Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

(yes, the classic Monty Hall problem).

yes

no

What are you going to do with this information?

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Ed (modelamac)

I think I will just put an OUT OF ORDER
sticker on my head and call it a day.

I pick the goat. Goats are cool, and I won't have to mow my grass in the back yard.

Quote
modelamac
What are you going to do with this information?

Yeah, exactly. Is this data being used by the CIA or something? Once we answer, should we all get VPNs and never type from anywhere except libraries?

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Yes.

[priceonomics.com]

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MacResource User Map: [www.zeemaps.com]#

Didn't check the above link, but I once read an article that explained that the odds are better if you change your original vote in that situation. Can't remember why though.

I think people that choose door #1 are not processing the operators in the correct order.

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In tha 360. MRF User Map

Well yeah. If you stick with your original choice, you’ve got a 1 in 3 chance of winning. The reveal doesn’t change that. When you switch, you improve your odds to 1 in 2.

The reveal does change the chances of winning. They become 1 in 2 if you stay and 1 in 2 if you switch.

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PeterW
The reveal does change the chances of winning. They become 1 in 2 if you stay and 1 in 2 if you switch.

1/2=1/2

Math says yes.

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rj
AKA
Vreemac, Moth of the Future

Here's another way to put it:

The host hands you a single six-sided die and says, "Rolling 1or2 selects door #1, rolling 3or4 selects door #2, rolling 5or6 selects door #3. Roll the die."

Your chances of rolling the 'right' door is 1 in 3, or roughly 33%.

If you roll a 1 (door #1), and the host says, "Congrats - door #3 had goats!," and hands you the six-sided die again, saying, "Do you want to roll again?"

This time, rolling 1or2or3 selects door #1, rolling 4or5or6 selects door #2.

Now, your chances of rolling the 'right' door is 1 in 2.

You should roll again.

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rj
AKA
Vreemac, Moth of the Future

Should I use PEMDAS or "left to right" for this decision?

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“Live your life, love your life, don’t regret…live, learn and move forward positively.” – CR Johnson
Loving life in Lake Tahoe, CA

Can you listen to the sounds coming from behind the doors?

Quote
space-time
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

When only doors 1 and 2 remain, there's an equal chance the car may be behind either one. Unless you glean some information from the way the host asks "Do you want to pick door No. 2?", you still have a 50% chance of winning by choosing either door. Why change your choice?

Edit: As Jimmy D shows above.



Edited 1 time(s). Last edit at 08/02/2019 11:39AM by neophyte.

still have a 50% chance of winning by choosing either door. Why change your choice?

wrong... your initial chance was 1/3. the fact that the host opens another door does not change the odds.

Maybe.

Quote
space-time
still have a 50% chance of winning by choosing either door. Why change your choice?

wrong... your initial chance was 1/3. the fact that the host opens another door does not change the odds.

Hmm, the original choice had 3 doors. By asking if you want to change, you now are choosing between 2 choices. Suppose you decided to change to door 2, then before the host could react you changed back to your first choice, door 1. Would you then be changing from a 1 in 2 chance back to a 1 in 3 chance when only 2 doors are left? It seems to me that by introducing the choice to change, the statistics change to match the choices available. Obviously, I am not a statistician by training.

Quote
space-time
still have a 50% chance of winning by choosing either door. Why change your choice?

wrong... your initial chance was 1/3. the fact that the host opens another door does not change the odds.

Let's change this game a little, and make it Russian Roulette. 1 bullet in a six shot revolver. Your initial odds of getting shot are 1 in 6.

And lets say you pull the trigger 5 times and it doesn't fire.

For that 6th shot, would you bet your life that the odds of getting shot are still 1 in 6?



Edited 1 time(s). Last edit at 08/02/2019 12:48PM by GGD.

Of course not. But that is a different scenario. Don’t feel bad because you don’t understand tho Monty Hall problem, a lot of smart people don’t.

