Understanding the Monty Hall Problem

Buxton

It appears I was correct, your English comprehension skills ARE as poor as your maths skills, that would make you a ‘double dullard’ I guess.

"Monty: “Do you want to switch to doors 2 AND 3 "
Since Monty doesn’t say that it’s not the MHP, a variation perhaps. You can make up different problems if you want to, but if you want to understand the correct answer to the standard MHP you need to understand the problem in the first place - you clearly don’t.

Now come on Richard, humour me, what about 2 goats and a car in a great big bag. How does your bizzare theory work here? You keep refusing to answer - we all know why you refuse to answer - but what’s your explanation?

Perhaps it’s time for you to ‘leave the field of play’ again, you’ve more than outstayed your welcome, if I was the ref I’d have you sent you off months ago - after your first few posts of incoherent gibberish.

Bond / Buxton, it is only TWO doors if you believe the probability of the first door opened concealing a car remains at 1/3 even after it has been shown to conceal a goat.

Buxton has refused to explain how a door that yields ZERO cars no matter in how many games it is selected can have a probability of 0.33 of yielding a car.

Buxton has refused to explain because it is inexplicable and irreconcilable with his view of the MHP. Put another way, the karma has run over his dogma.

Mr Johnathan…

this from you…
Richard, I note you steadfastly refuse to explain how you reconcile your 1/3 probability of winning the car if one choose the open goat door with your statement that the goat door holds a 1/3 chance of concealing the car.

It would be silly to say that a door has a 1/3 chance of the car ONLY if it has the car behind it - the 1/3 chance obviously allows for there not being a car - therefore - when there’s a goat - the 1/3 chance of the car is maintained. The actual result does not retrospectively adjust a previous calculation of what might happen.

Dismiss all considerations of balls in bags and multiple trials with various numbers of doors - just consider and argue the previous paragraph.

I arrive at the - best to swap for a 2/3 chance - conclusion by adding together two individual 1/3 chances - one from the goat-door and the second from the closed door. This is not rocket science - it’s really very easy… 1/3 + 1/3 = 2/3.

I notice that people are still unable to explain the mystery of the flying 1/3 chance - when challenged their response is along the lines of… Don’t be tiresome Richard - it’s already been explained.

Buxton, why are you unable to explain how a contestant who chooses the goat door with, you say, a 1/3 chance of concealing the car will never win the car if they choose that door in an infinite number of games.

The long run average is supposed to become apparent after 100 or more games and yet the sucker following your 1/3 prediction will never win a single car.

Clearly your 1/3 theory is disproved by the reality of the RESULTS.

" therefore – when there’s a goat – the 1/3 chance of the car is maintained."
Where do you get this absolute nonsense from Richard? It certainly doesn’t appear in any maths books. I was going to suggest ‘the back of a Cornflakes box maybe’, but that would be an insult to Kelloggs.

“It would be silly to say that a door has a 1/3 chance of the car ONLY if it has the car behind it” Yes it would, and nobody has said that, it makes no sense.

"The actual result does not retrospectively adjust a previous calculation of what might happen."
That sentence also doesn’t make sense but never mind: nobody has suggested that prior probabilities are being adjusted (retrospectively or otherwise), posterior probabilities are calculated once a goat has been revealed. Prior and posterior probabilities - look up what they mean Richard, you never know you might actually learn something new for a change

“Dismiss all considerations of balls in bags”. Why? it is after all the EXACT MHP (not some bastardised version that you keep prattling on about).
You can’t explain it can you? When the questions get tough Buxton bottles it.

"I arrive at the conclusion by adding together two individual 1/3 chances – one from the goat-door"
How many times do you have to be told that a door with a goat doesn’t have a 1/3 chance of being a door with a car? In fact how many times do you have to be told you have no fucking idea what you’re talking about?

“it’s already been explained.” And it has many times, re-read the earlier posts. Of course you won’t because you have absolutely no interest in learning the explanation. Or look it up on the web, there are 1000’s of sites that also explain it, but you can’t be arsed can you? You’re happy as a pig in shit to remain ignorant.

Richard writes:

I notice that people are still unable to explain the mystery of the flying 1/3 chance …

This has been explained to Buxton a number of times, including by me.

By way of example, three identical robots run a race along an obstacle course. Each has a 1/3 chance of winning the Grand Prize awarded to the winner of the second and final race between the two fastest qualifiers.

