Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Kuildeous]

One thing I've read (without citation) is that in practice, Monty only revealed a goat when the contestant correctly chose the car. In this regard, it would be using a person's knowledge of probability against him. But if you watched enough of the show to know for certain that Monty reveals a goat only when the selected door has a car, then that has its own power.

It also goes against the premise of the puzzle which presumes that Monty always reveals n-2 goats.

[sudgy]

That choice was 1 in 3 to begin with, and that probability does not change because it's not a new problem.

This is probably the most intuitive description I've ever seen describing the solution to the problem.  +1

[scott_pilgrim]

Say you have a million doors, with a prize behind one. You pick one door, the host opens 999,998 of them (not the one you picked), making sure not to open the one with the prize behind it, and then you have the opportunity to switch doors.

You'd have to be crazy not to switch. The host almost certainly picked the door he didn't want to open because it had the prize behind it. There was a 1 in a million chance you originally picked the correct door, but the other 999,999/1,000,000 times, the door that the host went out of his way not to open is the one with the prize behind it.

This is phrased slightly misleadingly. The premise of the problem is that the host of opens remaining doors without prizes, so there is no guessing of host's intention.

edit: sorry, I should say remaining up to all but one that are without prizes. If all are without prizes the opening is selected with equal probability.

I'm not sure what you mean but I agree I didn't say that clearly. What I meant was that in almost all cases, the host is forced to open all but the door with the prize behind it.

I guess another way you can think about it is that when you switch, you are betting that the prize is behind any one of the doors you didn't originally pick; if you don't switch, you are betting it is behind the single door you originally picked.

[Witherweaver]

Say you have a million doors, with a prize behind one. You pick one door, the host opens 999,998 of them (not the one you picked), making sure not to open the one with the prize behind it, and then you have the opportunity to switch doors.

You'd have to be crazy not to switch. The host almost certainly picked the door he didn't want to open because it had the prize behind it. There was a 1 in a million chance you originally picked the correct door, but the other 999,999/1,000,000 times, the door that the host went out of his way not to open is the one with the prize behind it.

This is phrased slightly misleadingly. The premise of the problem is that the host of opens remaining doors without prizes, so there is no guessing of host's intention.

edit: sorry, I should say remaining up to all but one that are without prizes. If all are without prizes the opening is selected with equal probability.

I'm not sure what you mean but I agree I didn't say that clearly. What I meant was that in almost all cases, the host is forced to open all but the door with the prize behind it.

I guess another way you can think about it is that when you switch, you are betting that the prize is behind any one of the doors you didn't originally pick; if you don't switch, you are betting it is behind the single door you originally picked.

I just mean the way you said it sounds as if the host has choice in the matter. Really the host is only flavor; the game runs algorithmically based on only the player's decision.

[silverspawn]

Yes, the key insight here is that it depends on whether the door revealed has been randomly chosen and could have also been the one with the car. If so it's 1/2 to 1/2; if not 1/3 to 2/3. Occasionally the question is posed without specifying this, in which case it does not have a correct solution, because information is missing.

If specified, it can be proven with Bayesian updating. But the 3 cases SP outlined are the easiest way of looking at it.

[Kuildeous]

Why do I keep reading this thing?

The 66% chance of winning if you switch doors is only half of the equation. The other half, which was pointed out by another commentator herein is: "what are the odds of winning, if you do not switch doors?" Your calculations have very elegantly shown us the answer to both questions. The odds of winning the car are as predicted, 50/50.

Basically it means that your odds of winning if you switch are 67% AND the odds of you winning if you DO NOT switch are 67%.

Switch / don't switch, your odds of winning are the same.

#nerdrage

[silverspawn]

It is deeply saddening if people present logically false things in this kind of tone.

[Kuildeous]

The thread has shown me new ways for people to misunderstand math. And I can understand people screwing up things. I'm not perfect in math either. But the unwillingness to examine one's concepts is rather disheartening.

I try to do my part. One guy in the 50-50 camp threw me for a loop when he insisted on differentiating between the two goats (and included a pig to mark the difference). He pointed out that there are six permutations of how the prizes are arranged. But then he says that the six outcomes become eight:
1a. car pig
1b. car goat
2a. car goat
2b. car pig
3. goat car
4. pig car
5. goat car
6. pig car

He did explain that options 1 and 2 are where the car was behind the first door, so each of those two permutations had two different outcomes depending on which door the host opened. And if you look at those eight, you could conclude that it's 50%. I had to point out that the probability of 1 and 2 are still the same, but now you have 1a at 1/12 and 1b at 1/12. It alarmed me because it made sense at first. But again, you reason through it and find the flaw. These people aren't willing to do that. They are sticking with the 1/2 conclusion and are crafting their arguments to fit it.

