Help Guy Pefect (the Sentinel/Sentinel+ Guy)

[FatherXmas]

Reminds me of my first apartment but without the 20 ft ceilings and windows (it was in a converted textile mill).

And what a lovely beige.  :roll:

[Aggelakis]

And suddenly all of the discussions of the Monty Hall problem I've watched Guy and Tony have over the years are brought into focus.

I had to immediately go searching for this oldie but goodie.
http://www.cohtitan.com/forum/index.php/topic,4187.0.html

[eabrace]

I had to immediately go searching for this oldie but goodie.

Man, was it that long ago?  I feel old.

[GuyPerfect]

And suddenly all of the discussions of the Monty Hall problem I've watched Guy and Tony have over the years are brought into focus.

According to Tony, then after the first door is eliminated, the likelihood of the laundry units being behind the door I selected decreases. (-:

[eabrace]

https://images.weserv.nl/?url=weknowmemes.com%2Fwp-content%2Fuploads%2F2012%2F03%2Fnot-listening-otter-meme.jpg

[TonyV]

The only relevant post is this one, in which everyone except Guy agreed that I was right.  I even wrote a little program that exactly duplicates the steps I outlined, and it demonstrated beyond any shadow of a doubt that I was right.

[eabrace]

https://images.weserv.nl/?url=weknowmemes.com%2Fwp-content%2Fuploads%2F2012%2F03%2Fnot-listening-otter-meme.jpg

[Aggelakis]

Good idea.

https://images.weserv.nl/?url=weknowmemes.com%2Fwp-content%2Fuploads%2F2012%2F03%2Fnot-listening-otter-meme.jpg

[Captain Electric]

I just wasted valuable time and brain space reading through that entire argument. Wow. Wow.

Seriously? I mean, come on.

The best option is obviously the goat door that has already been opened.

Also, it's obvious that changing observational data causes the scenario to evolve from a 1/3 chance to a 1/2 chance.

Observational data is everything when it comes to quantum mechanics.  :P

[Arcana]

That game is a classic example of people misapplying probability.  It happens all the time, but rarely with something so pernicious.  The fundamental issue from a mathematical perspective is that Monty Hall is not making a statistically independent choice.  Which means his opening one door is not a statistical change, its a magician's force.

Since Monty *knows* which door has the prize, his opening one of them is a red herring.  He can always do that no matter what, so doing so doesn't add any information to the situation.

Here's one way I've explained that problem to others.  Suppose its three boxes instead of three doors.  And suppose you pick one.  Now Monty asks if you want to switch: you can keep your box or take *both* of the other two boxes.  Switching is obviously the better choice because you get two shots instead of one.  Now suppose Monty takes both boxes into another room and then comes back out with one of the boxes.  He asks again: switch or stick.  Its still obviously better to switch.  And now he shows you the box he brought out is empty.  That changes nothing.

If it still seems like opening that box changes things, lets try again: this time there are one million boxes.  And you get to pick one.  Now Monty takes the other 999,999 boxes into another room and then asks: keep the one, or switch to the other 999,999.    Of course you're taking the other 999,999.  Now he goes back into the room, comes back out with one box, opens it, and says "keep or switch?"  Its obvious that he can do that at least 999,998 times, but no matter how many times he does that it should not affect your choice.  You should switch.  The fact that Monty can show you a bunch of open boxes is irrelevant: he can always do that because there's 999,999 empty boxes and he has at least 999,998 of them.

Still not convinced?  Ok.  Still one million boxes.  You pick one.  Now Monty takes the other 999,999 into another room.  Once there, he tells you all the boxes in the other room have magically vaporized.  In that room is either a prize, or nothing.  Do you keep your one box, or pick the room that contained the other 999,999?

Isn't it obvious that there's an overwhelming chance the prize was in one of the boxes that Monty took into the room, and whether some of the boxes are opened, removed, or vaporize into thin air changes nothing?