3, the slowest in the first race is eliminated. 3 now has 0% chance of winning the Grand Prize. If you bet on No 3, it is 100% certain you will lose your money, even though at the start of the competition No 3 had a 1/3 chance of winning.

As for Robots 1 & 2, they each had a 1/3 chance of winning at the start of the competition but they now have a 1/2 chance of winning the Grand Prize by being fastest in the final race between the two of them.

Similarly, the open goat door in the MHP has zero chance of concealing the car. The first choice has a 1/3 chance (this is proved by never switching and observing that, over time, the win rate will get closer and closer to 1/3).

The other closed door has a 2/3 chance (this is proved by always switching to this door and over time the win rate will get closer and closer to 2/3.

And in case it needs to be said, if the contestant irrationally decides to switch to the open door with the goat, they will never win the car because this door has a 0/3 chance of making the contestant a winner.

There is no “flying” of a 1/3 probability from one door to the other, it is simply a process of elimination by which the probability of the only other possible location of the car equals 1 - the probability of the contestant’s first pick, ie 1 - 1/3 = 2/3, or if you insist, 1 - 1/3 - 0/3 = 2/3.

In the case of the robots, the initial probability for each outcome is calculated by dividing 1 by the number of equally probable outcomes, ie 1/3, and in the case of the final, 1/2, as there are only two robots in the final.

I’m in B.C. Canada and would say say “hello” if we ever crossed paths, to be civil. Also, I would bet against PalmerEldritch.

Next, I will consider Hilbert’s Hotel paradox and infinity.

Hello Bond in Canada. What would you bet against me on, the MHP? (I won’t be offering you 2 doors though - that’s not the MHP :slight_smile: )

Monty is making you a good offer to switch. He confuses the issue by opening a door. There will always be a goat behind at least one of the two doors you did not choose. You should know that and not be bothered when Monty shows you.

Your original choice is a set of one door. Switching allows you to chose a set of two doors. It is a set of two doors even if when you know what is behind one. It is a better choice because it is two doors.

A good deal is staring you in the face.

Your first choice is to chose a set (containing 1 door) that has a 1/3 chance of containing a car. You can switch to a set that has a 2/3 chance of containing a car. The set has a 2/3 chance regardless of the doors you think it contains.

  1. ‘For a single door to have a 2/3 chance of something – there needs to be two somethings.’

Richard, consider three games where the contestant switches:
Monday - contestant chooses the left-hand door, Monty opens the middle door, contestant switches to the right-hand door and wins the car.
Tuesday - contestant chooses the middle door, Monty opens the right-hand door, contestant switches to the left-hand door and wins a goat.
Wednesday - contestant chooses the right-hand door, Monty opens the left-hand door, contestant switches to the middle door and wins the car.

Over the three games, the contestant achieved the 2/3 chance of the switch door having the car. A car was put in position three times over the three games (there’s your ‘two somethings’ plus one extra for the show to keep).

Johnathan… good maths :slight_smile: - but bad logic

This from you…
There is no “flying” of a 1/3 probability from one door to the other, it is simply a process of elimination by which the probability of the only other possible location of the car equals 1 – the probability of the contestant’s first pick, ie 1 –1/3 =2/3, or if you insist, 1 – 1/3 – 0/3 = 2/3


Might fool somebody who is not paying attention - it nearly fooled me.

The maths is good - Unity (the car) minus ⅓ (the chance of the car associated with each door) does indeed leave ⅔ - Which you assign to the third door.

All good stuff…

Now this ⅓ that gets subtracted - is it from the originally selected door or is it from the goat-door?

I ask because - if it’s taken from the originally selected door - the one we give up if we swap - then the remaining ⅔ applies to the other two doors.

But if the ⅓ chance we subtract comes from the goat-door - then the ⅔ remainder applies to just one specific door - the so-far unopened door - the potential target of the swap. Are you following?

I suspect you favour the second of these scenarios - the ⅓ chance associated with the goat-door is applied entirely to one specific door - the door we would open if we elect to swap - am I right?

Why?

Why does it not get assigned to the selected door? Why not the other two doors in equal part? - Well Richard we can’t split the ⅓ chance equally between two doors or else it would be a 50/50 deal and that would never do.