[silverspawn]

I think this problem in particular is so confusing that relatively few people will be able to really distinguish a true from a false explanation with intuition and logic alone. You can make false explanations seem really convincing.

But if you write down the probability space formally, define events, and use the rules of Bayes, there is no room left for disagreement.

[Kuildeous]

But if you write down the probability space formally, define events, and use the rules of Bayes, there is no room left for disagreement.

Except for people who don't know enough about math and will only go with their gut reaction that two doors mean 50%.

It took me a while to accept the math too. I felt like somebody did the math incorrectly, but eventually I had to accept it.

[silverspawn]

the best intuitive argument I think is looking at different cases. The problem with the way scott presented the cases is that you could argue ...

B. The prize is behind door B. The host eliminates door C, you switch to B and win.
C. The prize is behind door C. The host eliminates door B, you switch to C and win.

... that either case B or case C is impossible because we know now that the prize is not behind that door. So there are only 2 cases hence 1/2...

I'd explain it like so (this will also explain why how the door was revealed matters):
Case 1: The door you picked originally was the correct one. In that case, switching loses.
Case 2: The door you picked originally was not the correct one. In that case, switching wins.

This is always true, regardless of how the revealed door was chosen. Now all we need to know is the probability that the door you originally picked was the correct one.

If the moderator purposefully revealed a goat, then it is 1/3, because him revealing a goat tells you nothing about your original pick, because whether you pick the car or a goat, he can always reveal a goat. Again, Bayes.
If the moderator revealed at random, then it is no longer 1/3. It is not immediately obvious why it is 1/2, but it is logical that it has to be different.

Now if we assume he revealed a goat on purpose, then there remains a 1/3 chance that you picked right initially... and therefore a 2/3 chance that you are right if you switch.

But if someone presents you with a reason why it is 1/2 then you can't usually respond immediately because it takes some time to carefully think through every step and figure out where it goes wrong. And if the discussion is offline, then whelp.

[ashersky]

I get the math arguments, I think.  Thanks.

Part of my confusion stems from how the switch supposedly guarantees the win.  It seems it doesn’t, it just raises the possibility of winning by increasing the odds of the second door.

Here’s the other part.  Assume A is the car and I chose it.  The arguments so far say that because Monty removes C, it causes B to more likely be the car (than A).  But Monty had a 0/2 chance there to begin with, right?  It seems like he didn’t increase the chances of B becoming the car at all.

Why is B now more likely than A to be the car when we know it isn’t?

[ashersky]

The more I think about it, the more I’m convinced you are wrong.

Switching is only the right call 2/3* of the time, not 100%, because 1/3 of the time you chose right anyway.

*This number would change depending on the total number of doors**.  Taking the game to the absurd million door example, there’s still the 1 in a million time that you chose right to begin with, so switching is only the right call in 999,999 of out 1,000,000 times.

**I don’t know anything about math, so if there’s some sort of infinite numbers argument or something with space-time, I am not taking that into account.

At this point, I have still not been convinced by your posts that switching 100% is correct, as there are definitely situations (as pointed out by others) in which the original door was the correct choice.  Switching does increase your odds of winning more often than not, but it is by no means a guarantee, thereby disproving the theorem or whatever it is.

[silverspawn]

Part of my confusion stems from how the switch supposedly guarantees the win.  It seems it doesn’t

Yea, it definitely does not. No guarantees.

Here’s the other part.  Assume A is the car and I chose it.  The arguments so far say that because Monty removes C, it causes B to more likely be the car (than A).  But Monty had a 0/2 chance there to begin with, right?  It seems like he didn’t increase the chances of B becoming the car at all.

Why is B now more likely than A to be the car when we know it isn’t?

It isn't. If you presuppose that your original choice was correct, then just like you said, the probability of A is 100% before the reveal and 100% after the reveal.

The reveal only increases the probability of B if you don't already know where the car is.

