"Stay or switch" is relative to your first choice...

"Stay or switch" is relative to your first choice. When the exact same scenario of door openings happen but the question posed is "pick a door" - ie, not relative to your first choice, the probability is 50/50.

Your second decision is not dependent on your first decision, as the outcome of the first decision is always the same (a goat door is opened). The only thing making your first decision relevant to probability calculations is the fact that 'stay or switch' requires it. When the exact same events happen but the question posed is "pick a door", your first choice is not relevant to the probability calculations which makes it 50/50

You fuckers are still doing this shit?

Fuck off OP there are countless explanations and examples for this.

No its 2/3

Read the post again user
The monty hall problem - ie, stay or switch - is 1/3 to 2/3. This is empirically proven.

But the exact same series of events - with the only difference being the second question is "pick a door" (which is not dependent on the original decision in any way) - is a 50/50 chance because your first decision is not relevant to the probability calculations.

The outcome of the first decision is always one goat door being opened. You have no influence over this outcome. No matter what door you pick, this outcome will always be the same. Your decision was not important.

The only reason the first decision is counted in the probability calculations is that "stay or switch" is relative to the first decision.

When you eliminate any link to the first decision in the question (ie "pick a door"), there are no longer any ties to the first decision and it should be disregarded entirely from the probability calculations.

This is no longer the monty hall problem, but an identical series of events with a different question being asked

>The only thing making your first decision relevant to probability calculations is the fact that 'stay or switch' requires it. When the exact same events happen but the question posed is "pick a door", your first choice is not relevant to the probability calculations which makes it 50/50
Wrong. If you chose the car door then choosing the other door will always lead to a goat. If you chose a goat door then choosing the other door will always lead to a car. It's more likely you chose a goat door since there are two of them and only one car door. Making the host say "choose between the two doors left" changes absolutely nothing, because the fact that the host opened a door which he knew to be a goat makes it more likely that the door he didn't choose contains a car. So if you really wanted to, you could just say that choosing the door which the host decided not to reveal is the best strategy. Oh look, it has nothing to do with "switching." Happy now moron?

All of this of course assumes you are an absolute retard and not just pretending to be retarded. But given the number of retards in the world this is more likely to be true.

>A new person enters the room with no knowledge of the previous events
>The judge asks this new person if he'd like to keep the door the contestant chose, or switch

Does he have a 50/50 chance getting a goat?

Each door has a 1/3 chance of having the car at the beginning. Note that doors 2 and 3 together must have a 2/3 chance. The goat is revealed to be behind door 3. So door 3 has a 0/3 chance of having the car, which means door 2 must have a 2/3 chance by itself.

No, he has a 2/3 chance as long as he knows that the host knowingly revealed a goat door.

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>no knowledge of the previous events

You walk into a room, and the host says one door has been selected and you see a closed unselected door and a third with a goat behind it. Just about anyone would be able to figure it out.

But to get at what i think you're implying.

Imagine all that happened previously occurred, then the goat was taken off stage and all that was left was two unopened doors. The contestant is told to simply select one of the doors. Still there would be a 1/3 and 2/3 chance for each respective door.

The contestant is entirely pointless in this contest, and even what they choose. All that affects the probability is the conscious action of the host which still exists in your modified example.

>Still there would be a 1/3 and 2/3 chance for each respective door.
but there wouldn't because the second person's decision is in no way dependent on the first decision

Fun fact: If the host is speaking English, switching gives you a 2/3 chance; if the host is speaking Spanish, switching gives you a 1/2 chance; and no matter what, OP is a fag who doesn't understand basic semantics.

>but there wouldn't because the second person's decision is in no way dependent on the first decision

Either you're trolling or not listening. The dependency has nothing to do with the contestant, but rather the conscious choice of the host. The contestant could be blind, deaf, and dumb with only three buttons in front of them to feel and choose from. Then one taken away from them as the host shows a goat, then given the option to choose from two buttons they can only feel. All this time never know what at all is happening except they have buttons to press.

The original choice will still have a 1/3 chance and the switched door to would be 2/3.


Again it has nothing to do with the contestant, their knowledge, or their ability to solve the problem.

Statistics is all about knowledge. Given perfect knowledge, there's actually a 100% chance that the goat is behind one door, and a 0% chance it's behind the other.

Given no knowledge whatsoever, one can assume the goat is equally likely to be behind either door. A new person entering would be of this latter kind.

Another way to say this would be to make the host the contestant after a door was removed. His odds of selecting the right door are significantly better than 2/3 given that he knows which door it's behind.

It's not some inherent property of the door that provides the 2/3 chance, it's the information that a goat was removed that gives this chance.

0/10

sorry i cared to try educate someone who even mildly cared

Sorry you don't understand statistics.

There are 1,000,000 doors, 1 with a car behind it and 999,999 with donkeys. You choose 1. Then someone with no knowledge of where the car is opens 999,998 of the doors you did not choose. Miraculously, the car is not behind any of the doors opened. You now probably feel very confident that the door you initially chose has the car behind it. There is actually only a 50-50 chance.

The game is reset and you choose a door. But now someone who knows where the car is opens 999,998 of the doors you didn't choose, purposefully not opening any doors with the car. You now probably feel like there is a 50-50 chance that the door you chose has the car behind it. There's actually a 1/1,000,000 chance.

Think on these examples and you will understand the Month Hall problem.

sorry you dont realize that OP's original questions repeatedly uses the word probability.
Which makes sense because the monty hall problem is one of probability, not statistics..

statistics =/= probability

Monty Hall problem is usually introduced as an anecdote before the probability tree because there were plenty of mathematicians at the time who screwed it up since it's not an intuitive problem, hence an example to show how intuition is usually wrong.

these math problems are retarded any way
how could i use this shit to make something that someone will buy
protip you cant

The first decision is important because they have to open a goat door that is also not a door that you have picked.

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