Do you know the correct answer?

[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)

[Jimmy W]

Nick seems to be ignoring me. Someone ask him if he would prefer to be able to pick 2 doors than to pick just one even although the two doors are known to contain at least one goat.

[Keith Wilson]

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?

Yes, I know something about the other door. The reason that it's different from mine is history. My door has a 1/3 chance of hiding a car. The other door was half of a set of two, which in the aggregate had a 2/3 change of having the car, and one of which certainly hid a goat. The one with the goat was opened. This does not change the odds for the set of two; it's still 2/3.

[Tom Montgomery]

Nick seems to be ignoring me. Someone ask him if he would prefer to be able to pick 2 doors than to pick just one even although the two doors are known to contain at least one goat.

I understand Nick to be saying that he would do so IF the presence of the goat is not actually revealed.

I also understand Nick to be saying that THE ACTUAL REVEAL changes the probabilities to 50% for the Contestant and 50% for Monty.

[Keith Wilson]

If there were three doors, and Monty opened one showing a goat, then asked you to pick one of the other two, the odds would indeed be 50-50.

But that's not what happened. You picked one; he eliminated a goat from the remaining two. Switching therefore gives you the best of two out of three.

[Jimmy W]

Nick chose to ignore this after asking for it.

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.

[David G]

Nick chose to ignore this after asking for it.

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.

I offered up the mathematical formula - using Bayes' - a long while back. No one seemed interested. Here's hoping that it registers this time.

[leikec]

Jeff C

[L.W. Baxter]

"the cows can tape something by now"

[leikec]

"the cows can tape something by now"

Jeff C

[L.W. Baxter]

yeah, that was perfect jeff

[peb]

We did, several times.


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.

33% is not affected by the reveal. Indeed, the reveal tells us nothing regarding the door picked by the contestant. There are three doors, one has a car, 2 do not. When I pick a door, I already know at least one of the other 2 doors has a goat. Him showing that to me does not change anything at all. It only adds information about the closed door I did not pick.

[birlinn]

33% is not affected by the reveal. Indeed, the reveal tells us nothing regarding the door picked by the contestant. There are three doors, one has a car, 2 do not. When I pick a door, I already know at least one of the other 2 doors has a goat. Him showing that to me does not change anything at all. It only adds information about the closed door I did not pick.

The information being that, like your door, the closed door can either have a car or a goat.

[Peerie Maa]

33% is not affected by the reveal. Indeed, the reveal tells us nothing regarding the door picked by the contestant. There are three doors, one has a car, 2 do not. When I pick a door, I already know at least one of the other 2 doors has a goat. Him showing that to me does not change anything at all. It only adds information about the closed door I did not pick.

That does not answer my question.

It is then a pick from one of two unknowns. Show me how that can be anything other than evens if both are unknown?

Remember' the prize was assigned at random, yes?

[Tom Wilkinson]

That does not answer my question.

Remember' the prize was assigned at random, yes?

If you know from the beginning, before making your choice, that one of his doors has a goat behind it ( that is a given), how can “learning” something you already knew change the odds? He gave you no new information. It’s either behind your door, or behind one of his two doors. His odds are better.

[Breakaway]

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

The chosen people?



Kevin


Sent from my iPhone using Tapatalk Pro

[Peerie Maa]

The chosen people?



Kevin


Sent from my iPhone using Tapatalk Pro

Yep, chosen by Universities.

[Boatbum]

I offered up the mathematical formula - using Bayes' - a long while back. No one seemed interested. Here's hoping that it registers this time.

I think the acceptance by Nick will be more visceral acceptance than a formulaic and somewhat esoteric proof. For some there is just a Gestalt where it suddenly makes sense. All the proofs in the world cannot overcome this.

[Decourcy]

Forget the reveal and forget the charts and numbers. It’s a card game. Two jokers and a king. The cards are shuffled and dealt with you getting one card and the host two. The game is played 100 times, with the host always getting two cards to your one. The odds are always 2/3 that the host will have the king. This is a constant.


Revealing a joker of the host, which he will always have, doesn’t change the odds. Thinking so means thinking it’s always a 50-50 game, which, since the host always has 2/3 of the cards, is clearly not possible.



No idea how to make a simpler explanation PM, so if you can’t see it, you’ll never see it.


At which point I’d suggest you spend more time reading up on it elsewhere.

[Decourcy]

Yep, chosen by Universities.

Again, this is why we are doomed as a species.

[George.]

Why. If we know that Monty's door has zero probability, why does the 1/3 go to the other door, and not to the contestant's door? If it can be added to one door, it can be added to the other. Please enplane.

Because the original probability spread is 1/3 to the door you choose, and 2/3 to Monty's two doors. When he shows you one of his doors which has a goat, the probability of that one having the car becomes zero (duh), so the probability of Monty's remaining door having the car has to be the original 2/3.

The sum of the two new probabilities has to equal the original probability (from before Monty opens one door).