Science/Environment The Monty Hall problem. Trippy or what?

[The Goon]

Thinking time

 

[The Goon]

Yes, that is an example of calculating the probability of a future event which is independent of past events.

Which has nothing to do with the Monty Hall problem.

Sorry, you are correct, was tired when I wrote that and now can't remember where I was going with it.

Have slept on it now and get it.
Thanks for the patience.

 

[Sloth]

People are tricked into thinking it's 50/50 because there are two doors. It's not. Say I have 100 balls and only one is red. I randomly put a ball behind each of the two doors. Do you think you have 50/50 chance of a red one? Of course not.

 

[radiojake]

I haven't read through everyone's responses, but my mum's partner and I actually went through and did a bit of a simulation of this paradox -

We took three boxes - I would hide a ball underneath one of them - He would then choose a box, I would reveal an empty box and then he would automatically change is response to the other box.

I kid you not, but we each did it 100 times and when you change your response after the empty box is revealed you end up winning pretty much right on 66% of the time (or 2/3) -

Both of us won between 60 + 70 times once we switched boxes after an empty one was revealed -

That's because the odds of picking the prize on your first go is only 33% - So by switching after the first empty door is revealed, you have a 66% chance of winning the prize... Our simulation confirmed such odds

 

[Admiral Byng]

I haven't read through everyone's responses, but my mum's partner and I actually went through and did a bit of a simulation of this paradox -

We took three boxes - I would hide a ball underneath one of them - He would then choose a box, I would reveal an empty box and then he would automatically change is response to the other box.

I kid you not, but we each did it 100 times and when you change your response after the empty box is revealed you end up winning pretty much right on 66% of the time (or 2/3) -

Both of us won between 60 + 70 times once we switched boxes after an empty one was revealed -

That's because the odds of picking the prize on your first go is only 33% - So by switching after the first empty door is revealed, you have a 66% chance of winning the prize... Our simulation confirmed such odds

Well done for doing the experiment.

I recall the Mythbusters did exactly the same thing a year or three ago and got the same result. They do the experiment so I don't have to try it at home

 

[fishardansin]

Marcus de Sautoy ran the experiment 40 times. 20 times he switched and 20 times he didn't. The results were pretty close with 13 right with the switch and 7 for not switching.

I can understand why people find this confusing at first but surely after you're explained that the host knows where prize is when he reveals a door this takes the randomness out of his decision and by choosing switching to the one you didn't choose and the host didn't open means you affectively have the prior probability of choosing both the doors you didn't.

 

[Illinois Nazi]

Try it this way:

You choose a door - probability it has the car = 1/3

Host chooses a door - probability it has the car = 0

Only one door left - what chance it has the car?

 

[Chief]

I once demonstrated the answer to this to a workmate using a dice.

It came unstuck when I guessed the correct number first up 6 out of ten times, thus "proving" that I was speaking crap....

 

[Malifice]

I once demonstrated the answer to this to a workmate using a dice.

It came unstuck when I guessed the correct number first up 6 out of ten times, thus "proving" that I was speaking crap....

60% of the time, it works everytime.

Made with bits of real panther so you know its good.

 

[Niximus]

Just had a read through of this thread for the first time and was surprised at how many people still didn't get it after it had been explained to them.

I'll try my hand at it.

The Car is behind Door #2

Switching
You pick Door #1, the host reveals that Door #3 is empty, you switch to #2 - YOU WIN
You pick Door #2, the host reveals that one of the other doors is empty, you switch to the remaining door - YOU LOSE
You pick Door #3, the host reveals that Door #1 is empty, you switch to #2 - YOU WIN

2/3 chance of winning

Not switching
You pick Door #1 - YOU LOSE
You pick Door #2 - YOU WIN
You pick Door #3 - YOU LOSE

1/3 chance of winning

 

[fishardansin]

Just had a read through of this thread for the first time and was surprised at how many people still didn't get it after it had been explained to them.

I'll try my hand at it.

The Car is behind Door #2

Switching
You pick Door #1, the host reveals that Door #3 is empty, you switch to #2 - YOU WIN
You pick Door #2, the host reveals that one of the other doors is empty, you switch to the remaining door - YOU LOSE
You pick Door #3, the host reveals that Door #1 is empty, you switch to #2 - YOU WIN

2/3 chance of winning

Not switching
You pick Door #1 - YOU LOSE
You pick Door #2 - YOU WIN
You pick Door #3 - YOU LOSE

1/3 chance of winning

Very well explained. This is true no matter which door the car is behind of course.

In effect when you switch you get twice the probability based on prior ignorance.

 

[jdisc]

Try it this way:

You choose a door - probability it has the car = 1/3

Host chooses a door - probability it has the car = 0

Only one door left - what chance it has the car?

100%

 

[jdisc]

/thread

 

[kickazz]

Monty Hall problem is getting a run on The Conversation at the moment. Interesting to read through the comments, the 'two children' riddle which got a good run here comes up there as well.

Someone posts another interesting conundrum:

A similar problem that has always confused me:
You offered to take one of two envelopes containing money, and told one contains twice as much money as the other.
You choose one.
You open it and find that it contains $10. You are then offered the chance to switch envelopes. You reason there's a 50% chance of the 2nd envelope containing $5, and a 50% chance of it containing $20. The weighted probability is a return of $12.50, so you decide to swap.
Seems reasonable enough.
Now, repeat the above scenario, but intead of opening your envelope, you just assume it contains $X. You make the same calculation as before, and deduce that swapping will give a weighted probability of returning 1.25 $X
ie. regardless of which envelope you chose initially, you then deduce that you will be better off swapping, without any additional information. This is crazy, but where is the fault in the logic?

