Do you know the correct answer?

[Breakaway]

The same old same old is because they are explaining what is really going on.

?

Right!
Two plus two is four and the explanation for that remains the same, regardless of how many times it is repeated or reiterated.


Sent from my iPhone using Tapatalk Pro

[jpatrick]

Nick is one voice for a particular point of view. And there are many voices with a contrary point of view. So tell me, which point of view is most responsible for keeping this thread alive?

Jeff

[Peerie Maa]

That fault is that you continue to insist a random hand, with one prize in three, has the two card hand statistically equivalent to a hand of one. No matter how you try to work numbers and proofs that result is a logical impossibility.

That reads like gobbledygook. Do you want to try again in clear English, even clear Canadian would do.
I am pointing out that two options, with the prize assigned at random has even odds.
The third option is shown to be a red herring. The important fact is that there is a choice from two to be made to conclude the game, and there is nothing to distinguish between them.

[Jimmy W]

Nick is the most consistent voice with the wrong answer and the one implying that his better thinking abilities and university training make him correct.

[Paul Pless]

nick is the only poster i have seen thus far calling for tom to close the thread

[Jimmy W]

There is something to distinguish between them and that is that one is the result of choosing one door and the other is the result of receiving two doors and showing a losing door.

[Peerie Maa]

nick is the only poster i have seen thus far calling for tom to close the thread

Like I said above. I am bored with the same old same old being repeated as if it adds anything to the thread.

I'll see you somewhere more interesting.
Providing it is not stunk up with flatus.

[coelcanth]

I looked Bayes rule up in Wiki.
But in this case, if event A is the opening of the door, it has no relationship to event B, as it tells us nothing about the probability that the prize is behind either of the closed doors. We knew before Monty opened the door, that it would be one of them, as Monty was not going to reveal the prize, but we still have no data about which, so event A has no effect on event B.

close, but again you've made a fatal error..

"Event A" is NOT the opening of any door, or the showing of the goat in Monty's hand..

that event has no effect on any outcomes, except to illustrate that one door is a NON-FACTOR in the final outcome of the game
opening the door is a red herring and the sleight of hand
the magician has taken you in !

the first event (Event A in reference to Bayes' theorem) would be the first player's choosing of their FIRST door
the choice when the POOL of possibilities was THREE DOORS (1 in 3)
that first choice, at that moment, is the event that sets the probabilities for the rest of the game.

whatever prize or goat was behind that door at that moment can never move across the aisle to Monty's side or behind the two doors.
the hand has been dealt and will remain unchanged until a winner is declared.


the second event (that's B) is the second half of the game, when the player is asked to choose their door or Monty's

[Paul Pless]

coelcanth, i don't think the basis of the game can be explained any more clearly

[Peerie Maa]

close, but again you've made a fatal error..

"Event A" is NOT the opening of any door, or the showing of the goat in Monty's hand..

that event has no effect on any outcomes, except to illustrate that one door is a NON-FACTOR in the final outcome of the game
opening the door is a red herring and the sleight of hand
the magician has taken you in !

the first event (Event A in reference to Bayes' theorem) would be the first player's choosing of their FIRST door
the choice when the POOL of possibilities was THREE DOORS (1 in 3)
that first choice, at that moment, is the event that sets the probabilities for the rest of the game.

whatever prize or goat was behind that door at that moment can never move across the aisle to Monty's side or behind the two doors.
the hand has been dealt and will remain unchanged until a winner is declared.


the second event (that's B) is the second half of the game, when the player is asked to choose their door or Monty's

So what is the relation ship between those two that affects the outcome? There is none. The choice is still between randomised events of equal probability.

[AndyG]

I'd just like to add that it's midnight here, and my tablet is on 12% battery.

And is this really day three?

Goodnight, all.

Andy

[Peerie Maa]

The probability is 66.66666666666666666666666666666666666666666666666 66666666666666666666666666666666666666666666666666 666666666667%


That's about as precise as I am willing to make it.

Same old, same old, same old, same old.. Booooring.

[Peerie Maa]

I'd just like to add that it's midnight here, and my tablet is on 12% battery.

And is this really day three?

Goodnight, all.

Andy

Gnight Andy.

[Ron Williamson]

So what is the relation ship between those two that affects the outcome? There is none. The choice is still between randomised events of equal probability.

Rien. Les jeux ont etait fait quand le gar a choisi une porte et Monty reste avec deux portes.
(My apologies to any Francophones)
R

[Mcjim]

Another way of looking at it is if Monty doesn't open either of "his" doors but asks "Do you want to change from your door to the best option of my two doors?" If you swap, there is a 2/3 chance that one of his doors has the prize. As at least one of the doors must be non-optimal, Monty opening the non-optimal door doesn't make any difference to the game.

