Probability question

[root]

To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.

You see this is exactly why this solution is flawed, it assumes that switching is increasing the amount of doors that you're looking behind.

At the start there is. 1/100,000 chance that you pick the right door.

Now. If the host asked you do you wan't to keep your door, or take the other 99,999 doors then you'd dramatically increase your odds, you'd be a fool it two switch. It's a 1/100,000 chance vs 99,999/100,000 chance...

But that's not what happens, the host only opens doors that are incorrect, whether you switch or not you have equal odds because you've already been given the doors,

The idea that there is more chance that the prize is behind door nu
Bed two because you picked door number 1 and it wasn't behind door 3- door 100,000 is just crazy.

Those that use the statistical proof that you're getting more doors to check ignore the fact that regardless of whether you switch you've seen behind 99,998 doors, there are only two doors left, you have a new desecrate decision, the prize is bhind 1 of 2 doors and the odds are 50:50

 

[berry120]

Those that use the statistical proof that you're getting more doors to check ignore the fact that regardless of whether you switch you've seen behind 99,998 doors, there are only two doors left, you have a new desecrate decision, the prize is bhind 1 of 2 doors and the odds are 50:50

So you honestly think if there's a hundred thousand doors, you've still got a 50/50 chance whatever one you pick?

If you like, all statistical proof aside I'll knock up a program that emulates this and shows otherwise

 

[root]

No, you entirely misunderstand what I'm saying.

If there are three doors. You have to pick one, if the door opens you win prize that's a 1/3 chance
If there are 100 doors, you have to pick 1 if it opens you win that's a 1/100 chance.

I'm sure that we agree so far.

Now take away a door. You're working on the premis that there are still three doors, there aren't you can no longer pick the open door, there are only two doors left. One definilty has the prize, one definitely doesn't.
The odds are 1/2 it's a new decision entirely delegate from the first decision. The odds of that first decision are not compounded.


Also above you say that there are only three possible outcomes
You pick the prize to start with and switching looses
You pick nothing and switching wins
You pick nothing and switching wins

But hat ignores the fact that the fourth outcome is that
You pick the prize and switching to the other door (assuming that the choice of door is random) then sit hint always looses.

This is because this has to be a part of any decision tree that you write.

If you pick the correct door to start there is an eq possibility that either door will open and thus in the end you have an equal chance of switching to door 2 or 3 (whichevr is unopened).

If you pick the wrong door to start with the the host is forced to ion a specific door, thus there is only the choice that switching will win twice, what there is also the option that switching will loose twice also.

The odds are equal.

Though if you had written a program you'd have noticed that when you were writing the logic to randomly open the remaining doors if he correct door is picked to start with, or to open a specific door if the wrong door is chosen to start.

 

[berry120]

If there are three doors. You have to pick one, if the door opens you win prize that's a 1/3 chance
If there are 100 doors, you have to pick 1 if it opens you win that's a 1/100 chance.

I'm sure that we agree so far.

Yup, we agree on that one.

Now take away a door. You're working on the premis that there are still three doors, there aren't you can no longer pick the open door, there are only two doors left. One definilty has the prize, one definitely doesn't.
The odds are 1/2 it's a new decision entirely delegate from the first decision. The odds of that first decision are not compounded.

You're treating this like two individual events, which they're not - this is not the same as flipping two 50p pieces for instance, which are separate events. What makes this one stand out is that the door the host opens isn't just random, he HAS to open a door that's not got the prize behind. In which case, if you pick a door without the prize behind (a 2/3 chance) you FORCE the host to open the other door without the prize behind. If you pick the prize first, then the host can open any other door (but there's only a 1/3 chance of this happening.)

In other words, the run up to the final event counts.

If you pick the correct door to start there is an eq possibility that either door will open and thus in the end you have an equal chance of switching to door 2 or 3 (whichevr is unopened).

If you pick the wrong door to start with the the host is forced to ion a specific door, thus there is only the choice that switching will win twice, what there is also the option that switching will loose twice also.

But the host can only open one of the incorrect doors after the 1/3 chance that the prize is picked, therefore giving each one of those options a 1/6 chance combining to form 1/3:

I'm not sure whether you're saying you disagree with the fact it's a 2/3 chance (which has been mathematically proven, is shown by playing the game and has been demonstrating by mathematicians all over the world) or you're saying you can't see why it's a 2/3 chance?

 

[root]

No, I can see statistically why it's a 2/3 chance.

But I still think that it ignores the fact that whatever door you pick you've see 1 of the doors,

The idea is that you pick door 1
So the host has doors 2 and door 3

The host has a higher probability of having the prize, bu also a Cretan toy of having a goat.

When the host opens a door it stops becoming his exclusive knowledge, it's no longer hi s door it's your door as well,

Now the host has 2/3 doors. But because you know what's behind one of the doors you also have 2/3 doors.

The decision trees all ignore that. Basically they only work if you pretty much ignore the problem. Or at least the facts that come into play.

Also, when the host opens a door, whether his hand is forced or not you only have two doors remaining in play.

The hot simply asks you to pick a door. (asking if you wan't to switch is the same as just asking you to pick again).
It's effectiyma new question and aside from the fact that it has the same doors it's a completely unrelated question.

In that sense it's very much like flipping a 50p, a the fact that a head came up last time has no effect on the outcome this time, it's a new question a new event.

