're thinking conceptually, it helps to point it out specifically. I only bring this up because:

1) I've tried explaining the concept to people before and my experience has always been that it's hard to grasp until it gets pointed out that after the contestant picks a door, one of the two remaining doors (if not both) has a goat, and if the contestant picked a goat, Monty will _always_ reveal the remaining goat.

2) I wanted to share the story about my baffling friend.

In other words, if you pick a door, and then Monty randomly opens one of the remaining two doors and there's a goat behind it, your odds wouldn't go up by switching doors (I think). The key is that he is prevented from revealing the car.

Or maybe they would. I think I've talked myself into no longer understanding the problem.

I'm a complete idiot, so I don't understand this at all. Once you know a goat is behind door number two, that door no longer exists as an option, as if you now always just had the two doors. Why include an irrelevant door in the math, you may as well throw in the one leading to the parking lot with it. You're looking for a door with a red car and not the other goat, you have two doors that matter, what are the odds?

I don't understand.

Slippery, I had to watch it twice. The explanation at 2:30 seems very obvious, but yet...

If I started with three doors, my choice of a door does not move anything around. I had a 1 in 3 chance. Now monty shows us the empty door. There is always at least one such door in the remaining two.

At this point, I'm tempted to argue that I am now faced with a 1 in 2 chance, if I pick again. Those odds being substantially better than 1 in 3, I should randomly pick again.

That's different from her conclusion, which is to always switch. How to reconcile the two...

You only ever have two doors, the third door, and the act of numbering the doors, is a distraction. It seems like it is never 2/3 because one door will always be revealed as a goat. You can't pick two doors from the get go, one will always be a goat, therefore never actually in play or relevant, leaving two doors. It is only the act of walking up to the doors, seeing the numbers, and given the initial illusion of choosing one door disguised as two (because they are the same result, goats and can only be picked one at a time) that gives the illusion of 2/3.

I know I'm wrong, but damn.

"...the door that Monty opened would always have behind it a Zonk"

She says right there that it's always a goat (or zonk, to remain show-accurate).

Well, the trick about him having to reveal a goat (as there will always be two doors the player didn't choose, Monty can always reveal a goat) is that it reduces that option.

For instance, let's say the car is behind door number two. You have three options if you don't switch:

1) Pick door number one, stay with that door (loss).
2) Pick door number two, stay with that door (win).
3) Pick door number three, stay with that door (loss).

Now let's assume you decide to switch.

1) Pick door number one. Monty is forced to reveal door three. You switch. (win)
2) Pick door number two. Monty can reveal either door. You switch (lose).
3) Pick door number three. Monty is forced to reveal door one. You switch (win).

And she is right, there is no use trying to outsmart her (sadly!). The probability is 2/3s on a switch, and her later example (the one with the 100 door problem) is a much clearer illustration of this principle at work.

Yeah, Homer. My problem with the wording of that is "would always have [to]" implies a probability outcome, not a forced outcome. To someone trying to understand the problem, they would think that the rules of probability are forcing the case, and is therefore something they need to understand, as opposed to the rules of the game set up before the probability analysis enters the problem.

I realize I'm nitpicking here, but when dealing with counter-intuitive ideas, wording means everything. I should drop this, but fuck you, I STICK WITH MY DOOR!

Oh... oh!

Thank you, Hooker!

I still don't understand how the odds, on paper, swing in favor of the other door once one of the two unpicked doors is eliminated (and shown to have one of the two goats). My brain tells me that odds are 50/50 throughout, even in her 100 door example. Monty would always be able to expose the remaining goat(s) at that stage of the game, since no matter what the player chooses, there will always be remaining goat(s) available for him to reveal.

I obviously have to watch this a couple more times. I'm not seeing it, unless the game was rigged to have the car behind the unpicked door 2/3 of the time?

Well, maybe if someone is absolutely clueless about probability and is really, REALLY struggling to make sense of the problem, but I think you'd have to try pretty darn hard to overthink her statement enough to make that sort of mistake!

Anyway, your breakdown of the choices is very good. Maybe you should start making your own math how-to videos?

Her argument is basically that I can in effect choose to open two doors out of the three by switching, because of the reveal. If I stay pat, I can only open one door out of the three.

The sneaky part is that his opening the door only helps if you're on the other side of "the divide".

I'm a long way away from being in a place where I'm comfortable putting my voice and/or face on YouTube.

Hooker's version of this video would have been 30 seconds and perfectly clear, so I think they should go for it.

Well, my strategy would be to have a guy on the inside who could relay to me information about which door to pick. Possibly Monty himself; I'd tell him before the show that I'd sell the car and split the profits 50/50 with him, so long as he told me via secret hand gestures which door to pick.

I would then reneg on my deal, because what's he going to do? Complain to the producers and tell them he helped me cheat?

Also, unsurprisingly, Khan Academy already has a Monty Hall video, so that rules out my doing the same thing anyway.

https://www.youtube.com/watch?v=Xp6V_lO1ZKA

Note how he labours over the necessity that Monty must reveal a goat, Homer!