Likewise if Monty opens both of his doors and asks "do you want to swap to the best of these doors" you would want to change 2 out of three times.

[ShorelineJohn]

So calling all STEM teachers, we need help! So this is a "word problem", I was taught the whole trick to "word problems" (real life stuff) is to construct the equation correctly. The proof is the equation itself, seem like a simple 2 stage, one line equation is all that is needed. The problem I have is how to express the subtraction of the choice that that is the less of the two remaining choices, I think what He really does is subtract the most valuable choice that is less than the grand prize. (So Bob Bubank could describe all the things America could just go out and buy). The odds will work out the same regardless which of the two algorithms are used as long as the grand prize is not subtracted from the remaining choices.
I can construct the whole thing in python with one line, but that really a computer command not a equation. The equation should express the odds of one or the other remaining choices as one can get the other by subtracting the one you have from 100.

[Beowolf]

Wow…. 20 pages.

I’ve been out for the last half dozen pages, so I’ll take a swing.

No one reads my sh!t anyways.

Nick,

If Monty was selecting from his two doors at random and happened to select a goat, then you would be correct, the odds of either door containing the car would be 50/50.

But he’s not. Monty knows the locations of the car and goats from the outset. The draw is only random for the contestant.

Once Monty makes the informed choice to remove a goat from his two, the remaining door now has the odds are locked in a 1/3 and 2/3.

[coelcanth]

So what is the relation ship between those two that affects the outcome? There is none. The choice is still between randomised events of equal probability.

what is the relationship that affects the outcome ?!??

don't you see it by now ?
by choosing the first door we have removed that one door from the pool of three !
that is exactly it !! that is the exact event that affects the outcome of the final choice

MONTY NOW ONLY HAS TWO DOORS FROM WHICH TO PICK THE PRIZE WHEREAS WE HAD THREE

that is why he has better odds than the first player

[coelcanth]

If Monty was selecting from his two doors at random and happened to select a goat, then you would be correct, the odds of either door containing the car would be 50/50.

But he’s not. Monty knows the locations of the car and goats from the outset. The draw is only random for the contestant.

Once Monty makes the informed choice to remove a goat from his two, the remaining door now has the odds are locked in a 1/3 and 2/3.

no that's not it

as long as the contestant has to pick first, their door will always have worse odds than Monty's two doors

Monty is not cheating at the game and he's not 'removing' the goat from play
even with a 1 in 3 chance and without switching doors, the player will still win the prize sometimes, there's just a smaller chance
it's never 50/50

[Canoeyawl]

Flatus has the distinct odor of our last Pratus and Flotus

[CWSmith]

I have a separate question:

The Monty Hall fans are legendary. The participants are rabid.

If there was a double probability of winning the car if you switch, and since this problem has been published on the internet for a long time, and since the fans watch the show religiously, wouldn't every participant always switch to the point that the show would stop running this game? I don't think they have because I don't think the viewers have seen the pattern that some claim is so clear.

[rgthom]

I have a separate question:

The Monty Hall fans are legendary. The participants are rabid.

If there was a double probability of winning the car if you switch, and since this problem has been published on the internet for a long time, and since the fans watch the show religiously, wouldn't every participant always switch to the point that the show would stop running this game? I don't think they have because I don't think the viewers have seen the pattern that some claim is so clear.

This puzzle is called the Monty Hall game, but I understood that the lets make a deal show had different rules.

[Decourcy]

This simply has to be leg pulling. No-one could be so intentionality blind……could they?

[Jimmy W]

no that's not it

as long as the contestant has to pick first, their door will always have worse odds than Monty's two doors

Monty is not cheating at the game and he's not 'removing' the goat from play
even with a 1 in 3 chance and without switching doors, the player will still win the prize sometimes, there's just a smaller chance
it's never 50/50

Correct.

So what is the relation ship between those two that affects the outcome? There is none. The choice is still between randomised events of equal probability.

Nick keeps saying it is between randomised events of equal probability and it isn't.
Monty opening one door showing a goat doesn't make any difference, but that could only be random if the the player picked the car to start with. Otherwise which door he opens is determined by the position of the car.

[Beowolf]

no that's not it

as long as the contestant has to pick first, their door will always have worse odds than Monty's two doors

Monty is not cheating at the game and he's not 'removing' the goat from play
even with a 1 in 3 chance and without switching doors, the player will still win the prize sometimes, there's just a smaller chance
it's never 50/50

Yes. It is.

The OP states that the host is aware of what’s behind all three doors. Thus, if he opened a door to reveal a goat, it’s fair to say that he intentionally removed that goat.