[en.wikipedia.org]

It's all about possible misdirection

if the referee in the game of russian roulette knew where the bullet was at all times and right after you pulled the trigger on an empty chamber asked, 'don't you think you might wanna roll that barrel again? or maybe not?? hey - it's your call [smirk/smile]' - might that change the math a bit?

Quote
space-time
Of course not. But that is a different scenario. Don’t feel bad because you don’t understand tho Monty Hall problem, a lot of smart people don’t.

I assume this response is directed to GGD's Russian Roulette scenario. His "different scenario" involves a static situation, that is, there is no reduction in odds because the odds of any 1 of 6 chambers coming under the firing pin is the same with each trigger pull (true Russian Roulette involves a random spin of the chamber between trigger pulls).

The door choice DOES change the number of doors available at each stage of choosing, so I would think the statistics change based on the number of available choices at that time, because you are given the opportunity to change your choice. If the host had turned all doors at once, your odds were 1 in 3. But in reality, the host changed the game to a choice between 2 doors.

Quote
neophyte

Quote
space-time
Of course not. But that is a different scenario. Don’t feel bad because you don’t understand tho Monty Hall problem, a lot of smart people don’t.

I assume this response is directed to GGD's Russian Roulette scenario. His "different scenario" involves a static situation, that is, there is no reduction in odds because the odds of any 1 of 6 chambers coming under the firing pin is the same with each trigger pull (true Russian Roulette involves a random spin of the chamber between trigger pulls).

I must admit I had the rules for Russian Roulette wrong, forgot about the spinning before each shot. Lack of practice, I've never actually played.

I want to switch to door #3.

I have 3 cars, but no goats.

Quote
rjmacs
Here's another way to put it:

The host hands you a single six-sided die and says, "Rolling 1or2 selects door #1, rolling 3or4 selects door #2, rolling 5or6 selects door #3. Roll the die."

Your chances of rolling the 'right' door is 1 in 3, or roughly 33%.

If you roll a 1 (door #1), and the host says, "Congrats - door #3 had goats!," and hands you the six-sided die again, saying, "Do you want to roll again?"

This time, rolling 1or2or3 selects door #1, rolling 4or5or6 selects door #2.

Now, your chances of rolling the 'right' door is 1 in 2.

You should roll again.

Thank you for trying to clarify using a single, six sided die.

EXCELLENT use of a well known gambling device to explain and illustrate probability.

Only to fit the problem in the logically correct way, the illustration should be thus:

Initially, two sides of the die have #1 on them.
Initially, two sides of the die have #2 on them.
Initially, two sides of the die have #3 on them.
#'s 1, 2, & 3 each represent the doors on the game show floor.
At this point, what's behind all doors is unknown.
Whichever door you choose has a 33.3% chance of being the Ferrari.

You choose door #1, with its 33.3% chance of hiding the prancing horse by virtue of its two sides on the die. The other four sides represent the house's, or Game Show Host's chances of winning, or keeping the expensive sports car.

The Game Show Host hands you the die, and just as you're about to roll it to find out your fate, the GSH stops you, and asks if you'd like to find out what's behind one of the two doors you didn't choose...

You can say yes, or no. Two choices.

Clearly, if you say no, nothing has changed, and your choice of door #1, and it's two represented sides of die still give you a 33.3% chance of riding off on the bright red Italian Stallion. The GSH retains the other four sides of the die.

If you say yes, which this problem is based upon doing, two things can happen. GSH chooses door #3. Since you chose door #1, GSH could only choose between doors #2 or #3. Before door #3 opens, all doors still have their 33.3% chance of revealing the Ferrari.... so, when door #3 opens an reveals the Ferrari. Game Over. Result one of two.

But that's not what happens. Door #3 reveals a goat when it's opened. Other possible result of two.

Now you still have that six sided die in your hand that hasn't been rolled yet. But now, two of the GSH's four sides that represented door #3 are essentially worthless, unless you really like goat stew more than strutting around in a free Ferrari. Remember, those two side of the die with #3 on 'em were/are the GSH's sides of the die.