What drives this (flying) ⅓ chance away from the goat door towards the place you want it to go to? What logic - force of nature - natural law takes the ⅓ and plonks it on a poor innocent door to make it carry a ⅔ chance of the car? Why that door and not the other one?

Your maths is excellent - but hardly difficult - the way you apply it is flawed.


And what do you mean - over time it will get closer and closer to xyz% ? - we discuss only ONE iteration of the MHP - people don’t get to come back week after week and stand the chance of winning multiple prizes - life isn’t like that.


I ignore your racing robots and their two races - too silly to contemplate.


Freddie - ditto your Monday - Tuesday - Wednesday scenario
Silly - contrived - and beyond ridicule.


In the MHP - to have a ⅔ chance of something (a goat for instance) behind a single door - you need two of them - But two individual ⅓ chances of the car from two doors - taken together - will give you a ⅔ chance of the car.


Finally - if you disagree - your case is weakened if you resort to personal criticism and the language of the gutter and the trailer-park social club - I know it’s difficult for you but you can at least pretend to have some class.

© RB – December 2014
All rights reserved…

Richard, you said:
660. ‘For a single door to have a 2/3 chance of something – there needs to be two somethings.’

In the case of the ‘switch’ door, we expect that single door to contain the car in two out of every three MHP games, a 2/3 chance.

Over every three games there are three cars made available, more than enough for that 2/3 chance.

I got bored after the 1st thousand words of your post, at which point I still hadn’t worked out what your argument was,
You said “both sets of people are both wrong”, which is incorrect. Richard was wrong in his explanation, Jonathon, Freddie, myself and several others are correct.

This is also incorrect "you were never under any circumstances going to be allowed to pick “Monty’s” door. " Which door is Monty’s door, which door can’t you pick? You’ver got it the wrong way round, Monty can’t pick your door.

Draw three rectangles on a piece of paper. Those represent doors. Above those rectangles put C, G, and G.

Put a giant X through one of the doors and one of the G’s. You are subtracting (1) from the supposedly possible (2) G’s and (1) from supposedly (3) options of doors.

Monty crosses off a door and a goat. You have to subtract 1 from the class of object “door” and 1 from the class of the type “goat”.

Goat and Door are two different variables. They exist independently but are combined in the same way we use containers in programming.

Example:

Door C = Goat (2)

You have two objects you’re dealing with.

Your goal is to get a car. There are only one of those.

You don’t have to worry about a second goat or a third door being an option. The problem lies in information transference because both groups are combining objects.

Simply put:

Stage 1:

3 Doors
2 object types (car or goat)
2 possible objects (car or 1 of 2 goats)

Stage 2:

2 Doors
2 object types (car or goat)
2 actual (now defined objects (car and 1 goat)

At stage 1 either goat is possible to end up with, as well as the car. By stage 2, you will have defined that 1 of the goats is not possible to pick. Monty removes it from the pool of options. In addition, he removes the door as well.

In fractions you don’t just change the enumerator, you change the denominator. The denominator represents possible choices and the enumerator represents quantity of choices or guesses.

Simply put: Stage 1 and Stage 2 are not the same. He is not there to reveal information. We already knew one of the remaining doors contained a goat. The location doesn’t help you solve anything whether its on the right or left side. He reveals nothing.

Why is he there?

Only to eliminate you having to choose from a pool of 3 doors, because prior to this each door had unknown contents.

2 total guesses (“stay” and “switch”), but only one of those guesses will be revealed to your eyes.

Monty is saying:

2 guesses (1 unrevealed) out of 3 doors you can pick…no, let me make this easier. Ill take a door and goat both and you choose from the remaining pool.

The problem is actually simple if you understand why we use fractions to begin with.

Amount of guesses/Possible prizes

Use that formula in each stage (as information is revealed). Then you will see that Stage 1 and Stage 2 are entirely separate fractions.

When I said both groups have a problem:

In Stage 1, you have to assume your first choice is a goat (2/3 chance). You should switch immediately according to probability logic.

In Stage 2, you no longer have to worry about those same odds because the goat & door which were previously half the quantity of your entire loss probability (1/2 x 66.6 percent) has nnow been eliminated from a possible decision.

That number is evenly distributed to the 2 remaining mystery doors. (16.67 percent to each)

You had 3 decisions originally = 3 doors = bottom denominator.

