Do you know the correct answer?

[AnalogKid]

Nick - try coming up with a scenario where opting for the switch doesn't reverse the original choice.

There'll always be a goat behind one of the two remaining doors, which Monty will reveal to you. If you chose a car, the other door also has a goat, so if you switch doors, you'll swap a car for a goat. If you chose a goat (doesn't matter which one) and Monty reveals the other, then switching doors will swap you goat for the car.

Therefore if the initial pick has a 1:3 probability of being the car, then switching is the reverse of that, 2:3. It's not adding the probabilities, is the probability of picking a goat and then switching (which will always result in a car).

[Peerie Maa]

This is why we are doomed as a species.

Nah, some are trained to think clearly, as their career depended on it.

[Tom Montgomery]

33% goes straight out of the window with the reveal.

It does not. This belief is erroneous. Try playing the game.

[Peerie Maa]

This is why we are doomed as a species.

Nah, some are trained to think clearly, as their career depends on it.

[Tom Montgomery]

Nah, some are trained to think clearly, as their career depends on it.

Everyone is capable of error no matter how well trained to think clearly.

I am asking you to actually test your hypothesis by playing the game as I did. I played 100 times. It goes quickly. And I'm sure it will take fewer plays for you to decide whether or not your hypothesis is correct.
.

[Jimmy W]

And when he opens the door, the game becomes pick one of two unknown doors. Why is that so difficult to grasp?
Can you provide a link that shows adding the two probabilities is a valid action in these circumstances, rather than "sharing" the probability between the two, or adding it to the contestant's door?
Don't make me post the Mrs Doyle clip again.

I have provided a link a couple of times. https://en.wikipedia.org/wiki/Monty_Hall_problem

From that link:

When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter.[9] Out of 228 subjects in one study, only 13% chose to switch.[21] In his book The Power of Logical Thinking,[22] cognitive psychologist Massimo Piattelli Palmarini [it] writes: "No other statistical puzzle comes so close to fooling all the people all the time [and] even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.[23]

Most statements of the problem, notably the one in Parade, do not match the rules of the actual game show [10] and do not fully specify the host's behavior or that the car's location is randomly selected.[21][4][24] Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated.[25]

Although these issues are mathematically significant, even when controlling for these factors, nearly all people still think each of the two unopened doors has an equal probability and conclude that switching does not matter.[9] This "equal probability" assumption is a deeply rooted intuition.[26] People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.[27]

The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as:

The endowment effect,[28] in which people tend to overvalue the winning probability of the already chosen – already "owned" – door.
The status quo bias,[29] in which people prefer to stick with the choice of door they have already made.
The errors of omission vs. errors of commission effect,[30] in which, all other things being equal, people prefer to make errors through inaction (Stay) as opposed to action (Switch).

Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition.[31][32] Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting.[33] There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability".

[rgthom]

Okay, Nick, I'll see if this old dog can learn a new trick:

Contestant Picks this Door	Prize is Behind This Door	Monty Shows This Door	Stays	Switches
1	1	2	Win	Lose
1	1	3	Win	Lose
1	2	3	Lose	Win
1	3	2	Lose	Win
2	1	3	Lose	Win
2	2	1	Win	Lose
2	2	3	Win	Lose
2	3	1	Lose	Win
3	1	2	Lose	Win
3	2	1	Lose	Win
3	3	1	Win	Lose
3	3	2	Win	Lose
		Win %	0.5	0.5

Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.

Edit - Sorry. I had a typo that I corrected.

I think I can explain the problem with this. After the contestant picks a door, there is a 33% chance of the car being behind it. That means there is a 33% chance that there will be some result, either win or lose, of a door with a car. This table shows there is a 50% chance of a result (again any result, win or lose) if the car is behind the door chosen. The chance of there being an output is greater than the chance of an input. This is impossible. The table must be wrong.

[Decourcy]

Nah, some are trained to think clearly, as their career depends on it.

You sound exactly like Aquinian

[Dave Wright]

Nah, some are trained to think clearly, as their career depended on it.

Oh really? Lot's of us have had technical careers and have seen awful mistakes come out of unquestioned technical competence.

Maybe this is what it eventually boils down to:


The doctor says: "I'm going to say 4 words, repeat them after me: elephant, umbrella, orange, bicycle. Now tell me, how are you feeling today Mr. Jones?