 

[Chief]

Try it this way:

You choose a door - probability it has the car = 1/3

Host chooses a door - probability it has the car = 0

Only one door left - what chance it has the car?

50%

 

[Chief]

Someone posts another interesting conundrum:

Do you know which has the most money? Like, are you only told one has 2x the other or are you told the possibilities are $5, $10 and $20?

 

[SJ]

Monty Hall problem is getting a run on The Conversation at the moment. Interesting to read through the comments, the 'two children' riddle which got a good run here comes up there as well.

Someone posts another interesting conundrum:

I don't really the see the problem here. Imagine it was $1, with a 50% chance of $1,000,000 and 50% chance of $1/1,000,000. Surely you would just do it again and again.

 

[kickazz]

Do you know which has the most money? Like, are you only told one has 2x the other or are you told the possibilities are $5, $10 and $20?

Basically you are told that one has twice as much as the other, but you do not know which is which.
Also, for the purposes of the problem I don't think utility comes into account (as in, what the amount means to you personally) its just a matter of maximising the expected return.

 

[kickazz]

I don't really the see the problem here. Imagine it was $1, with a 50% chance of $1,000,000 and 50% chance of $1/1,000,000. Surely you would just do it again and again.

That is kind of what the paradox is getting at - if you have envelope x in hand, and envelope y could be eithe x/2 or 2x - each with 50% probability. That leads to an expected outcome of y = 0.5(x/2) + 0.5(2x) or y = 1.25x

So going by this you would be better off swapping and taking envelope y. But then, the same as above would apply, and so the expected value of x (not in your hand anymore) would be x = 1.25y.

Going by this you should just continue to swap - this clearly being an absurd course of action - showing that obviously the modelling of the problem, whilst seemingly correct mathematically is obviouly flawed.

 

[Chief]

Basically you are told that one has twice as much as the other, but you do not know which is which.
Also, for the purposes of the problem I don't think utility comes into account (as in, what the amount means to you personally) its just a matter of maximising the expected return.

If you have no way of knowing the maximum, and each game is discreet with different amounts, I don't think you need to switch do you? You never know if the one you have is max, middle or minimum. Play it 1,000 times and you'll come out a (#games x middle amount) anyway won't you?

Actually, no.

Say it is $1, $2 and $4.

Average is $2.33

There is only ONE option better than the average return, two below the average.

Swapping after the first pick only works if one of the envelopes is taken away as a worse option than the one you picked first or if you are told the maximum beforehand.

 

[kickazz]

If you have no way of knowing the maximum, and each game is discreet with different amounts, I don't think you need to switch do you? You never know if the one you have is max, middle or minimum. Play it 1,000 times and you'll come out a (#games x middle amount) anyway won't you?

Actually, no.

Say it is $1, $2 and $4.

Average is $2.33

There is only ONE option better than the average return, two below the average.

Swapping after the first pick only works if one of the envelopes is taken away as a worse option than the one you picked first or if you are told the maximum beforehand.

Be sure to realise there are only TWO envelopes. So using your numbers above, I think a concrete example would be:

Envelope X and Y. All you know is one has double the other.

You open X, it contains $2. So you now know that Y has either $1 or $4. Is it in your best interests to swap or not?

 

[SJ]

That is kind of what the paradox is getting at - if you have envelope x in hand, and envelope y could be eithe x/2 or 2x - each with 50% probability. That leads to an expected outcome of y = 0.5(x/2) + 0.5(2x) or y = 1.25x

So going by this you would be better off swapping and taking envelope y. But then, the same as above would apply, and so the expected value of x (not in your hand anymore) would be x = 1.25y.

Going by this you should just continue to swap - this clearly being an absurd course of action - showing that obviously the modelling of the problem, whilst seemingly correct mathematically is obviouly flawed.

Ah, you mean the same envelope. I'm with you now. I thought you meant just an infinite stack of new envelopes.

Yes, that doesn't make sense.

 

[Malifice]

That is kind of what the paradox is getting at - if you have envelope x in hand, and envelope y could be eithe x/2 or 2x - each with 50% probability. That leads to an expected outcome of y = 0.5(x/2) + 0.5(2x) or y = 1.25x

So going by this you would be better off swapping and taking envelope y. But then, the same as above would apply, and so the expected value of x (not in your hand anymore) would be x = 1.25y.

Going by this you should just continue to swap - this clearly being an absurd course of action - showing that obviously the modelling of the problem, whilst seemingly correct mathematically is obviouly flawed.

But your formula above is prefaced on your existing knowledge of the value of x (you've already opened the envelope), whereas y has an indeterminate value.

Similar to the Monty problem, where Monty knows what door has the goat, and which one has the car, and will always open a dud door.

 

[Pie eyed]

You dont automatically win the car.

You could have picked right the first time (with your first guess) and already be standing in front of the 'car' door

Switching (in this case) would result in you losing the car.

And how is it not a 50/50? Two doors remain right?

One has a car, while the other is empty.

Switching shouldnt matter in a 50/50.

Although it appears that switching in fact increases your chances of being correct and winning the car!

I think you are on the right track, because the second choice is a totally new choice, not part of your original 1 in 3 decision.
You are simply making a second, unrelated 50/50 choice.
It's no different than the third option never having existed.

The "Monty Hall" only exists if the mathematicians ignore the fact the host has tampered, fatally with the original equation.

E=Mc2 fails miserably if you randomly subtract 1/3 of the universes energy.

 

[fishardansin]

It's a pretty simple problem. There is a 66% chance that the car isn't behind your door. With one of them opened and you picking it then you've used up all of the 66%.