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Those maddening "coincidences"!?
#26

Those maddening "coincidences"!?
The Monty Hall problem is named for its similarity to the Let's Make a Deal television
game show hosted by Monty Hall.   This still confuses me somewhat.  My (presumably)
faulty common sense tells me that your final choice of door is 50/50 for the car, so
there's no point in changing your choice.     My maths is obviously lacking LOL.

The problem is stated as follows. Assume that a room is equipped with three doors.
Behind two are goats, and behind the third is a shiny new car. You are asked to pick
a door, and will win whatever is behind it. Let's say you pick door 1. Before the door
is opened, however, someone who knows what's behind the doors (Monty Hall) opens
one of the other two doors, revealing a goat, and asks you if you wish to change your
selection to the third door (i.e., the door which neither you picked nor he opened). The
Monty Hall problem is deciding whether you do.

The correct answer is that you do want to switch. If you do not switch, you have the
expected 1/3 chance of winning the car, since no matter whether you initially picked the
correct door, Monty will show you a door with a goat. But after Monty has eliminated one
of the doors for you, you obviously do not improve your chances of winning to better than
1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3
chance you will win the car (counterintuitive though it seems).
I'm a creationist;   I believe that man created God.
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#27

Those maddening "coincidences"!?
(10-20-2024, 02:19 PM)SYZ Wrote: The Monty Hall problem is named for its similarity to the Let's Make a Deal television
game show hosted by Monty Hall.   This still confuses me somewhat.  My (presumably)
faulty common sense tells me that your final choice of door is 50/50 for the car, so
there's no point in changing your choice.     My maths is obviously lacking LOL.

The problem is stated as follows. Assume that a room is equipped with three doors.
Behind two are goats, and behind the third is a shiny new car. You are asked to pick
a door, and will win whatever is behind it. Let's say you pick door 1. Before the door
is opened, however, someone who knows what's behind the doors (Monty Hall) opens
one of the other two doors, revealing a goat, and asks you if you wish to change your
selection to the third door (i.e., the door which neither you picked nor he opened). The
Monty Hall problem is deciding whether you do.

The correct answer is that you do want to switch. If you do not switch, you have the
expected 1/3 chance of winning the car, since no matter whether you initially picked the
correct door, Monty will show you a door with a goat. But after Monty has eliminated one
of the doors for you, you obviously do not improve your chances of winning to better than
1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3
chance you will win the car (counterintuitive though it seems).

The outcome is a little counterintuitive because of the way that we typically state probabilities. Most people say something like you did above, that the odds are 1 in 3. Stated properly, the odds of picking the correct door purely at random are 1 in 3. Note the bit in bold, it's important.

In the show, your initial choice is entirely random, so you have 1 in 3 odds of having picked the correct door. That means that you have 2 in 3 odds that the car is behind one of the other two doors because the odds must sum to 1.

Next Monty opens one of the other doors to reveal a goat. The goat is vitally important. In no case does Monty ever open a door and go, "Wow! Looks like I just won a car! Lucky me!" The door that Monty opens always has a goat behind it because it is not picked at random. This seems pretty obvious, but it means that any further decisions that you might make aren't purely random either. You're acting on the information that Monty has provided to you and the 1 in 3 odds no longer apply.

Because the odds must sum to 1, the door that you selected has a 1 in 3 chance and the other two combined have a 2 in 3 chance. This doesn't change just because one of them is open revealing the goat behind it. That means that because you have been informed that the odds of the car being behind the opened door are nil, the door that you didn't choose now has the combined odds of 2 in 3 and you should really switch!

One way to look at the apparent paradox is to continue to choose purely at random. That way if you switch, you'd have a 1 in 2 chance of selecting the door that Monty has already shown has a goat. The game doesn't work that way and only an idiot would play that way, but it gets you the expected result that switching is no better than staying. It also demonstrates that purely random behaviour is a losing strategy in an informed scenario. Unless you really want that goat!

Another, more intuitive way to visualize it is to imagine a vast studio for the game with a million doors. You select one at random. Monty then opens 999,998 doors to reveal a herd of goats. Do you still feel like you guessed right at the start when the odds were one in a million or does your gut now say that the car is behind that last closed door?
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#28

Those maddening "coincidences"!?
(10-20-2024, 09:41 AM)SYZ Wrote: It's an interesting question to ask this of
a non-mathematically inclined person:

I flip a coin nine times, and it comes up
heads every time.  What are the odds of
it coming up tails on the tenth flip?