A better example is rolling a six sided dice, the odds of each number are 1/6 (you pick 1)
If I now say that numbers 456 are dud and start using a three sided dice the odds aren't better for 2 or 3 coming up simply because you originally picked 1 and 456 are now gone.

The odds of getting 1 are still the same as the odds of getting a 2 or a 3.

The fact that I roll the dice on the same table bears no relation to it either. New question, new decision tree.

The only time you wouldn't make a new decision tree is if the event really did bear ground to the other events.

Like vetting head head head in a series of coin tosses.

Th fact is,
You pick 1 there is a 1/3 chance you're right a 2/3 chance you're not
If you pick 2 there is a 1/3 chance you're right a 2/3 chance you're not
If you pick 3 there is a 1/3 chance you're right a 2/3 chance you're not

Say you pick 1, your odds are 1/3
The host has odds of 2/3 that it's bhind either door 2 or door 3
They open door 2, what are the odds that it's behind door three.

Basically. Because the door is open you now have that door too.
You now have the benefit of knowing what's behind that door. The host has two doors. 1 mystery 1 open, you have two doors, 1 mystery and the same one that the host has opened.

How many doors in play?
If you say three then the odds are
Door 1 .5
Door 2 (the open door) 0
Door 3 .5

If you say only two doors in play then it's 1/2 probability that you have either the right or wrong doors.


As I said above. When you have 1 door and the host has 2 the host is twice as likely to win as you.
When the host opens that empty door he given up his advantage.

Either through the fact that he has only 1 of three doors that could contain the prize and has reveals his duff door.
Or because he's given up his duff door. Which you now have. So you both have a mystery door and know it's not door 2
Or because you realise that there are only two doors inplay.

Mathematicians that have proved this ways write a decision tree such that it ignores that fact that one door is out o play and now you have a new distinct choice to make.

The same as the dice example I gave above.

I see I'll need to correct my posts when I have a computer and not a phone to write on.

 

[berry120]

The only time you wouldn't make a new decision tree is if the event really did bear ground to the other events.

But that's just it - it does bear ground to the other event, because the choice the host makes is limited by your first choice which will not be the same in all cases. That's where the statistical odds come into play. If, after opening the door, the host shuffled the two remaining doors randomly then that would make the choice unrelated to the run-up for want of a better word, but as it stands it's the run up that generates the biased result.

 

[root]

my point is that when the host asks if you want to swap doors the effectively are shuffling and starting again. it's only because of the phrasing of the question that you believe that they are related.

basically you're saying, pick door 1 2 or 3, there are equal odds that the prize is behind any one of the doors.

the host the opens another door that you did not pick.

then the host says would you like to swap doors.

you say the act of having a door to swap from makes the two events related.

but what I'm saying is what the host takes away a door.
then removes your choice and says pick a door.

the act of being able to swap pretty much is taking away the doors shuffling them up and asking you to pick again.

I'm not disagreeing with the decision tree's.

I'm just saying that they don't relate to the problem. because they assume that the choice of swapping is related to the first act of choosing. and it's simply not.

being able to swap is effectively letting you have another go at picking with a choice of only two boxes, so the odds are 50/50,

 

[berry120]

my point is that when the host asks if you want to swap doors the effectively are shuffling and starting again. it's only because of the phrasing of the question that you believe that they are related.

Not at all, if shuffling was involved you'd have no guarantee that the door you picked had the same prize behind is as it did originally. The difference here is that you have complete control over the door, which you can use to your advantage.

I'm just saying that they don't relate to the problem. because they assume that the choice of swapping is related to the first act of choosing. and it's simply not.

It's not the choice of swapping, but the door the host opens that's related to the first act of choosing - and there's no denying that since the host is either free to choose what other door he wants, or restricted to just one door.

If you're still having difficulty seeing then play the game by the above rules - seriously, do it 10 - 20 times and you'll most likely see the chances come much closer to 2/3 and 1/3 than 50-50!

 

[JogaBonito1502]

@Root, referring to the 1,000,000 door example:
There are 1,000,000 doors to pick from. You pick one. Your odds are 1/1,000,000 to pick the door with the prize.

Agree?

The host now clears 999,998 doors, leaving 1 door closed. When you picked your door, you didn't know this, your probability of having picked the right door is still 1/1,000,000. Just because you know the other doors don't have the prize doesn't mean your odds of picking the correct door have increased to 50%. That would be like saying you could have picked 500,000 doors before and you can open them all simultaneously. The host opening 999,998 doors would then have no effect, because 499,999 of those doors were already yours. The host just opens 999,998 doors that are incorrect after you've picked your door, which means you still have 1/1,000,000 chance of being right, but you just happen to know that the other 999,998 doors were not right. The odds that you picked the right door are super low because you didn't know this at the time you made your choice. The events are not independent.

If you can't believe this, then do as Berry says. Try it out yourself. Have someone hide a ball in 5 cups while you're not looking. Once you make your pick, have them reveal 3 empty cups. Then always switch to the other cup. You'll see that 80% of the time you'll get the ball.

I usually don't say things like this and I definitely don't mean to offend, but various mathematicians who are amongst the best with statistics have agreed that the events are not independent and your chances increase by switching.

 

[superman22x]

You have 3 doors, you pick one, whether or not you picked the right one, the host will pick the the only or one of the empty doors left. Your choice and his have no bearing on each other at all, and all we know is that the host will pick one of the empty doors. So there was a chance you picked the right one first really, and your first choice has no bearing on the host's choice.