Assuming you're on a goat free diet, when GSH offers you the opportunity to switch, what has happened to the die? By revealing #3 as the goat, he is converting his #3's to two more #2's, representing his remaining unknown door.... you haven't switched yet, so you still have your two sides of the die represented by #1, for your unknown door, and GSH still has his four sides of the die, only now they're all represented by #2. So, when he pulls out another six sided die with your two #1's, and his four #2's on its sides, and asks you if you want to swap doors before you roll the new die, you'd have to be freaking crazy not to swap.

That's how you use the six sided die to properly illustrate this problem. And yes, Marilyn vos Savant and I had a brief thing back in the 1980's.
==

"Now you still have that six sided die in your hand that hasn't been rolled yet. But now, two of the GSH's four sides that represented door #3 are essentially worthless, unless you really like goat stew more than strutting around in a free Ferrari. Remember, those two side of the die with #3 on 'em were/are the GSH's sides of the die."

Not worthless, rather meaningless. At that point, the game now has only two possible outcomes, and a two sided coin will represent them, not a three sided object. If all three doors are in play, and are all opened simultaneously, the host indeed holds odds on two of the three paired sides of the die. But once one door is opened and revealed to both you and the host, the game has changed, as have the statistics.

Or am I somehow being obtuse?

Or am I somehow being obtuse?

Not obtuse. Just wrong.
You're changing a problem that hasn't been changed the way you think it has.

You pick one door. House keeps other two. Just because house didn't win when goat was revealed doesn't change problem itself, it only changes odds relative to finding car if/when a choice is allowed to change. House had 2/3 chance of having car before goat was revealed. You had 1/3 chance. House still has 2/3 after reveal, if reveal is goat. You still have 1/3 chance if reveal is goat. If house wins upon first reveal, house has 100% of winning, you have 0%. Initial odds remain in place until game is totally won, or lost, depending on which side you're on. The event that changes the odds is that win or loss, not any intermediate choices that do not result in a win or loss. If house offers to swap you its 2/3 chance w/ door #2 for your still 1/3 chance w/ door #1.... take it.
==

Why did the skunk go to church?






He wanted to sit in his own pew.

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Ways to improve web conference image and sound quality. [forums.macresource.com]


Phony Stark's Robot

"House had 2/3 chance of having car before goat was revealed. You had 1/3 chance."

Agreed.

"House still has 2/3 after reveal, if reveal is goat. You still have 1/3 chance if reveal is goat."

This is counterintuitive to me. Only 2 doors remain; house and I both know this. As I see it, house LOST some of its odds advantage in that one of its 2 doors revealed a goat. I however have lost no odds because my door is still in play.

Another way I look at the original question (whether to take house's offer to change my choice after the first reveal) is that with 2 doors remaining, one hiding a car and one hiding a goat, the odds are equal that I could win or lose, so I see no compelling reason to change my choice.



Edited 1 time(s). Last edit at 08/02/2019 05:57PM by neophyte.

here is a very simple way to look at it

100 doors. You pick one. 99 other doors that you didn't pick

The house opens 98 doors with goats. There is one door left closed, and of course the door you picked initially. So 2 doors closed total, out of 100.

do you switch?



Edited 1 time(s). Last edit at 08/02/2019 06:45PM by space-time.

Maybe this illustration will help....

"Another way to look at this is to break down every door-switching possibility. As we’ve delineated below, 6 out of the 9 possible scenarios (two-thirds) result in winning the Ferrari":


==

Quote
GGD
Can you listen to the sounds coming from behind the doors?

Or go up and smell around the doors?

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Edited 1 time(s). Last edit at 08/02/2019 07:00PM by lost in space.

OK, let's go back to rolling the six sided die, and skip the goats and car for now. You win $1 from space-time if you hit your roll, I win the $1 from him if you don't.