Last thing: imagine this game being played with one catch.  The game is being executed by Penn and Teller, not Monty Hall.  And while you can't see inside the boxes the audience can.  Since the audience can see inside the boxes, they can obviously see that the odds you picked the correct box was one in three, and everything that happens after that point is just misdirection.  It doesn't change anything from their point of view.

In a sense, the game is over the moment you make your first pick.  The audience immediately knows if you win or lose.  And the audience, watching game after game, knows its better to switch.  Everything else, including when someone opens a box, is irrelevant to them. 

But the game can't be different just because an observer that isn't involved with the game in any way has x-ray vision.

[Rotten Luck]

The Myth busters tested this one as well.  Forgot what episode it was on, however they did a side by side test run of switching and non-switching.  Sure enough the one that switched won more times.

[GuyPerfect]

The problem can be demonstrated with two simple observations:

* If you pick the "right" door first, then one of the wrong doors is eliminated, and switching means you'll pick the other wrong door.
* If you pick a "wrong" door first, then the other wrong door is eliminated, and switching means you'll pick the right door.

Basically, the "pick then switch" strategy fails if and only if you select the "right" door first. And there's only a 1/3 chance of that happening, so it is therefore the most advantageous approach.

The issue that occurred in conversation is that every single time Tony presented the scenario, he did so in the context of, as an example, "door number three". At the beginning of the game, door number 3 either has the good prize behind it or it doesn't. That doesn't change whether I select a different door, one of the doors is eliminated, Monty or the audience knows which is the right door, what Donald Trump had for breakfast, or anything. The prize doesn't move, so the likelihood of it being behind door number three never changes.

[TonyV]

Okay, I defy anyone to read the above and try to justify it, how the scenario I presented or the context in which the question was asked makes any difference whatsoever to the correct final answer, which is that you should switch doors.  Again, here is an exact transcript of the scenario I presented, just to make sure it's not taken out of context:

TonyV: Well, here's the set up.  You're on Let's Make a Deal with Monty Hall.  He gives you a choice of three doors to pick.  Behind two of them are "goat" prizes--worthless stuff you don't want.  Behind one door is something really nice worth a lot. So you pick a door.  Go ahead, pick a door.  Which one do you want?  Door number one, door number two, or door number three?

Guy Perfect: Let's go with door number one.

TonyV: Okay, now Monty, like me, knows which door the prize is behind.  But like any good showman, he wants to build up the drama and suspense. So when you pick door number one, he says, "Okay, let's open up one of the other doors first.  How about we see what's behind...  Door number two!" Obviously, he's going to open up one of the goat doors. So now there are two doors left.  Monty says, "Okay Guy, I'll tell you what.  Are you still feeling good about door number one?  Because if you want, I'll let you change your mind and go with door number three.  Of course, if you like door number one, you are more than welcome to stick with that one." The question is, statistically speaking, if you want to win the nice prize, should you stay with door number one, or switch to door number three?  Or does it make a difference?

Because it sounds to me like what Guy is saying that in the abstract, the probability is 1/3 for the door you choose and 2/3 for the door you didn't, but that since I gave him a concrete example with actual doors and an actual prize instead of just asking about it in the abstract, the probability for some weird reason breaks down to 50/50.  It's kind of like saying, "Hypothetically speaking, if there were a meteor in this spot out beyond Neptune and travelling in that direction at that speed, the odds that it would hit the earth on its next pass are 0.0000179%," but then if there were an actual meteor there, the odds would suddenly be 50/50 because either the meteor will hit the earth or it won't.

[Blondeshell]

Soooooooo, Guy, have you put any furniture in your place yet, or are you just camping out on the floor?

</threadjackjack>

;D

[Codewalker]

He said something a few days ago about 'inflating the mattress', so I think he has a bed at least. As someone who slept on an airbed for around 2 years I can attest that the modern generation of them are not only affordable, but actually really nice, resilient, and surprisingly comfortable.