No it doesn't!

While we all enjoy the ASMR soothing voice of Khan Academy, we need videos with a particular poeTV sensibility. Can you sing, and do you own multiple cats?

Also, wow, I haven't been to Khan Academy in a while, and boy does its layout now completely suck. Talk about the most bloated site possible to embed a YouTube video in. We should all really appreciate how clean poeTV is.

***** for Hooker. I was able to "prove" this using his enumerated explanation, testing from 3-10 doors, 1,000,000 attempts each:

int mindoors = 3;
int maxdoors = 10;
int attempts = 1000000;
srand(0);
for(int doors = mindoors; doors

I wouldn't want to add anything that already exists. I could make a fairly authorative video series about the nonsense universe that is the collected works of Godfrey Ho (which I call the Godfreyverse), and then I guess I could present it using that crude animation thing that futurebot used to use whenever he wanted to make slanderous videos about other people.

Comment fields don't like code. Trying again:

int mindoors = 3;
int maxdoors = 10;
int attempts = 1000000;
srand(0);
for(int doors = mindoors; doors <= maxdoors; ++doors)
{
int switchwins = 0;
for(int ix = 0; ix < attempts; ++ix)
{
int car = rand() % doors;
int choice = rand() % doors;
switchwins += (car == choice) ? 0 : 1;
}
printf("%d doors: won %d out of %d times when switching.\n",
doors, switchwins, attempts);
}

Results:

3 doors: won 665840 out of 1000000 times when switching.
4 doors: won 749994 out of 1000000 times when switching.
5 doors: won 800078 out of 1000000 times when switching.
6 doors: won 832549 out of 1000000 times when switching.
7 doors: won 857750 out of 1000000 times when switching.
8 doors: won 874947 out of 1000000 times when switching.
9 doors: won 888779 out of 1000000 times when switching.
10 doors: won 899979 out of 1000000 times when switching.

Ok. Hooker gave a good explanation, but the Khan Academy video helped a lot as well.

HA! I was wondering if one of you fuckers would attempt to experimentally prove the thing out. God bless you, Matlock, you magnificent bastard.

After putting it into code, I realized that the entire problem was simplified to your chances of being right being 1 out of n doors, no matter what information you get afterward. When he tells you it wasn't behind door #2, then asks you if you want to trade door #1 for door #3, he may as well have just said "so, do you think you chose correctly?" There's a 2/3 chance you didn't, so switching is always the smart option.

Though I still have a tough time wrapping my drug addled brain around it if I think about it too much.

Given that I can't drive for shit, I'd just take the goat.

Milking a goat'd give you an infinite supply of chevre laits.

Hooker's walk through the options is a Must Read

Try it this way:

Initially the odds are 2:3 that you picked wrong.

When you switch doors, what you're really picking is "all the other doors", and if you switch "all the other doors", there's a 2:3 chance you'll be right to switch. Yeah, Monty did remove a door in there somewhere, but he made sure it was a goat, so "all the other doors" is just as likely to contain a car whether or not Monty removed anything.

For you, Bort. Much more understandable via the way that common humans think.

Still another way to look at it is, your odds go up on a switch because Monty did you the favor of removing a losing option. Since the odds were pretty good you were sitting on a losing option to begin with (2/3 of the time), there just aren't that many arrangements where the remaining door is also a losing option -- the only way it could happen is if you initially guessed right (i.e., 1/3 of the time).

Or even a slightly different take. There are exactly two directions the show can go:

- 2/3 of the time, you initially guess a goat, Monty takes away a goat and you have the option to switch to the car.

- 1/3 of the time, you initially guess the car, Monty takes away a goat and you have the option to switch to the other goat.

Those are the only two ways things can go. You get either the one situation or the other, and the former case happens twice as often as the latter.

Or yet still another take: when it's at the point where you can either stick or switch, there is one goat and one car, and you might think that it's a 50/50 chance -- except that there's a 2/3 chance you initially guessed a goat, so there's a 2/3 chance that switching means switching to the car.

Even simpler: after the reveal, there is 1 winning and 1 losing door. If you picked the winner, switching is an auto loss. If you picked the loser, switching is an auto win. There is a 2/3 chance you picked a loser.

I never watched the show though, so there's another factor: did Monty ALWAYS offer this choice? If not, did he tend to offer it more often when the contestant initially picked the car? Because then that would throw all this fancy MATH out the window.

Still more rumination. Most of this door business seems random, but there's one non-random thing going on: Monty Hall knows where the winning door is, and he can be counted on to not pick that door. That means that any door that he could have picked, but didn't, has a somewhat better chance of being the winning door. (In this case, a MUCH better chance, because there are so few doors to choose from.)

... even simpler: "Why didn't Monty Hall open door #3? Does he know something I don't, say, that there's a car behind it?" That won't work all the time, but it will work 2/3 of the time.