50 minutes later the doctor asks: "Can you tell me those four words we discussed at the start of your appointment, Mr.Jones?"

[Peerie Maa]

You sound exactly like Aquinian

You really have a nice line in insults.
Whatever did I do to you?

[Jimmy W]

I must be on ignore.

[Ron Williamson]

OK, explain the mechanism that makes 1/3 become 2/3, based on the rules of probability. Where is it written down?

Monty has two doors(2/3)(one with a goat,for certain).
You have one door(1/3)
If you switch, you get his door (with no goat)(2/3)
End of story.
R

[Peerie Maa]

Monty has two doors(2/3)(one with a goat,for certain).
You have one door(1/3)
If you switch, you get his door (with no goat)(2/3)
End of story.
R

If you think that is a proof, think again.

[lupussonic]

You have one door(1/3)

End of story.
R

No no.

[Ron Williamson]

So with two doors, he doesn't have an advantage over you with one door?
When you switch you don't get his advantage?
R

[ron ll]

Looks like there is never going to be an end to this story. At this point it seems like you all have to ask yourselves who would you accept as an authority on this. I still think it is Schrodinger’s goat. The goat is both behind the door you pick and not behind the door you pick until you open the door.

[Decourcy]

You really have a nice line in insults.
Whatever did I do to you?

Didn’t mean it as an insult. Just that you are defending the indefensible. Which I think is a good lesson for everyone. You are obviously educated and intelligent, but you’ve got yourself caught in a loop. A good lesson to keep in mind when someone that is defending the indefensible elsewhere leads to frustration.

[Peerie Maa]

So with two doors, he doesn't have an advantage over you with one door?
When you switch you don't get his advantage?
R

After he reveals the door with the goat, you have to pick between two unknowns. What are the odd on betting on a choice of two, flipping a coin for example?

[bob winter]

Waste of band width. Nick is clearly correct.

[AnalogKid]

Your post 469 shows that win/lose is always reversed when you switch.

That's all there is to it.

The reveal doesn't do anything to the odds, because the odds only matter in the initial pick, one of three doors.

Form there, you can keep your 1:3 choice, or swap it for the opposite.

[Tom Montgomery]

The reveal doesn't do anything to the odds, because the odds only matter in the initial pick, one of three doors.

Form there, you can keep your 1:3 choice, or swap it for the opposite.

Correct. But the human brain seems to be wired to believe that the reveal changes the odds. It does not. I thought it did at first. But when I was told I was wrong I looked into why I was wrong. I learned something in the process. It is a well-known problem and there are a lot of sources of information on the statistics underlying the solution.

[coelcanth]

And when he opens the door, the game becomes pick one of two unknown doors. Why is that so difficult to grasp?
Can you provide a link that shows adding the two probabilities is a valid action in these circumstances, rather than "sharing" the probability between the two, or adding it to the contestant's door?
Don't make me post the Mrs Doyle clip again.

you are so close to grasping this !

understanding that the 2/3 probability is shared between BOTH of Monty's doors is key to understanding the puzzle..

the probability that one of his two doors has a goat is 100% (and it always was)
the fact that he shows us behind this door only confirms that
yes, the probability that it hid the prize is now 0
but when one of the two doors is open the probability that one of his two contains the prize is still 2/3 !!
the fact that one door is open and flapping in the breeze does not change nor affect that outcome !

think about it !

you cannot adjust the probability of the first pick after the fact because that pick was already made in the past and Monty is not still shuffling the prize around behind the doors.
the game is set
if that first pick was made from the original pool of three then the chances will always stay 1 in 3.
likewise, Monty's two doors together will always have a 2 in 3 probability of containing the prize and that cannot be adjusted either !
opening one door does NOT reduce the choice to two doors nor does it reduce the probability to 1 in 2
that one door has a 0 in 3 chance of containing the prize and we all know it because we can see the goat.
1/3 + 0/3 + 2/3 = 3/3

[oldcodger]

img008.jpg

[coelcanth]

let's try the K.I.S.S. method:

if we baked a pie with one whole plum in it,
and we sliced that pie into three equal shares in which only one slice contained that sole whole plum,
and we put two slices on your table and one slice on my table,
and then you ate one whole slice and told me you didn't have one bit of plum in it,
and, now, left with one uneaten slice on my table and one on yours,
whose slice do you think would be more likely to contain the plum ???