You get any two sides, and I get the remaining four sides.
You have 33.3% chance of winning, and I have a 66.7% chance of winning.
After picking your two sides, but before rolling the die, the Gambling Commissioner jumps in and says, "That's not fair, Buzz has twice as many sides as you. Since Buzz tried to screw you, the Gambling Commission would like to offer you the opportunity to trade your two sides of the die, for Buzz's four sides of the die. Do you want to trade?"

The fact that there is such a disparity between a goat and a Ferrari, and we remember Monty Hall haggling over buying the door, or box, or whatever alternative, clouds the basic logic of the problem being presented. I'm pretty sure that's a big part of why the "Monty Hall problem" has survived as long as it has. The roll of the single, six sided die for a buck just simply isn't as sexy, but the problem always boils down to who gets the 1/3 chance, and who gets the 2/3 chance. The haggling GSH, braying goats, and sensational sports car, are just smoke and mirrors for who gets two sides of the die, versus who gets four sides.
==

Quote
jimmy d

This graphic that I posted previously remains the simplest proof that changing doors results in winning the car 2/3 of the time while sticking wins 1/3. The graphic includes all possible moves and results. What could be clearer?



Edited 1 time(s). Last edit at 08/02/2019 11:00PM by jimmy d.

Clear, but totally wrong.

Once one of the choices is removed, the odds change. Whatever door is revealed is no longer a choice. The odds of the car being behind your door is 50% and 50% for the unchosen door. The initial number of doors are meaningless as long as you only have two choices at the end.

The “Monty Hall” question is so delicious. I used to bring it in up class when teaching adults adults.

I remember being confused by it in Marilyn vos Savant’s “Ask Marilyn” column in “Parade” magazine in 1990. It took me (and many others) multiple attempts at explaining to see how one increases one’s odds from 1:3 to 2:3 by switching.

It also became an excellent study in sexism. The Wikipedia article cited by jimmy d (above) explains some of this. Here is another article that goes a bit more in depth.

Kevin Spacey (playing an MIT math professor) also used the question as a screen to pick his protogés to try to beat the odds against Las-Vegas casinos in the 2008 movie “21”.

It’s amazing how hard it is for people to change their minds about this question. Once people understand that (in the original scenario) Monty does not randomly open a door, but is forced by the rules to always show a goat, they can more easily understand how switching increases (but does not guarantee) one’s chance of winning.

My personal experience has been to emphasize this aspect. The rules force him to NEVER open the door with the prize. That effectively changes the situation from asking if the player would like to switch (after exposing a door he ALWAYS knows does NOT have the prize) to not exposing a door, but asking, “Great, you picked door X. Would you like to stay with that door, or would you like to switch to the TWO other doors?“

The two scenarios are identical. If framed that way, most people see that staying with the original choice maintains the odds at 1:3, but switching to the new option increases the odds to 2:3.

A similar confounding question (though easier for most people to finally come to the right conclusion) is determining the least number of rounds with a balance scale to determine the one heavy ball in a set of eight billiard balls.

Todd’s always-stumbling, always-learning keyboard

Quote
PeterW
Clear, but totally wrong.

Once one of the choices is removed, the odds change. Whatever door is revealed is no longer a choice. The odds of the car being behind your door is 50% and 50% for the unchosen door. The initial number of doors are meaningless as long as you only have two choices at the end.

This is my thinking too. And clearly stated. Thank you.

"here is a very simple way to look at it
100 doors. You pick one. 99 other doors that you didn't pick
The house opens 98 doors with goats. There is one door left closed, and of course the door you picked initially. So 2 doors closed total, out of 100.
do you switch?"

So 2 doors closed out of 2 possibilities now in play, not 100. Thus 50% chance either hides the car.

The key idea in my mind is that every reveal removes a contending door from the original game, and a new statistic is thus required to calculate the odds for the situation that is now in play.

Lies, damned lies, and statistics.

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Todd's keyboard
The “Monty Hall” question is so delicious. I used to bring it in up class when teaching adults adults.

