The Monty Hall Problem: Help!

James_H

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I've read tons of explanations for this, on and off this thread, and I still don't get it at all. Trying to do maths can be like banging my head against a brick wall.

I find this kind of thing very frustrating. I never got very far with computer programming as a hobby because I literally can't get my head around how recursion works, and it's a fundamental and basic part of program logic.
 

INT21

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The answer is that the three should have paid one third £10 = £3.33 each, (the new price for the meal) not one third of £15 pound, ( the old price).

So when the waiter gave each one £1 back, he should have given them £1.66. They thought they were getting a refund, but in act the waiter was bilking them out of £0.66 each. And that adds up to the £2 in his pocket.

The key is that the customers didn't know the manager had reduced the price.
 

EnolaGaia

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I've read tons of explanations for this, on and off this thread, and I still don't get it at all. Trying to do maths can be like banging my head against a brick wall. ...
James:

Are you referring to the Monty Hall Problem or the more recently-discussed 3 men in a cafe puzzle?
 

INT21

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Once the host has revealed a goat door then you are down to a 50/50 choice; Assuming you didn't pick the door he opens.

So changing makes no difference. It is one of the remaining two, so you may as well stay where you are.

I don't see why people make this to difficult.
 

blessmycottonsocks

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"So when the waiter gave each one £1 back, he should have given them £1.66. "

Irrelevant! The fact remains that each customer payed £5 and got £1 back. The dishonesty of the waiter doesn't alter the fact that the meal therefore cost the men £4 each.
Or, to put it another way, they originally paid £15, but finished by paying only £12.
 

INT21

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The aim of the exercise was to find the missing £1.

They payed £4 for a meal that should have cost them £3.33
 

Peripart

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Once the host has revealed a goat door then you are down to a 50/50 choice; Assuming you didn't pick the door he opens.

So changing makes no difference...
Except it does make a difference. You are better off swapping, which is why this problem is so counter-intuitive.
 

INT21

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Except it does make a difference. You are better off swapping, which is why this problem is so counter-intuitive.
If it's got to be one or the other, why are you better off ? You have no idea which of the two remaining doors has the car.
 

Bad Bungle

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Five trillion to one! Punter scoops second lottery win with the same numbers By David Wilkes
The odds against being so lucky were mind-boggling - more than five trillion to one.

Just read on the Mail Online that someone got 5 numbers and the bonus ball in June, scooping £194,501. He kept playing the same numbers and got another 5 numbers and bonus ball last Saturday (£121,157).
Quote:
"The odds of picking five correct numbers and the bonus ball at random are 2,330,636 to one (14 million divided by six)
The odds of the same six correct numbers coming up at random again are 5,400,000,000,000 to one, because that is 2,330,636 squared. This is because the two different draws are independent events and the chances of them happening each time are the same. So the odds are 2,330,636 multiplied by 2,330,636.
'Thinking of something that is as unlikely as this is almost impossible,' said Professor Simon Cox, of the University of Southampton"


If the chances of two independent events are the same then why square them ? I'm wondering what Prof Cox teaches at Southampton.

https://www.dailymail.co.uk/news/ar...cond-lottery-win-numbers-By-David-Wilkes.html
 

INT21

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Surely the odds are the same as they were first time. If the selection is truly random.
 

Peripart

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If it's got to be one or the other, why are you better off ? You have no idea which of the two remaining doors has the car.
It took me a while to accept it as well, even though my degree was maths-based, but it's true. As was mentioned before, Mythbusters have even done the legwork to demonstrate the truth of it the hard way, for those (like me) who took a while to believe the stats!
 

INT21

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, Mythbusters have even done the legwork to demonstrate the truth of it the hard way,
Mythbuster also were supposed to demonstrate that you could survive pissing on the third rail. But, if I remember correctly, they downgraded to an electric fence, then appeared to fudge the actual act. Giving some cock (if you'll pardon the expression) and bull story about there being no solid line of pee from dick to rail.