You had 1 guess originally. Monty never substantiates the results of this guess. He only gives you the option of making a new guess, which is the same as this guess being the first one. You never satisfied the results of this original guess, so it was only a placeholder to get you to Stage 2. It wasn’t even a real guess because nothing is final until it is compared against the opposing information.

Trying to link the two guesses together is where people are having a problem. The paradox is that Monty is changing the game into a “new” game each time.

To better illustrate, imagine you are on “Who wants to be a Millionaire” and you have (4) possible answers. It doesn’t matter how many times you change your guess until Regis says:

“Is that your final answer?”

None of those original guesses will ever be submitted against the correct answer. Only your final selection has any bearing on whether you win.

In Stage 1 of Monty’s game (we should actually call it “Game 1”) our supposed “guess” doesn’t matter. It’s not a real guess, it only exists if and when we follow through with it. We can change our mind 1,000 times but only the “opening of the door” seals that conceptual guess into an actual guess…in the same way that Regis revealing an answer seals our guess on his show.

The revelation is the point of no return. Assuming that the “two” Stages are “one” game and are interrelated is only complicating solving the problem.

The math is this problem is very basic math which we all learned in elementary school. However, the conceptual science is advanced and thus we are manipulating the wrong digits to achieve our answer.

Put another way: 10 coin tosses are 10 quantities of (1) coin toss each. There is no need to group them together. Humans created this grouping by concepts such as “best out of 10”. Each is independent and the only thing achieved by comparing to another flip is “averages”.

Averages in themselves only exist once we define the quantity of however many things we are going to compare. All of these concepts are interrelated.

The maths supports the logic, Richard, that’s what proves it is logical and not purely philosophical.

How can Buxton be wondering where the various integers came from? I stated quite clearly that the 2/3 is the result of deducting the probability of the FIRST PICKED DOOR (1/3) from 1 (1 being the sum of probabilities for all POSSIBLE outcomes - the goat door concealing the car not being a possible outcome, it is disregarded after being opened and shown to be incapable of yielding a car).

In terms of maths being sound but logic not so, How do we prove the theory that 12/4 = 3?

The proof is that 4 x 3 = 12.

No “force of nature” required. Just arithmetic.

One credible approach to probability is that the relative frequency of occurrence of an event, observed in a number of repetitions of an experiment, is a measure of the probability of that event.

If we accept that, how do we prove the theory that the goat door has a 0/3 chance of providing the car?

Proof: Choose the goat door X number of times (where x = any number from 1 to infinity) and you will win the car zero times.

How do we know the probability of the only other closed door equals 1 - probability of door first picked or 1 - 1/3 = 2/3?

Proof: Switch to the other door n number of times and if n is large enough the number of wins will converge on 0.67 x n.

Theory: Probability of open goat door concealing a car is 1/3.

Disproved because no matter how many times the open goat door is chosen, the car is not yielded once.

Theory: Probability of door first picked concealing a car is 1/3

Proof: Stick with the first door picked n number of times and if n is large enough the number of wins will equal 0.33 x n.

A number of us have explained the logic of how the other closed door has a probability of 2/3 all by itself and the goat door has a probability of 0/3. For his part, Richard has put up no evidence let alone proof to show how the goat door retains a probability of 1/3 of concealing the car. All he has expressed is an opinion that probabilties are forecasts that are fixed and never vary. Not very Bayesian or anything elsian. Reminds me of that scene from “Something about Mary”

Hitchhiker: You heard of this thing, the 8-Minute Abs?
Ted: Yeah, sure, 8-Minute Abs. Yeah, the excercise video.
Hitchhiker: Yeah, this is going to blow that right out of the water. Listen to this: 7… Minute… Abs.
Ted: Right. Yes. OK, all right. I see where you’re going.
Hitchhiker: Think about it. You walk into a video store, you see 8-Minute Abs sittin’ there, there’s 7-Minute Abs right beside it. Which one are you gonna pick, man?
Ted: I would go for the 7.
Hitchhiker: Bingo, man, bingo. 7-Minute Abs. And we guarantee just as good a workout as the 8-minute folk.
Ted: You guarantee it? That’s - how do you do that?
Hitchhiker: If you’re not happy with the first 7 minutes, we’re gonna send you the extra minute free. You see? That’s it. That’s our motto. That’s where we’re comin’ from. That’s from “A” to “B”.
Ted: That’s right. That’s - that’s good. That’s good. Unless, of course, somebody comes up with 6-Minute Abs. Then you’re in trouble, huh?
[Hitchhiker convulses]
Hitchhiker: No! No, no, not 6! I said 7. Nobody’s comin’ up with 6. Who works out in 6 minutes? You won’t even get your heart goin, not even a mouse on a wheel.
Ted: That - good point.
Hitchhiker: 7’s the key number here. Think about it. 7-Elevens. 7 dwarves. 7, man, that’s the number. 7 chipmunks twirlin’ on a branch, eatin’ lots of sunflowers on my uncle’s ranch. You know that old children’s tale from the sea. It’s like you’re dreamin’ about Gorgonzola cheese when it’s clearly Brie time, baby. Step into my office.
Ted: Why?
Hitchhiker: ‘Cause you’re fuckin’ fired!