[Dave Wright]

Looks like there is never going to be an end to this story. At this point it seems like you all have to ask yourselves who would you accept as an authority on this. I still think it is Schrodinger’s goat. The goat is both behind the door you pick and not behind the door you pick until you open the door.

This forum is notorious for asking: "give me a link to substantiate what you say."
An "authority" is not necessary when the application of basic principles is involved. If basic principles are not self evident then a lot of education money has been wasted. Basic principles may be difficult to grasp, momentarily confused, or forgotten, that's understandable. But if a point is reached where obstinance and pertinacity prevail over technical competence, then we have an unresolvable situation.

[Tom Montgomery]

img008.jpg

A very good illustration of the game. The numbers of the doors do not matter and are not provided.

Above the first box write 33.33% Probability. Then write 33.33% Actual above the door within the first box.

Above the second box write 66.66% Probability. Then write 0% Actual above the first door within the second box (Goat) and either 0% Actual or 66.66% Actual above the second door within the second box (Car). Monty will ALWAYS open one of his 0% Actual doors and then offer to switch the other door with the contestant's door.
.

[Keith Wilson]

After he reveals the door with the goat, you have to pick between two unknowns.

But the two unknowns are not the same. Staying with your original choice means you have one of the original three doors. Switching means you get two of the original three doors. One had a goat at the beginning, but knowing which of those two it was makes no difference.

[Peerie Maa]

But the two unknowns are not the same. Staying with your original choice means you have one of the original three doors. Switching means you get two of the original three doors. One had a goat at the beginning, but knowing which of those two it was makes no difference.

How are they not the same? What do you know about your door that is different from the other door? What information have you been given about those two specific doors? Anything?
Think it through, don't just post a knee-jerk.

[Peerie Maa]

let's try the K.I.S.S. method:

if we baked a pie with one whole plum in it,
and we sliced that pie into three equal shares in which only one slice contained that sole whole plum,
and we put two slices on your table and one slice on my table,
and then you ate one whole slice and told me you didn't have one bit of plum in it,
and, now, left with one uneaten slice on my table and one on yours,
whose slice do you think would be more likely to contain the plum ???

Nobody knows.
In a sort of sepulcher tone of voice.
Really, it could be in either slice.

[coelcanth]

there is a vast difference between the two sets of doors ! (contestant's vs. MONTY'S)

one was a single door chosen from a set of THREE
one is a door chosen from a set of TWO out of THREE

! ! ! ! ! !

[coelcanth]

Nobody knows.
In a sort of sepulcher tone of voice.
Really, it could be in either slice.

yes, we all agree, it could be in either slice
but one of them is more likely

[Tom Montgomery]

.
All the skeptics who believe the reveal changes the odds to 50/50 should test that hypothesis by actually playing the game. I did so 100 times. I believe it will take you fewer plays to decide if your hypothesis is correct or not. If you find it is incorrect, investigate where the error in your reasoning lies. If you find it correct, I will cry "Uncle" and close the thread.

Deal?

[coelcanth]

How are they not the same? What do you know about your door that is different from the other door? What information have you been given about those two specific doors? Anything?
Think it through, don't just post a knee-jerk.

you have been taken in by the red herring
seeing a goat is no new information

all the information was laid out in the very beginning of the problem
you get to choose one door,
and Monty will have two doors
there is only one prize
that is all the information you need to calculate the probabilities of the whole game
nothing more
knowing that at least one of Monty's doors does not contain a prize was known before a single door was chosen or even opened

[Tom Montgomery]

you have been taken in by the red herring
seeing a goat is no new information

all the information was laid out in the very beginning of the problem
you get to choose one door,
and Monty will have two doors
that is all the information you need to calculate the probabilities of the whole game
nothing more

And this is easily verified by actually playing the game a number of times. But the skeptics appear extremely reluctant to do so.

[Ron Williamson]

After he reveals the door with the goat, you have to pick between two unknowns. What are the odd on betting on a choice of two, flipping a coin for example?

Opened door is a red herring.
Monty has two doors and therefore the advantage.
Switch with Monty and you improve your odds.
R
(GD wonky effing old ipad can't process my gd effing slack assed brainfunction fast enough to post in real time)