I remember being confused by it in Marilyn vos Savant’s “Ask Marilyn” column in “Parade” magazine in 1990. It took me (and many others) multiple attempts at explaining to see how one increases one’s odds from 1:3 to 2:3 by switching....

I remember reading about it, too. And it confounded then and still does, and further confounded me because I always thought of myself as having a strong ability to deduce logic and math. Even with jimmy d's graph, it still confuses me how the odds improve simply by changing my choice.

Part of my confusion came from imagining if I were in Monty's position: If the contestant chose Door 1, and I/Monty new that Door 1 contained the car, and I did NOT want the contestant to win, then one of my strategies would be to expose one of the door goats, and then say, "You sure you don't want to reconsider and perhaps choose another door?"

In any case, I love this example because it's very humbling. I'm not as smart as I think.

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I think the most confusing parts of the question are:

1) People don’t immediately realize that Monty is not randomly choosing a door to open. If Monty’s reveal were random, then there would not be any advantage to switching one’s choice after the reveal. Monty is bound by the rules to never open the guesser’s door and never reveal the grand prize.

2) An extrapolation from 1 (above) is that Monty has any free will in the matter. For example, neophyte wrote (above):

“ ... Unless you glean some information from the way the host asks (italics added) ‘Do you want to pick door No. 2?‘,... “

Monty’s tone of voice (and/or interpreted opinion on the matter) is already there in the rules. Monty always has to reveal the goat, AND he can never reveal the prize.

This really is such a great question. I’ve also used neophtye’s situation of making it 100 doors (instead of 3) to show that it increases one’s chances to switch after the reveal.

Imagine picking 1 door out of 100. Monty is required to reveal a door after you choose. The reveal can never be the prize and can never be the guesser’s current choice. (Remember, Monty is not revealing a door randomly; he knows more than the guesser.)

Let’s say the guesser chooses door 27. Monty knows the prize is behind door 59 (and will remain behind door 59 for the entire game).

This version of Monty is methodical. He reveals doors one at a time, in numeric order. He reveals door 1 and gives the guesser the choice of staying with 27, or switching to any other door. Guesser switches to 20.

Monty then reveals door 2 and gives the option to stay with #20 or switch to any other unopened door (doors 1 and 2 are not out of the game). To build suspense, the pattern of reveal and option to switch to any remaining door continues until only two doors are left.

The pattern continues. In turn, Monty reveals (in order) all the doors to 58. In the next iteration he HAS TO skip 59 (which has the prize). The rounds continues with doors 60 through 100 being revealed (minus the guesser’s current, latest choice).

Wouldn’t it be better to make your final choice the door that Monty skipped in the progression (59)?

Viewed another way: The way the rules work for Monty is that he is (without saying it clearly) always offering the guesser the choice between sticking with his/her original set of one door, or switching to the set that includes ALL of the other doors. In the original question, switching increases the odds from 1:3 (guesser’s original set) to 2:3 (set of all other doors). In the example of 100 doors, the odds change from 1:100 (guesser’s set) to 99:100 (set of all other doors).

What a great question.

Todd’s still-amazed-after-several-decades keyboard

space-time's OP is not a pure chance game because the host knows what's behind each door. Thus, the first reveal is the host's choice and thus will reveal a goat (else the game ends with you losing because the host knew you had not picked the door with the car, and the host then chooses that door with the car). This is not the same as jimmy d's graphic which shows you choosing the first door, not the host. So now the host in this rigged game knows that only 2 possibilities remain: you correctly chose the door with the car, or you didn't. By deduction, you know this too. The odds are now even that you chose the car or a goat. Either you read something into the way the host asks you if you want to change your choice, or you second-guess yourself for no evident reason, or you stick with your original choice.

I choose to stick because I see no reason to change.

Or I overlooked another possibility, which y'all are trying to point out to me and I just don't comprehend because your dice analogies and graphical representations don't seem to fit with the actual (rigged) game presented by space-time.