[GuyPerfect]

It's a Select Comfort Sleep Number. Of course it's comfortable!

I went to ye olde thrift store a few days ago to buy a couple of folding chairs and something to put my monitor, keyboard and mouse on. Sitting on the floor to use the computer is quite uncomfortable.

[Arcana]

The problem can be demonstrated with two simple observations:

* If you pick the "right" door first, then one of the wrong doors is eliminated, and switching means you'll pick the other wrong door.
* If you pick a "wrong" door first, then the other wrong door is eliminated, and switching means you'll pick the right door.

Basically, the "pick then switch" strategy fails if and only if you select the "right" door first. And there's only a 1/3 chance of that happening, so it is therefore the most advantageous approach.

The issue that occurred in conversation is that every single time Tony presented the scenario, he did so in the context of, as an example, "door number three". At the beginning of the game, door number 3 either has the good prize behind it or it doesn't. That doesn't change whether I select a different door, one of the doors is eliminated, Monty or the audience knows which is the right door, what Donald Trump had for breakfast, or anything. The prize doesn't move, so the likelihood of it being behind door number three never changes.

The question Tony kept asking, though, is after the first selection was made and two doors remained, "Which of the remaining doors is the better choice?" What this does is funnel down the scenario to the context of two doors, behind one of which is a prize. That means 1/2 chance for either door, and in fact you can flip a coin and be just as likely to pick the prize either way.

Suppose someone starts playing the game, and gets to the point of Monty asking if they want to switch.  At that point another person is led into the studio, and independently asked which box they want to pick.  With no other information, that person would be forced to conclude the choice is 50/50.  But if they were watching the game up to that point, then asked to pick one of the two boxes without having initially participated in the game, the information available would compel that person to pick Monty's box, for the same reason the first player should switch.  The odds are only 50/50 if you know nothing about how the game evolved to that state.  But if you know, it doesn't matter how the choice is presented.

The initial pick itself is also a smoke screen.  Imagine a game where Monty Hall puts a prize in one box, mixes that box up with two other boxes, then divides the boxes up into two groups: one box to the left, two boxes to the right, randomly.  Then Monty Hall looks inside the two boxes, and opens an empty box.  Now which group do you pick: the left or the right.  Answer: the right.  The initial pick, the switching choice, and Monty opening empty boxes are all distraction.  There's only one statistically important fact: the left side has one in three chances of having the prize and the right side has two out of three.  No matter when you're given the opportunity to choose a box, that fact never changes.  As long as you are aware of the initial split in boxes: one this way and two that way, and *that* choice is effectively random, everything else about the game's mechanics is irrelevant to the optimal strategy.  You should pick the side that had the most shots at the prize.


So here's the real test of probability mastery.  You're looking for the bathroom, but you've forgotten which door leads to the bathroom.  You really, really need to use the bathroom and you can't wait long enough to open all three doors.  You mentally pick the door you guess is the right door, but as you run up to that door the wind blows open one of the other two doors, showing its the washing machine.  Should you switch doors?

[Mister Bison]

So here's the real test of probability mastery.  You're looking for the bathroom, but you've forgotten which door leads to the bathroom.  You really, really need to use the bathroom and you can't wait long enough to open all three doors.  You mentally pick the door you guess is the right door, but as you run up to that door the wind blows open one of the other two doors, showing its the washing machine.  Should you switch doors?

Depends on wether or not a god exists and played a Monty Hall on you at that time or not :P

In the two problems, the solution is the same, you should change your door. That's not because the opener knows the solution. It's because he opened a wrong door. If he opened the good door, you wouldn't have to choose anymore, it wouldn't matter.

To solve the problem picture the global outcome. In absolute terms, because it's pure random, you had two chances to be wrong, and 1 chance to be right when choosing your door, out of three chances. Then the wind (assuming the opening was random) also had 2 chances to open the bad door and 1 to be right. You being right and the wind being wrong has probability 2/3*1/3, and you being wrong and the wind was wrong is 2/3*2/3.

