Classic question I'm sure everyone's heard by now but like airplane on a conveyor belt I'm curious if there will always be proponents of other answers

The version that originally (more or less) appeared on Massassi in ~2002:

You are playing a game show of which Monty Python is the host. Before you are three doors -- one leads to innumerable riches, and the other two lead to yaks. You want the riches and not the yaks. You choose one of the doors, and Monty Python opens another (not the one you chose) door to reveal a yak. He then gives you the option to either switch your choice to the other remaining door, or to stay firm on your original door choice. What should you do (if anything can be done) to maximize your chances of getting the riches, and why?

Switch.

Monty HALL.

Anyone who argues against the fact that switching gives you a 2/3 chance of winning is :downswords:.

Hell, it's easy as **** to model this problem and run simulations that show it.

Monty Hall had goats Monty Python has yaks

If the real answer is, in fact, to switch, I can't say I understand why it would matter.

The real answer, of course, is that Pommy is misinformed, and that yaks are far better than riches.

Many years ago, when I was taking intro stats, this came up on my first midterm. Luckily I'd already seen this discussion on Massassi, so I knew what the answer was supposed to be.

Your first choice gives you a 1/3 chance of winning.

Your second choice gives you a 1/2 chance of winning.

But, surely since theres a yak door shown, that is not your door, that has reduced your first choice to 1/2, same as if you switch?

Actually, in the normal formulation of the problem, switching gives you a 2/3 probability of winning.

Of the entire problem, yes. Isolating each choice, no.

Please explain to me how it's not a 50/50 chance whether you switch or not.

Yeah I don't get it.

Theres 3 doors, 2 of them have yaks behind them.
You know one has a yak, therefore you have a 50% chance to pick the right door.

Where did I go wrong?

people still don't understand this one?

Kinda like flipping a coin - you have a 1/2 chance of getting heads on your first flip. You have a 1/4 chance of getting heads again on the second flip.

I think.

It's been a while since I've had to think about math stuffs.

It's not a 50/50 because when you first pick, there's a 2/3 chance you picked a yak.

So when Monty shows you the other yak, there's a 2/3 chance that the last one is *not* a yak.

[ninja'd] It's an elimination game, not "pick the right one". You likely pick a yak, you're shown the other, therefore the last one is probably right.

i'll try and explain this without excessive numbers and as simply as possible.

if you had initially picked the riches (1 in 3 chance), the host revealed one of the yaks, meaning the remaining door is the other yak.

If you had initially picked a yak (2 in 3 chance), the host revealed the other yak in which case the remaining door is the car.

Therefore, if you always follow a strategy of switching, your 2 in 3 chance of originally picking a yak gives you a 2 in 3 chance of winning the riches instead. If you follow the strategy of not switching, the best you can hope for is your original 1 in 3 chance of guessing correctly.

Good explanation, thanks!

I'll throw in my method of understanding it:

Before any doors are opened, realize the only way you can lose by switching is by picking the car at the beginning. Picking either yak door will result in you winning the car by switching, after the other yak door is opened.

Although the opening of one of the yak doors is significant to the problem, I think it's also the part most people put far too much focus on.

I understand that the probability increases after one of the doors/yaks is eliminated, but I still don't see how switching maximizes my chances of winning (which apparently has swapped to a yak, riches and cars -- what?).

As for the coin example, I was under the impression that the probability was ALWAYS 50/50, no matter how many times it was flipped, hence the gambler's flaw.

EDIT: I guess I'm HURR because I still don't understand how my probability changes to 2 out of 3 by switching. I feel like Detty's explanation should clarify things but the last sentence still doesn't make sense to me (nevermind the inclusion of a car).

It's 50/50.

This is exactly why we are in the economic state that we're in, people get confused on simple statistics and try to make things more complicated then they are.