Part of my confusion came from imagining if I were in Monty's position: If the contestant chose Door 1, and I/Monty new that Door 1 contained the car, and I did NOT want the contestant to win, then one of my strategies would be to expose one of the door goats, and then say, "You sure you don't want to reconsider and perhaps choose another door?"

My last reply should have clarified that... all the Monty stuff is irrelevant smoke and mirrors.

The key idea in my mind is that every reveal removes a contending door from the original game, and a new statistic is thus required to calculate the odds for the situation that is now in play.

My last three replies should have clarified that... all the Monty stuff is irrelevant smoke and mirrors, including revealing any farm animals. I explained it, borrowed a clear demonstration to show all possibilities of it, etc., in several different ways; all with the same provable point, that if you stick w/ your initial 33.3% choice, it remains a 33.3% choice if you do nothing. I showed/explained logically identical scenarios confirming that switching under any of them doubles your chance of winning from 33.3% to 66.7% by using 2 out 3, 4 out of 6, and 6 out of 9 illustrations, hoping that at least one illustrations would sink in. Monty and the goats, and the fancy car, the possible incentivizing to switch and/or sell your choice, all have nothing to do with the math; their **ONLY** purpose is to confuse you. And by "you", I of course mean anybody that remains confused.

A similar confounding question (though easier for most people to finally come to the right conclusion) is determining the least number of rounds with a balance scale to determine the one heavy ball in a set of eight billiard balls.

In my sheltered, though mostly fulfilling life, I've never come across that problem before, but because you said "heavier", the logician in me says I could do it in two rounds w/ a balance scale:

Round 1: Put three balls on each side, and retain the two remaining balls.

If the scale then reaches equilibrium, put the six weighed balls away, and;
Round 2 = Place the retained balls on the scale to see which is heavier.

If the scale was heavier on one side, put the lighter side's three balls with the two previously retained balls, and;
Round 2 = Place two of the three balls from the heavy side in Round 1 on the scale, retaining the third ball. If the scale reaches equilibrium, the retained ball is the heavy one, otherwise it's the ball on the heavy side of the scale.

OTOH, by switching to physics instead of math, there are numerous ways to find the heavy ball in a single step.

So again, please believe me when I say if Monty shows you a goat, and offers to let you swap doors, you should swap.
==

If Buzz’s clarifications still aren’t doing for anyone...

I not only love the Monty-Hall question, I love how different people approach it. I’ve only met one person that initially got it wrong, and then “logicked” it out by himself. (in a method similar to how Buzz explains it.)

I remember one friend advising me to give up ever trying to explain it to anyone who does not grasp it at first. His experience was that any further explanations were fruitless.

How about this? Imagine if Monty asked the same question in a slightly different way. Instead of revealing one door and asking if the guesser would like to switch, Monty said, “Okay, you picked door X out of three. You can stick with that pick. Or, you can choose BOTH of the other two doors. Not only that, before you decide to stick or switch, I’ll even open up one of the doors for you.”

If that scenario still doesn’t make it clear that the guesser increases the odds of winning by switching, make it 100 doors again. Same scenario. After the first choice, the guesser is given the option to keep that choice, or switch to ALL of the other 99. If the prize is behind any one of those other 99 doors, the guesser wins. Oh, and by the way, Monty will eventually reveal all of those 99 doors (save for the last one), one at a time.

Another way to think of it is to understand that Monty is not randomly opening doors. It’s a classic case of information asymmetry. One of the reasons most people dislike buying cars so much is because of asymmetrical information. The average car buyer gets one round of experience and information every few rounds. Auto salespeople have multiple rounds of practice (and lots more information) on a daily basis.

BTW, congrats, Buzz, on the billiard-ball question. Most people (including me, at first) need three rounds with the scale. Even after being told it was possible in two rounds, someone had to explain it to me.

I think the title of the book I found that one is, “Are you smart enough to work at Google?” Turns out I’m (maybe) smart enough to work the fast-food restaurant across the street from Google.

