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[Neonivek]

I still don't understand the Monty Hall problem.  It's not that I doubt it - I've followed it step by step and sure enough switching is better.  I understand it's better to basically pick a door *not* to open, since that gives a 2/3 chance of success.  It still seems really weird that switch accomplishes that, mechanically.

Think of it this day.  Instead of three doors, you have one hundred.  You pick one, and the host opens 98 doors.  Your door still has a 1% chance to be correct, but the one other door still there has a 99% chance.

That is temporal because that one TECHNICALLY has a 99% chance if you use that mindset. That is the major hurdle when understanding the Monty Hall Problem.

The major issue that most people have with it is this...

You have three doors you chose one... it has a 1 in 3 chance.

One of the doors is opened and it is the false one and you are given the ability to switch your answer.

But... Both of the doors should have equal chance to work right?

It had a easier explanation but I cannot remember what it was exactly.

I know that no matter which of the three doors you chose you will ALWAYS have a false door... But... well...

In Scenario 1 you keep your answer:

The combination at which you succeed is that both unopened doors are false.

Scenario 2 you change your answer

The combination which you succeed is just that

O XX
X OX
X XO

But let me see... if you chose the first one... you succeed 1/3rd of the time

OX X
XO X
XX O

If you switch you succeed 2/3rd of the time.

[Darvi]

Basically. It's really just a matter of deduction and probabilities. When you choose a door at the beginning, you have a 2/3 chance of picking a dud. When the second door gets opened, you know that one's a dud. So if you know one door is the wrong one, and know there's a 2/3 chance of the other one also being wrong, that implies that for those 2/3, the third door is the only one left that can win.

On the other hand, not switching makes everything afterwards irellevant, meaning you're stuck with the 1/3 win chance that you had when you picked the door.

[Neonivek]

Basically. It's really just a matter of deduction and probabilities. When you choose a door at the beginning, you have a 2/3 chance of picking a dud. When the second door gets opened, you know that one's a dud. So if you know one door is the wrong one, and know there's a 2/3 chance of the other one also being wrong, that implies that for those 2/3, the third door is the only one left that can win.

On the other hand, not switching makes everything afterwards irellevant, meaning you're stuck with the 1/3 win chance that you had when you picked the door.

Temporal Darvi, which isn't to say your wrong but the hard part is to understand where the change is taking place. Since logically why isn't it a 50/50 chance?

Most people think of it in this

OXX

Hmm but if I take away a fake number I get...

OX

So then why does it matter if I change?

But lets look at my first set of numbers

O XX
X OX
X XO

Notice something there? This was where I kept the first number... all the possible combinations... but

O
X
X

Is my outcomes (as in all the possible outcomes are "Right", "Wrong", "Wrong")... But my unused sets are

XX
OX
XO

Which we know for sure is all the possibilities.

But lets not change and see what sets we get while including the freebie door which as we know is always wrong

OX
XX
XX

So not changing gets us that set... because we are given the same 1/3 chance with a door that is always wrong.

While changing gets us the

XX
OX
XO

Set. Where we get the right answer 2 out of 3 times.

That is my best attempt.

Interesting to note that 1/3rd and 2/3rd when averaged makes 1/2.

You get 1/2 if when you got the option to stay or switch... you picked purely at random. Getting this set

OX
XX
XX

and also combined with

XX
OX
XO

Essentially you are right half of the time... because the full set of possibilities (as in all outcomes possible) you get a correct number half the time

But I am sure that doesn't help.

[Darvi]

It would be a 50/50 chance if the second door being opened was independant from the first door opened. But it isn't, so it's not.

[Neonivek]

It would be a 50/50 chance if the second door being opened was independant from the first door opened. But it isn't, so it's not.

Or if you chose whether or not to switch purely at random. As that gets you 6 sets of possibilities as I demonstrated last page.

[Darvi]

Well, no. Because the door you have chosen at first always has that 2/3 chance of being the wrong one, no matter what happens afterwards.

If you had no knowledge about the two doors that are left after one has been opened, you would have an even chance. But since you are 2/3 certain that one of them is the wrong one, you are better off picking the other one, i.e. switching.

[Dutchling]

Ah this one. It's easier to imagine there being 100 doors.

After you pick your door, 98 are removed from the equation, one of the two that are left have the prize.

Are you switching doors?

[Rolan7]

Thanks for all the good explanations.  It does seem a bit more natural now.  I think a helpful part is recognizing that the player isn't trying to guess the prize door at first.  The only thing they're doing is choosing a door which won't be opened by the game host.  *Then* they're trying to guess which door holds the prize.  So the two remaining doors aren't equal at all.  One was exempt from the host's selection and still has the original 1/3 chance.  The other door has the remaining 2/3 chance.

I already understood it mathematically, it just *felt* wrong because I was stuck in the mindset of a game show where the player would try to guess the correct door (both times).  Which isn't how the game works.

[ChairmanPoo]

The trick in the Monty Hall problem lies in what the presenter knows: He never opens a door with a car. Always opens a door with a goat.

After your first choice, you already determined his own actions: either he opens one in two doors with a goat, or he opens the only door with a goat (2/3). Therefore, you should switch, because it's more likely that you picked the other goat in the first place, than that you picked a car.

[KingofstarrySkies]

MATH

[hops]

The one thing I don't get with the Monty Hall problem is that.... what if I switch from door 1... to door 1?

Theoretically, I'd be selecting an option from 2 options, which means that I have a 1/2 chance of getting the right door, right?

[MagmaMcFry]

The one thing I don't get with the Monty Hall problem is that.... what if I switch from door 1... to door 1?

Theoretically, I'd be selecting an option from 2 options, which means that I have a 1/2 chance of getting the right door, right?

No, because the showmaster opened a door, giving you information, and information changes probability distributions.

[hops]

But what if I ignore the fact that there were originally 3 doors and focused on the fact that I can now choose between two doors?

I don't understand how knowing something will cause reality to rearrange itself.

[Dutchling]

It's not knowing something. It's because the showmaster opened a door without the prize behind it.

The chance that you chose the right door is 1/3 (leading to switching doors later being incorrect)
The chance that you chose the wrong door is 2/3 (leading to switching doors later being correct)

[MagmaMcFry]

But what if I ignore the fact that there were originally 3 doors and focused on the fact that I can now choose between two doors?

I don't understand how knowing something will cause reality to rearrange itself.

No. Reality was fixed in the first place. Uncertainty is only in your knowledge, never in reality. For example the showmaster knows exactly which door the car is behind, so his probability distribution was 100% to 0% to 0% in the first place.