Exceptionally poor if you're sitting in a stats class and the prof hasn't explicitly stated that it's a fair coin.
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#29

Those maddening "coincidences"!?
(10-20-2024, 06:52 PM)Paleophyte Wrote: ...The outcome is a little counterintuitive because of the way that we typically state probabilities. Most people say something like you did above, that the odds are 1 in 3. Stated properly, the odds of picking the correct door purely at random are 1 in 3. Note the bit in bold, it's important.

In the show, your initial choice is entirely random, so you have 1 in 3 odds of having picked the correct door. That means that you have 2 in 3 odds that the car is behind one of the other two doors because the odds must sum to 1.

I thank you for the detailed explanation mate.     Thumbs Up

It's at this point (above) that I get lost!

With only two doors left—one has the car, the
other has the goat.

Therefore the odds of getting the car (or, sadly
the goat) for each of those doors is 1/2.  So it
wouldn't matter whether you change your choice
or not.  

If the car's behind your initial choice of door #1
and you stick with it you win.  If you change your
choice to door #2—as is apparently advised—than
you lose.

  —WTF am I missing?    Huh
I'm a creationist;   I believe that man created God.
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#30

Those maddening "coincidences"!?
Don't look a gift goat in the mouth.
Being told you're delusional does not necessarily mean you're mental. 
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#31

Those maddening "coincidences"!?
(10-21-2024, 11:20 AM)SYZ Wrote:
(10-20-2024, 06:52 PM)Paleophyte Wrote: ...The outcome is a little counterintuitive because of the way that we typically state probabilities. Most people say something like you did above, that the odds are 1 in 3. Stated properly, the odds of picking the correct door purely at random are 1 in 3. Note the bit in bold, it's important.

In the show, your initial choice is entirely random, so you have 1 in 3 odds of having picked the correct door. That means that you have 2 in 3 odds that the car is behind one of the other two doors because the odds must sum to 1.

I thank you for the detailed explanation mate.     Thumbs Up

It's at this point (above) that I get lost!

With only two doors left—one has the car, the
other has the goat.

Therefore the odds of getting the car (or, sadly
the goat) for each of those doors is 1/2.  So it
wouldn't matter whether you change your choice
or not.  

If the car's behind your initial choice of door #1
and you stick with it you win.  If you change your
choice to door #2—as is apparently advised—than
you lose.

  —WTF am I missing?    Huh

You're changing the probabilities, and you can't do that short of actually finding out what's behind the doors, which is precisely what Monty does for you.

When you first select your door your odds are 1 in 3, and that's immutable. Those odds don't change when Monty opens a completely different door. They don't suddenly go from 1 in 3 to 1 in 2. Intuitively that might be what your gut is telling you, but that just goes to show that your digestive system isn't very good at math or reasoning. It would be just as reasonable to suppose that your odds had changed because Monty had the doors repainted. The door that you initially selected has odds of 1 in 3 and that won't change for anything until you open it and find out if it's a goat or a car.

Think of it this way. After you make your first selection, Monty blows up the door and whatever prize may have been behind it. It's gone. Disintegrated at a subatomic level without you ever knowing whether it was a car or a goat. The odds are 1 in 3 that Monty just vaporized the prize car. That means that the odds must be 2 in 3 that he didn't. We now return to your regularly scheduled program and Monty opens one of the remaining doors to reveal a goat. The odds were 2 in 3 that the car hasn't been vaporized before he opened that door and because there's been no further Monty Hall/Michael Bay cross-over those odds remain unchanged. That means there was a 1 in 3 chance that Monty blew up the car behind the first door that you chose, no chance that the goat that Monty revealed is actually a Porche wearing a clever disguise, and thus 2 in 3 odds that the car is safely located behind the surviving unopened door. So Monty's pyrotechnics did you a big favour by forcing you to switch and doubling your odds of winning. And that means that there's a 2 in 3 chance that the SPCA wants to have a stern word with Monty for his mistreatment of the former goat behind the door that he annihilated.
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#32

Those maddening "coincidences"!?
I looked at this on wiki. It told me that I should stay away from math.
Being told you're delusional does not necessarily mean you're mental. 
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#33

Those maddening "coincidences"!?
(10-21-2024, 02:54 PM)Paleophyte Wrote: ...You're changing the probabilities, and you can't do that short of actually finding out what's behind the doors, which is precisely what Monty does for you.

When you first select your door your odds are 1 in 3, and that's immutable. Those odds don't change when Monty opens a completely different door. They don't suddenly go from 1 in 3 to 1 in 2. Intuitively that might be what your gut is telling you, but that just goes to show that your digestive system isn't very good at math or reasoning. It would be just as reasonable to suppose that your odds had changed because Monty had the doors repainted. The door that you initially selected has odds of 1 in 3 and that won't change for anything until you open it and find out if it's a goat or a car.