I used to open up a new class of advanced ESL students with that question. After most of them choosing 3, I left the question open. (with the promise I would explain it at the end of class)

If any student figured out the two-step solution, s/he was free to interrupt the class with the explanation. Someone usually figured it out in 20-30 minutes.

Not to sound prejudiced, Asian ESL students (visiting Vancouver) were usually quicker at solving the “puzzle” than students from other parts of the world.

Todd’s humbled keyboard

"My last three replies should have clarified that... all the Monty stuff is irrelevant smoke and mirrors, including revealing any farm animals. I explained it, borrowed a clear demonstration to show all possibilities of it, etc., in several different ways; all with the same provable point, that if you stick w/ your initial 33.3% choice, it remains a 33.3% choice if you do nothing. I showed/explained logically identical scenarios confirming that switching under any of them doubles your chance of winning from 33.3% to 66.7% by using 2 out 3, 4 out of 6, and 6 out of 9 illustrations, hoping that at least one illustrations would sink in. Monty and the goats, and the fancy car, the possible incentivizing to switch and/or sell your choice, all have nothing to do with the math; their **ONLY** purpose is to confuse you. And by "you", I of course mean anybody that remains confused."

I appreciate the time and effort you have spent trying to educate me. Thank you. My mental block revolves around this (as I stated above): you state "by using 2 out 3, 4 out of 6, etc", yet after the first reveal only 2 doors remain, not 3, and in effect, the host has reduced his odds by taking one door out of play. Thus at the beginning of play (3 doors to choose from), I may or may not have chosen the door with the car, but the host knew which of the other 2 doors to throw out of contention. By revealing that door, the play is reduced to 2 doors, and although I may or may not have chosen correctly, I have a 50% chance of being correct at this point in the game, vs my 33.3% at the beginning of the game.

"How about this? Imagine if Monty asked the same question in a slightly different way. Instead of revealing one door and asking if the guesser would like to switch, Monty said, “Okay, you picked door X out of three. You can stick with that pick. Or, you can choose BOTH of the other two doors. Not only that, before you decide to stick or switch, I’ll even open up one of the doors for you.”

This makes no sense as written. "Instead of revealing one door" means we are at the beginning of the game, no? Say I pick door 1, the host then says "you can choose BOTH of the other two doors. Not only that, before you decide to stick or switch, I’ll even open up one of the doors for you.” So the host opens door 2 or door 3 (and of course reveals a goat), meaning the car is behind my pick or the other unopened door (2 or 3). So there is no "both" doors to choose from (unless you are giving me the chance to claim both remaining doors, in which case you will most likely be fired from the game host position), or you are again giving me the chance to switch from one door to the other, again replicating the 50% odds .

Another possibility: I am too obtuse to ever understand your explanations ("I remember one friend advising me to give up ever trying to explain it to anyone who does not grasp it at first. His experience was that any further explanations were fruitless."). But none of your explanations have shown me where my logic is flawed vis a vis the first reveal that resets the statistics (2 doors remain, with a car or a goat behind either).

Again, I thank all my fellow MRFers who have devoted their time to patiently explaining their reasoning.

Thanks Todd, while I am sorta ESL, mostly by virtue of being LFL (Logic as First Language), I presumed the billiard ball answer before finishing the sentence, or giving it any actual thought. After finishing the first part of my last reply, I wanted to at least address something you had said in your reply, so I copied and pasted the scale problem. It took maybe 10 to 15 seconds to logically confirm my presumption of two rounds in my head, then a whole lot longer to dictate, and edit it the message window/box.

Several weeks ago, pRICE had a thread about MRF'er's personal Mt. Everest's that we've conquered. In my response, I alluded to having, after much struggle, and mostly fruitless help from some of the country's sharpest minds, I figured out how to calculate a value that theretofore always had to be iterated. The solution i devised, was to figure out a way to calculate the iterations using a creative way to look backwards thru a series of calculated arrays, without getting busted for using circular logic, or getting caught in a loop.