Mine looks pretty solid when I unload a bladder full.
 

Bad Bungle

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If it's got to be one or the other, why are you better off ? You have no idea which of the two remaining doors has the car.
There are twice as many goats as there are cars so there is twice as much chance of you picking a goat with your first guess.
What's behind the doors stays put.
You have no idea which of the two remaining doors has a car but Monty does, which is why he choose to open a door with a goat behind it (and not randomly).
The odds that you picked the goat with your first guess is still larger than that you picked the car - time to switch doors.
 

INT21

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I think what is happening here is that I am misunderstanding the rules of the game.
 

INT21

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Perhaps someone can explain just how this game works.

I get the impression that there are three door, let us say a,b and c. Goats behind two and a car behind the other.

You have to select a door. I assume you have to declare the door you select.

Is this the stage where Monty asks you if you want to switch your choice. Before any door is opened ?

Then what happens ?
 

gordonrutter

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Perhaps someone can explain just how this game works.

I get the impression that there are three door, let us say a,b and c. Goats behind two and a car behind the other.

You have to select a door. I assume you have to declare the door you select.

Is this the stage where Monty asks you if you want to switch your choice. Before any door is opened ?

Then what happens ?
You specify your door. Monty then opens one of the other two doors revealing a goat. You are then asked if you want to change.
 

Ringo

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@INT21

3 doors: A, B and C. A car is behind one of them and will not move.
Let's say you choose Door B. The chances that you picked right from the beginning (Door B) is only 1 in 3.
Therefore the chances that it is somewhere else (Door A or Door C) is a combined 2 in 3.
You tell the host "I choose Door B".

The host now opens a losing door. Let's say Door A is to reveal NO WIN.
So you now know that Door A is empty. The chances say that the car is behind the door you originally picked is still 1 in 3 as there are still 3 doors. It just so happens that you can see behind one now. So the chances that it is elsewhere is still 2 in 3 only now you know that it isn't behind Door A. So it must be behind DOOR C. So you have a better chance of winning of you swap doors.

HOWEVER, let's say that after the host opens Door A he was allowed to swap position of the car around, maybe moving it from where it was to the other remaining closed door. That would revert your chances to 50/50 so it wouldn't be of benefit to swap.
 

INT21

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Ringo,

Yes, there are still three doors.

But you know, in your example, that Door A is a loser.

So the car has to be behind B or C.

With only two doors to pick from. how can you still have three effective doors. One of them has been eliminated.

You actually only have one choice. Stay or move. The can has to be behind one of the two doors.

INT21.
 

maximus otter

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Mythbusters (Episode 177: Wheel of Mythfortune) demonstrated that the paradox worked, i.e. that switching gave better results.

The result was discussed in the aftershow here:


Adam states (paraphrase): "Nobody has yet [broadcast date 23.11.11] written a peer-reviewed mathematical proof of why it exists."

maximus otter
 

blessmycottonsocks

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Mythbusters (Episode 177: Wheel of Mythfortune) demonstrated that the paradox worked, i.e. that switching gave better results.

The result was discussed in the aftershow here:


Adam states (paraphrase): "Nobody has yet [broadcast date 23.11.11] written a peer-reviewed mathematical proof of why it exists."

maximus otter
His explanation is to treat the doors as two groups. Group one is the door you picked with a 33.3% chance of winning, whereas group two is the other two doors, with a 66.7% chance.
Surely though, once one of the doors in group two is opened to reveal it's a loser, the remaining odds cannot still be 66.7%?
 

Bad Bungle

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15 years on, the thread that just keep giving.