Think of it this way. After you make your first selection, Monty blows up the door and whatever prize may have been behind it. It's gone. Disintegrated at a subatomic level without you ever knowing whether it was a car or a goat. The odds are 1 in 3 that Monty just vaporized the prize car. That means that the odds must be 2 in 3 that he didn't. We now return to your regularly scheduled program and Monty opens one of the remaining doors to reveal a goat. The odds were 2 in 3 that the car hasn't been vaporized before he opened that door and because there's been no further Monty Hall/Michael Bay cross-over those odds remain unchanged. That means there was a 1 in 3 chance that Monty blew up the car behind the first door that you chose, no chance that the goat that Monty revealed is actually a Porche wearing a clever disguise, and thus 2 in 3 odds that the car is safely located behind the surviving unopened door. So Monty's pyrotechnics did you a big favour by forcing you to switch and doubling your odds of winning. And that means that there's a 2 in 3 chance that the SPCA wants to have a stern word with Monty for his mistreatment of the former goat behind the door that he annihilated.

Thanks again for trying to overcome my failed attempts       Thumbs Up
to rationalise the laws of odds!    

But Monty (and the contestant) each know what's behind
the door that Monty initially opens—which is always one
of the goats.  

The odds that this particular door aren't 1 in 3, but rather 1 in 1.

Therefore, with the remaining two doors—from the perspective
of the contestant, the odds are now 1 in 2 for either of the
remaining two doors, particularly now, for all practical purposes,
the third door no longer exists.

The "odds" of flipping a head after flipping tails 9 times has
nothing to do with the actual fall—or even after 999 tails flips.

And yes...  I failed mathematics in Year 12 at high school,
particularly involving calculus and applied math LOL.

   Colour me embarrassed!     Blush
I'm a creationist;   I believe that man created God.
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#34

Those maddening "coincidences"!?
(10-22-2024, 10:02 AM)SYZ Wrote: Thanks again for trying to overcome my failed attempts       Thumbs Up
to rationalise the laws of odds!    

But Monty (and the contestant) each know what's behind
the door that Monty initially opens—which is always one
of the goats.  

The odds that this particular door aren't 1 in 3, but rather 1 in 1.

Yup. The door that Mony opens has a zero chance of being a car. That's how this problem produces an apparent paradox. The 1 in 3 odds are for probabilities that are independent of one another. These probabilities aren't independent, so the 1 in 3 odds doesn't work. If you redesign the game so the probabilities are independent of one another then yes, in that case there's no advantage to switching. It gets pretty weird though because there's a 1 in 3 chance that Monty will open a door to reveal a car and a 1 in 3 chance that he'll open the door that you picked. If he does both (1 in 9 odds) who wins what?!?

Quote:Therefore, with the remaining two doors—from the perspective
of the contestant, the odds are now 1 in 2 for either of the
remaining two doors, particularly now, for all practical purposes,
the third door no longer exists.

Nope. That's odds for independent, unrelated events. For related events you can't redistribute the odds like that. Effectively what you're saying is that there are 1 in 3 odds of the car being behind any one door, but by opening one of those doors you can change those odds to 1 in 2. Try this:

These are the probabilities before anything happens:

1/3 1/3 1/3

Now I'm a lazy bugger, so I'm going to just pick door number 1:

(1/3) (1/3 1/3) <--- 2/3 combined

Now Monty opens door number 2 to reveal a goat:

(1/3) (0 2/3) <--- 2/3 combined

Door number 1 has to remain with 1 in 3 odds, so by telling us where the goat is Monty has also effectively told us where the car is more likely to be. The probability of the car being behind door number 3 hasn't actually changed, it was always 2 in 3, we just didn't know that until Monty opened door #2. There were 50/50 odds that he was going to open door #3 to reveal a goat and if that had happened we'd know that the 2 in 3 odds were behind door number 2.

Unlike most statistical problems, we don't have to take this one on theory. There's a pretty trivial number of arrangements of doors, cars, and goats. You can write them all out on a single sheet of paper and examine your chances for the two strategies. The "Switch" strategy produces twice the winnings that the "Stay" strategy does.

Quote:The "odds" of flipping a head after flipping tails 9 times has
nothing to do with the actual fall—or even after 999 tails flips.

And those are independent probabilities. The 543rd flip isn't influenced by any of the prior results. With the Monty Hall problem, Monty's choice of doors is dependent on your initial choice and your final choice is dependent on Monty's. It's pretty clear that they aren't unrelated events, so we can't treat them as if they are for statistical purposes.
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