In one of my earlier responses herein, I joked about having a thing w/ Marilyn vos Savant back in the 80's. Since you also brought Marilyn up, she was one of the sharpies I brought into my Everest morass, and that is what I was alluding to in my "80's thing" reference. Like everyone else, Marilyn was unable to figure out a way to calculate the desired solution. She continued to pursue the challenge, though when I had gotten to the point of using an event trigger, of having a certain field's value change, after values were present in several other fields, to run the scripted iterations which ended by pasting the desired iterated value in the desired field, Marilyn agreed that was pretty slick. But by continuing, she never gave me the accidental clarity that Midnight Mike did by categorically declaring, in a convincing fashion, that we were all barking up the wrong tree.

Back at the Old Place there was some extended dialog on my Everest solution. This whole Monty Hall thing makes me think about how lots of people look at, and try to solve problems. A day, or so, after figuring out how to calculate a series of arrays on the fly, and use them to therefore calculate the desired solution to my Everest, I got w/ Marilyn and could visualize, if not hear her facepalm. I imagine Midnight Mike only beat Marilyn by a day, or two, into giving me the iteration only conclusion. By being in it up to my neck, I couldn't see the forest from the trees. That's the way it is w/ a lot of these problems. Sometimes you need someone that doesn't have a dog in the fight, to provide a touch of sanity and direction.

I don't know how much longer I would've pounded on my Everest from within the box I was trapped, but there was something that kept telling me the solution could, in fact, be calculated. Marilyn thought so, too, as did Midnight Mike initially... it was just that Midnight Mike was the first to kick me out of the box in order to start looking for the solution from outside the box.

Monty's doors, and boxes, and goats, and cars, and pocketfuls of crisp $100 bills, aren't anywhere near as complex IRL, but I can sure appreciate how people can get caught up in trying to understand the solution in essentially the exact same, frustrating fashion.

After sharing my Everest way back when, it's been fun watching how others have expounded on the premise.
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neophyte
So there is no "both" doors to choose from (unless you are giving me the chance to claim both remaining doors...

Yes, that is exactly the situation. Monty is (in a very obscure way) giving the guesser the opportunity to switch to BOTH doors, or stay with the original choice of just one door. By revealing one door (always a goat, never by chance), Monty‘s offer essentially is saying you can stick with your initial set of one out of three, or you can switch to the set of two out of three (or the set of ever how many are left, excluding your latest pick). He hides that generous offer by always revealing a door he knows has a goat, but making it seem like he is opening a door at random.

Buzz, thanks for the story about Marilyn vos Savant. She handled the Monty situation with more grace than most would have. As I recall, she invited math teachers all over the world to have her students play multiple rounds and record how many times they won when sticking as opposed to how many times they switched. It was great to see the letters from astonished teachers expressing how much fun their students had with this question.

Todd’s taking-another-break-from wrestling-with-a-piece-of-Ikea-furniture keyboard

(Turns out it’s quite challenging to fix up a vacation/retirement home when one has not yet moved one’s tools to the island.)

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Todd's keyboard

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neophyte
So there is no "both" doors to choose from (unless you are giving me the chance to claim both remaining doors...

Yes, that is exactly the situation.

Respectfully, no that is NOT the exact situation. In your new scenario, say I choose door 1, and the host then offers to allow me to switch to both doors 2 and 3, and BEFORE I switch, the host reveals either door 2 or door 3 (unless I misunderstand your scenario). The host then reveals, say, door 2 with its goat, THEN asks me if I want to switch from door 1 to door 3. The situation now is only 2 doors left (doors 1 and 3), and there is a 50% chance the car is behind either one. Again, why would I switch?

In jimmy d's graphic, in a purely unbiased game with random door reveals (his graphic shows that I turn a door, not the cheating host), the odds favor switching. In space-time's OP, the rigged game does not favor switching, because the host ALWAYS reveals a goat in the first reveal, so in effect the game REALLY starts with only 2 doors, and the odds are 50% for either.