Edited to add: Wait… aren’t we in agreement, here?

[Jimmy W]

I have a separate question:

The Monty Hall fans are legendary. The participants are rabid.

If there was a double probability of winning the car if you switch, and since this problem has been published on the internet for a long time, and since the fans watch the show religiously, wouldn't every participant always switch to the point that the show would stop running this game? I don't think they have because I don't think the viewers have seen the pattern that some claim is so clear.

As rgthom said, the problem is called the Monty Hall Problem. I really don't know, but do you know if this ever really happened or if it is a hypothetical problem based on the format of the show.

[L.W. Baxter]

monty hall is dead.

long live the monty hall problem!

[coelcanth]

Yes. It is.

The OP states that the host is aware of what’s behind all three doors. Thus, if he opened a door to reveal a goat, it’s fair to say that he intentionally removed that goat.

Edited to add: Wait… aren’t we in agreement, here?

no, you and Nick are both stuck on the same point..

revealing a goat in the second round does nothing to affect the probability of a pick that was made in the past

thinking about the reveal as 'removing' that goat from play is probably the wrong way to go about it and that's what's confusing you.
revealing the specific location of the goat really makes no difference (we already knew Monty had to have AT LEAST ONE goat, right?)
the particular door that it is in has no effect on whether Monty has the prize or not

if we think of Monty's two doors as his entire hand we can say that his entire hand has a 2 in 3 chance of containing the prize
opening the one door means that entire 2 in 3 chance now rests on the only unopened door in Monty's hand

that is why his last door has a 2/3 chance and your one earlier pick has only a 1/3 chance of winning

[rgthom]

I have a separate question:

The Monty Hall fans are legendary. The participants are rabid.

If there was a double probability of winning the car if you switch, and since this problem has been published on the internet for a long time, and since the fans watch the show religiously, wouldn't every participant always switch to the point that the show would stop running this game? I don't think they have because I don't think the viewers have seen the pattern that some claim is so clear.

CW, if you are back could you please take a look at my post 667. I think this shows there is a problem in your table.
Nick is a lost cause, but I still hope to convince you.

[ShorelineJohn]

The show has known about this for most of the show, I watched a show years ago where Monty Hall himself talked about it. And the internet was decades away! The idea the show has a problem with this is silly, they have loved it. The show is for advertisers to highlight their products while the audience is watching a show. This is the stuff dreams are made of for those guys!
I do think the equation is to subtract the highest priced prize that is less than the grand prize from the two remaining choices.

[Beowolf]

no, you and Nick are both stuck on the same point..

revealing a goat in the second round does nothing to affect the probability of a pick that was made in the past

thinking about the reveal as 'removing' that goat from play is probably the wrong way to go about it and that's what's confusing you.
revealing the specific location of the goat really makes no difference (we already knew Monty had to have AT LEAST ONE goat, right?)
the particular door that it is in has no effect on whether Monty has the prize or not

if we think of Monty's two doors as his entire hand we can say that his entire hand has a 2 in 3 chance of containing the prize
opening the one door means that entire 2 in 3 chance now rests on the only unopened door in Monty's hand

that is why his last door has a 2/3 chance and your one earlier pick has only a 1/3 chance of winning

Go back and read my post. Our last sentences are nearly word-for-word identical.

I am in not in any agreement with Nick in this.

So it’s either my piss-poor writing skills or your reading comprehension.

I’ll tell you what, how ‘bout we split that 50/50?

[George Jung]

33%, best offer!

[Beowolf]

33%, best offer!

Oh no…. I’m switching to the other door!

[Nicholas Carey]

The show has known about this for most of the show, I watched a show years ago where Monty Hall himself talked about it. And the internet was decades away! The idea the show has a problem with this is silly, they have loved it. The show is for advertisers to highlight their products while the audience is watching a show. This is the stuff dreams are made of for those guys!
I do think the equation is to subtract the highest priced prize that is less than the grand prize from the two remaining choices.

Also bear in mind that in the real Let's Make a Deal!, there's a whole lot more going on.

Monty sometimes offers cash up front to entice the contestant to switch (or not switch).

The lovely Carol Merrill pops up out of a trap at random with a cart carrying a box that might or might not contain something of value, and that gets figured into the equation, too: "Stick with the door you selected? Or switch to the box that the lovely Carol Merrill has for you?"

Sometimes, it's even, "You switch doors? Great! I've got another deal for ya -- the lovely Carol Merrill has just brought this box on stage. Would you be interested in switching to that?"

The "Monty Hall Problem" is the stripped-down minimal example for analysis.

[lupussonic]

How can the probabilities of choosing between 2 different doors be 30% and 50%?