Bayes' rule gives that probability you were right knowing wind is wrong (which is the probability we are interested in... I guess), is the probability that you were right and the wind was wrong divided by the probability the wind was wrong, which gives 2/3*1/3 / (2/3) is 1/3.

The (null) difference in the Monty Hall's problem, is that the probability of Monty being right (in his good prize guessing) is 0, whatever the conditional probability. So, in the beginning, the probability of you being right and Monty being wrong is 1/3. Formula gives that probability of you being right knowing Monty was wrong, is probability of you being right and monty being wrong divided by his chance to be wrong, hence, 1/3 / 1.

Also, the magic happens when the outcome, in the end, where magically the only other choice you have, is to switch being right and being wrong. Since a wrong door was opened, that's why "changing door" transforms 1/3 to 2/3 in both cases.

But in your case Arcana, I wouldn't count on chance of being right being only 1/3. It may be in the concious' mind eye, but the unconcious mind has the troublesome habit of skewing you own choice to the one it thinks is the good one, and generally it has a far better memory and statistics.

[TonyV]

So here's the real test of probability mastery.  You're looking for the bathroom, but you've forgotten which door leads to the bathroom.  You really, really need to use the bathroom and you can't wait long enough to open all three doors.  You mentally pick the door you guess is the right door, but as you run up to that door the wind blows open one of the other two doors, showing its the washing machine.  Should you switch doors?

I'd probably switch because, hey, it couldn't hurt, and it beats standing there trying to work out whether the wind opening up one door is a statistically significant event....

But statistically speaking, it doesn't matter.  Since the the wind doesn't know which door is the bathroom, the 1/3 probability of it being the door it opens is shifted equally to the other doors, which now makes it 50/50.  That's why in the Monty Hall problem, it's critically important that you make it crystal clear that Monty knows which door has the prize behind it, and he deliberately opens a door that has a goat.  If you mess that part of the scenario up, then the person you're presenting it to has a legitimate complaint in that you presented it wrong.  But in spite of *ahem...* certain people acting like it's voodoo that affects the probability, it REALLY DOES affect the probability.  If Monty doesn't know which door has the prize and randomly opens one of the other two doors, you have the wind situation above and it makes no difference if you keep your original door or switch.

[Mister Bison]

But statistically speaking, it doesn't matter.  Since the the wind doesn't know which door is the bathroom, the 1/3 probability of it being the door it opens is shifted equally to the other doors, which now makes it 50/50.  That's why in the Monty Hall problem, it's critically important that you make it crystal clear that Monty knows which door has the prize behind it, and he deliberately opens a door that has a goat.  If you mess that part of the scenario up, then the person you're presenting it to has a legitimate complaint in that you presented it wrong.  But in spite of *ahem...* certain people acting like it's voodoo that affects the probability, it REALLY DOES affect the probability.  If Monty doesn't know which door has the prize and randomly opens one of the other two doors, you have the wind situation above and it makes no difference if you keep your original door or switch.

Tony, I think you're wrong. Even if the wind didn't know which door was the good one, it doesn't matter. What matters, is: 1/ You have a flat chance of 1/3 of being right 2/ the opened door is not yours, and 3/ The opened door is not the good choice.

Obviously, if any of the 1/ to 3/ changes, the problem is fundamentally different (you don't have any choice left, or the probability is different, but if they all hold, the result is the same.

Let's take the problem with 4 doors. You have one chance to be right, in the beginning. In the second try, you then have 3/4 chance that one of the remaining doors is the good one, and that is the same as being distributed evenly among the remaining unopened doors. What's important here is that the initial choice of being right is a set probability, and that the remaining set of choices gets reduced.

Or, said in another manner, you can't go against maths. Unless they are badly applied (which you have to prove).

In fact, it's the choice that sets all hells loose. Just the choice.