"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
Carson Brooks
You fuckers are still doing this shit?
Fuck off OP there are countless explanations and examples for this.
Joshua Lewis
No its 2/3
Samuel Foster
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
Levi Robinson
>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.
Brody Scott
>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?
Ian Hernandez
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.
Andrew Ramirez
No, he has a 2/3 chance as long as he knows that the host knowingly revealed a goat door.