Choose one of the doors and you have a 1/3 chance of a car behind it and a 2/3 chance the car is behind one of the other (not specified) doors.
These odds are demonstrable if Monty were to open both remaining doors after your pick and you repeat the process a few times.
Monty opens one of the remaining doors. If it was a random choice and revealed the car, there'd be no point switching doors.
But Monty knows what's behind the doors and always opens a remaining door with a goat behind it (makes for better Game Show ratings)
The car and goats don't move around behind the doors, nor are they in a quantum state.
If the odds of the car being behind one of the other doors was 2/3 and one of those door reveals a goat, then the odds must be 2/3 that the car is behind the remaining other door.
You may have picked the car on your first choice but it's twice as likely you didn't. So switch.
The counter-intuitive bit comes from assuming (incorrectly) that the odds in any two-option choice must be 50:50
 

AgProv

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Hmm. Do something to make the goats bleat and go for whichever door appears to be silent with an absence of baa. Or else go for the door that doesn't reek of goatshit and piss. The smell is hard to mask.

Or...

"Contestant Esmerelda Weatherwax from Lancre - what State is that in again? Congratulations, you've chosen the car!"

"Drat. I can't be doin' with cars, nasty smelly things. But another goat would have been useful."
 

JamesWhitehead

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The Wikipedia page on the problem has a conditional probablity table, to show the odds of switching versus sticking in 300 cases.

It is quite the most lucid way to present the problem. In the two-out-of-three cases where you are initially wrong, the host has no choice which door he can open. If you switch in these cases, you must now win. In the one-out-of-three cases where you are were right, any switch will now cause you to lose. So the odds are 2/3 in favour of the switch.

I would still hate to have to explain this in a maths class! :frust:
 

Ringo

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Ringo,

Yes, there are still three doors.

But you know, in your example, that Door A is a loser.

So the car has to be behind B or C.

With only two doors to pick from. how can you still have three effective doors. One of them has been eliminated.

You actually only have one choice. Stay or move. The can has to be behind one of the two doors.

INT21.
Because you don't pick once there are one two doors in play. You pick before any doors are opened. Your choice is 1 of 3 doors. Probability says that the car is more likely to be elsewhere.

If the car/goat were shuffled once one door was opened, and then you got to change your door if you wanted, then it would be 50/50. But as the car never moves, then probability says that it is more likely to be behind the remaining door.

Try it yourself with a friend as the host ans 4 bits if paper, one marked with an X.
 

INT21

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Because you don't pick once there are one two doors in play. You pick before any doors are opened. Your choice is 1 of 3 doors. Probability says that the car is more likely to be elsewhere.

If the car/goat were shuffled once one door was opened, and then you got to change your door if you wanted, then it would be 50/50. But as the car never moves, then probability says that it is more likely to be behind the remaining door.

Try it yourself with a friend as the host ans 4 bits if paper, one marked with an X.
sorry, I simply don't agree.

You say above 'if the car/goat were shuffle once one door was opened'. I don't see any reference to the doors being shuffled. even if they were, it wouldn't matter as there would still be only two doors, and so it is still the same choice; stay or move.

Let's assume you picked the correct door. Monty opens a goat door (as he is always going to do as he is simply removing a losing door from the equation).

So, you have picked the correct door. Why should moving improve your chances, you have already picked the winner.

Also, why four pieces of paper ? there are only three doors. surely you mean three pieces, one marked.

The way I see it, he is playing two games. In the first one, your odds are 1 to 3.

Then there is a new game in which your odds are 1 to 2. The result of your choice in the first game has no effect on the new game.

Anyway, I will try the idea of using paper. I had considered it already.

Maybe play it one hundred times. Fifty I don't change, fifty I do. I'll get back to you with the results. May be a few ays though. Rather busy at the moment.

INT21.
 

INT21

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The Wikipedia page on the problem has a conditional probablity table, to show the odds of switching versus sticking in 300 cases.

It is quite the most lucid way to present the problem. In the two-out-of-three cases where you are initially wrong, the host has no choice which door he can open. If you switch in these cases, you must now win. In the one-out-of-three cases where you are were right, any switch will now cause you to lose. So the odds are 2/3 in favour of the switch.

I would still hate to have to explain this in a maths class! :frust:

...the host has no choice which door he can open..

Of course he has a choice. Which ever door you pick he can always open a goat door. which he will as that is the whole point of the game.
 
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