Do you know the correct answer?

[Paul Pless]

This thread is just circling the plug hole.
No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

Tom, you may as well close the thread.

lmfao

'

[Peerie Maa]

If you switch, you only lose if you had chosen correctly. You had a 33% chance of that. It's is as simple as that. Cannot be explained better.

To repeat #619 in slightly different words:

If you don't switch, your odds are one in three.

If you switch, you are effectively getting what's behind two doors out of three. At least one of those two was certain to have a goat, and revealing that makes no difference.

No true
After the reveal, we have two unknown doors and one with zero probability. No other data. Nothing to bias the result. So the odds that the two closed doors are winners are now evens.

[Tom Montgomery]

No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

Tom, you may as well close the thread.

Nick, you simply misunderstand the statistics. I will close this thread if you first play the game for yourself 50 times and report the results to us. I trust you will not cheat.

[Jimmy W]

This thread is just circling the plug hole.
No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

Tom, you may as well close the thread.

From the wikipedia link that I posted back somewhere in this thread.

Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

[Keith Wilson]

No true
After the reveal, we have two unknown doors and one with zero probability. No other data. Nothing to bias the result. So the odds that the two closed doors are winners are now evens.

No. Before opening anything, Monty's two doors have a 2/3 chance of containing a car, and are certain to contain at least one goat. Monty knows where his goat is from the beginning, and opening that door does nothing to change the 2/3 chance.

[Tom Montgomery]

.

Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

As I said, the contestant holds door 1 (say) with a probability of 33.33% while Monty holds two doors - door 2 (say) with with a probability of 66.66% (twice that of door 1) and door 3 (say) with a probability of 0%. Those probabilities will never change. Monty will always offer to switch his door with a 66.66% chance of a car with the contestant's door with a 33.33% chance of a car.
.

[Peerie Maa]

At this point, I think it might be better called "stubbornness". "Best of two" has better odds than "One of three."

Just so. One of two is even odds, best of three is a lot less than evens. You got that bit right.

p.s. I'd really like to see a comment from you or CWSmith about Peb's observation first made in post 123 and repeated or copied a couple of times since. It's pretty hard to reconcile his observation with your viewpoint.

We did, several times.

If you play my the rule to always switch
You only lose if you picked the right door. That's a 33% chance of losing. It is very, very simple.

33% goes straight out of the window with the reveal. It is then a pick from one of two unknowns. Show me how that can be anything other than evens if both are unknown? When the game reaches that stage, it is equivalent to flipping a coin.

[Decourcy]

This is why we are doomed as a species.

[Peerie Maa]

.
As I said, the contestant holds door 1 with a probability of 33.33% while Monty holds two doors - door 2 (say) with with a probability of 66.66% (twice that of door 1) and door 3 with a probability of 0%.

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.

[Keith Wilson]

33% goes straight out of the window with the reveal. It is then a pick from one of two unknowns.

But those unknows were not created at random; see 663 and 664. One of the unknows, the door you picked first, has a 1/3 chance of hiding the car. The other has a 2/3 chance.

[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.