[Awaclus]

Switching does increase your odds of winning more often than not

It increases your odds of winning every time. It doesn't matter whether or not the prize is actually behind that door because you don't know where it actually is, and you can't do math based on information that you don't have. What you know is that there's a 1/3 chance it's behind the door you chose and 2/3 chance it's behind the other one, and 2/3 is better than 1/3, so switching is better.

It doesn't guarantee winning though.

[silverspawn]

The more I think about it, the more I’m convinced you are wrong.

Switching is only the right call 2/3* of the time, not 100%, because 1/3 of the time you chose right anyway.

Yes, you're entirely correct. What someone who says "switching is 100% right" means it that "if you are in this situation, you should always switch because it is correct 2/3 (i.e. most) of the time." It's a confusing way of saying it, but it does not contradict what you think.

[ashersky]

Switching does increase your odds of winning more often than not

It increases your odds of winning every time. It doesn't matter whether or not the prize is actually behind that door because you don't know where it actually is, and you can't do math based on information that you don't have. What you know is that there's a 1/3 chance it's behind the door you chose and 2/3 chance it's behind the other one, and 2/3 is better than 1/3, so switching is better.

It doesn't guarantee winning though.

If your odds of winning go from 100% to 0%, that is a decrease, not an increase.
That happens 33% of the time.

Are those two correct statements in the context of the example? 

If so, it appears that your odds of winning do not increase 100% (every) of the time.  Hence my (possibly mistaken, but still unchanged) belief that switching is only correct X-1 of out X (most) times.

[ashersky]

The more I think about it, the more I’m convinced you are wrong.

Switching is only the right call 2/3* of the time, not 100%, because 1/3 of the time you chose right anyway.

Yes, you're entirely correct. What someone who says "switching is 100% right" means it that "if you are in this situation, you should 100% switch because it is correct 2/3 (i.e. most) of the time." It's a confusing way of saying it, but it does not contradict what you think.

If you were to say “you should definitely switch every time because you are more likely to win by switching because mathymath” then no one would be arguing, would they?  Seems pretty straightforward.

[silverspawn]

The more I think about it, the more I’m convinced you are wrong.

Switching is only the right call 2/3* of the time, not 100%, because 1/3 of the time you chose right anyway.

Yes, you're entirely correct. What someone who says "switching is 100% right" means it that "if you are in this situation, you should 100% switch because it is correct 2/3 (i.e. most) of the time." It's a confusing way of saying it, but it does not contradict what you think.

If you were to say “you should definitely switch every time because you are more likely to win by switching because mathymath” then no one would be arguing, would they?  Seems pretty straightforward.

Yes. You should say it that way.

[Awaclus]

If your odds of winning go from 100% to 0%, that is a decrease, not an increase.
That happens 33% of the time.

That doesn't happen. Your odds of winning aren't 100% until you've actually opened the door and learned that the prize was there. As long as you don't know where it is, your odds can't be greater than 2/3 and the way to have 2/3 is by switching.

[Watno]

If you were to say “you should definitely switch every time because you are more likely to win by switching because mathymath” then no one would be arguing, would they? 

They would.

[silverspawn]

If your odds of winning go from 100% to 0%, that is a decrease, not an increase.
That happens 33% of the time.

That doesn't happen. Your odds of winning aren't 100% until you've actually opened the door and learned that the prize was there. As long as you don't know where it is, your odds can't be greater than 2/3 and the way to have 2/3 is by switching.

Please don't have this discussion (at least not like this). It is just frequentism vs Bayesianism and it doesn't have anyhing to do with the Monty Hall Problem.

Frequentists say probabilities of things that are already decided (like "the car is behind door A") are either 100% or 0%. Bayesians say, no, probability is a way of expressing your subjective uncertainty, so it may first be 1/3 and then 2/3.

If we discuss this we should make it clear that we're arguing about Frequentism vs Bayesianism and not about Monty Hall.

[ashersky]

If you were to say “you should definitely switch every time because you are more likely to win by switching because mathymath” then no one would be arguing, would they? 

They would.

I guess you are right about that.

[ashersky]

It is just frequentism vs Bayesianism and it doesn't have anyhing to do with the Monty Hall Problem.

I know/knew nothing about any of this.  Super neat!

https://en.m.wikipedia.org/wiki/Frequentist_probability

Thanks for introducing some cool philosophy stuff.

[ashersky]

Also, to get back on random, when I copy/pasted that Wikipedia link, almost pasted an entire Role QT from my upcoming RMM.  That would have been unfortunate.