Initially you have a 1/3 chance of picking the correct door. However, because one of the doors is revealed and then you are essentially given a whole new opportunity to choose, your chance of picking the right door is 1/2 on the second chance. You gain nothing by switching and your odds are just the same. There could just as easily be a yak behind the other door as the riches, so again, switching does not increase your odds at all. People let the first chance confuse them with the second chance, which are totally separate.

This goes for the people who believe flipping a quarter and it landing on heads the first time will greatly decrease your chance of it landing on heads with each subsequent attempt. The reality is that it's totally possible to flip a quarter and it always land on heads. You have an ~50% chance of it landing on heads or tales every flip and an extremely low probability of it landing on it's side. This all depends on physics and has nothing to do with numbers or statistics.

After all, there's statistical predictions and then there's reality. Look at any kind of forecast (financial, weather, or anything else) and there's virtually zero percent probability that the statistical forecast will match what actually occurs. Brokers on Wall Street have some of the most complex statistical analysis of the market in real time then probably any other field. Even they will tell you that those numbers are just a reference and that they still have to rely on their intuition to make a decision on what they think the market will actually do.

Alco/Geb: Regarding the coin flip example, there is a difference between the probability of getting subsequent heads or tails (e.g. probability of getting heads+heads) vs. getting heads on a specific flip at any point.

Alco: It is in fact not 50% because this is a conditional probability; e.g. the two decisions are NOT separate and DO influence each other.

Alco's makes a lot more sense to me, but I have a feeling that'll still put me in the HURR catagory.

.

Prove it. I argue that they do not. The revealing of the first yak eliminates it. If you were not given a choice to change which door you pick, I would agree.

Again, I have to agree with Alco on the choice factor. I might be able to take it on faith if there were a great number of studies to show me otherwise, but it still wouldn't make sense to me at this time.

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

There's an entire Wikipedia thread that says you're wrong, complete with proofs.

Sucks to be you.

hahahahahaha


hahahah

oh man

pot calling kettle black

hahah

We'll see.

We'll see? It's been proven. Just because you don't understand the proofs doesn't make it incorrect.

No, we won't. Proofs have already been provided, please educate yourself before making yourself out to be more :downswords:. At least everyone else was just trying to understand, you just went out there on a limb, insisting we're all dumb, and then got it wrong.

Let me rephrase the problem to illustrate:

Three cups are presented to you upside down. You are told that one of the cups has a ball underneath it. You chose the cup that you think the ball may be under. One of the other cups is then lifted to reveal that the ball does not exist under it. What is the probability that you chose the cup with the ball under it?

The answer is 50%. At that point it's totally irrelevant that there was initially a third cup.

.

There's a difference between statistical probability and reality, as I've already pointed out.

Er, no, it's 1/3. Pretend you didn't show the other cup. If you lifted your cup, your chance is still 1/3.

http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayesian_analysis

There is also a visual diagram further up on the page

Initially, yes, but the question is at the end AFTER one of the cups was revealed.

It's not a probability. It a definibility!

If you bothered to read the Wikipedia article, they even give you nice pretty pictures to help you understand.

You don't understand what probability or statistics are at all.

Saying that a coin will land heads 50% of the time does not mean to suggest that it MUST happen 50% of the time, only that the trend is towards 50%.

It seems you are trying to suggest that even if the "statistical probability" as you call it is 2/3, the "reality" is 50%. That makes no sense at all.

As stated on the Wikipedia page which you have obviously not read, you can perform an empirical experiment to demonstrate this, using cards. You will find that, after enough trials, that average chance came out to about 2/3. This is not because of some "real world error" but because that's what the chance actually is.

Knowledge of what's under a cup AFTER choosing does not affect the probability that you chose correctly BEFORE revealing the cup.

One again, being provided a visual under the popular solution has made the probability clear to me, even if I still need to spend time understanding it fully.

However, from the looks of it, this problem was posted incorrectly, so I call shenanigans.

EDIT: For the record, I would have still been tricked before had it been posted correctly, but at least now I understand the importance of the host behavior, which I assumed was an ignorant host, which is not the case in this condition.