Plus, as Freddie said, Richard Buxton did say:

“For a single door to have a 2/3 chance of something – there needs to be two somethings.”

If this is so, it must also be so if the 2/3 is spread across two doors through the addition of another door, ergo (according to Buxton):

“For TWO DOORS to have a 2/3 chance of something – there needs to be two somethings.”

However, Richard acknowledges there is only ONE car. And yet his view of needing to be two of something before one can have a 2/3 chance of it being so is patently incorrect as there is only one car.

So Richard’s statements, once again, are irreconcilable with one another and reality.

If both goat doors are opened, there is a single door with a 3/3 chance of concealing the car (proved as per above) and yet only one car. According the Richard’s statement there must be 3 somethings, or cars in this case, for a door to have a 3/3 chance of concealing the car.

But your “chance probability” is not predictive, Richard. And probability is nly of any use if it predicts the result.

Your 1/3 theory for the goat door says that picking the open goat door will result in the car being won, on average, 1 in 3 times if it is selected every time.

The alternative view of 0/3 says the car will never be won.

Which theory better predicts the result?

The ‘somethings’ are chances, not cars or goats. Cars and goats are the meanings we apply to the chances. In this case you have 1 in 3 chances of winning and then, if you switch, 2 in 3 chances.

A 2/3 chance of something is just that, 2 chances in 3. This does not imply there must be a total of three physical somethings that pair with each chance. Similarly, a 66.7% chance of something does not imply there must be 66.7 of some physical objects and 33.3 of others.

I could enter a car race with a 2/3 chance of winning. That doesn’t mean there will be two of me racing simultaneously against only one other driver. One racer (or door) can have a 2 in 3 chance of something, anything you want. In the MHP the door that Monty did not open and that we did not initially select has a 2/3 chance of concealing a car.

For the first choice it just so happens that a door is paired with each chance. For the switch option a door has been eliminated but the chances remain the same, the pairing is gone.

Your first choice in the MHP has a 1 in 3 chance of winning. The set of doors you did not select must have the remaining 2 in 3 chances of winning. Monty can show a goat behind one of the doors you did not choose. The set of doors you did not select still must have a 2 in 3 chance of winning.
Even when you see a goat revealed that set has a 2 in 3 chance of winning because it must include everything you did not select in your original 1 in 3 chance, that is, the other 2 in 3 chances.

Switching does guarantee a win but increases your chances from 1/3 to 2/3, even when that 2/3.

Monty makes a good offer then confuses the player by opening one of the doors you did not chose to reveal a goat. Since Monty knows where the car is he can always show you a goat.

‘We see a goat? Well that was the most likely outcome but the door had (and still represents) a less likely chance of the car.’

Richard, it is with a heavy heart that it becomes my solemn duty to demand that the bridge authorities in Reading strip you of the titles you claim to have won. You, Sir, are a deceitful knave. Each time you played what you claimed was a winning hand, it had (and still represented) a greater chance of containing a losing combination of cards. Say you showed the King of Spades and it was required to win the game. When you were dealt it, face down from a randomly-shuffled pack, it had a 51/52 chance of not being the King of Spades. What logic – force of nature – natural law takes the 51/52 chance the rest of the pack had of being the King of Spades and plonks it on that poor innocent card to make it carry the winning 1/1 chance you claimed? Surely it still carried a 13/52 chance of being Clubs for example. And the same goes for every other card or hand you played. Prepare to bow and grovel in abject humiliation before the regulators of gaming in Berkshire - you are about to be unmasked.