In this example, there were 100 choices in the beginning, and later you reduced to 2 choices. Clearly an advantage. Does the same apply to the 3 door problem?
Let’s take this question in another angle. Instead of 3, there are only 2 doors. I am to choose one out of 2, which has a prize. After I choose one, you show me a third door which is empty. Now, should I change my option?
Yes, it’s the same concept. The same math/logic behind it doesn’t change. You’re choosing 1/3 or you are choosing 2/3 and I’ll tell you which of the two is incorrect. It’s just easier to visualize with 100 doors instead.
I’m not sure I’m following the other angle…there are 3 correct possibilities at the start but I can only choose 2? Or there are 2 possibilities and then you introduce a 3rd door that is never correct?
Do you know the third door is never correct? Because then the probability doesn’t change.
Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn’t a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)
Scenario 2: If you think it’s possible that 3 could be correct (but it actually never is) then, no, you wouldn’t want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there’s no way to know this ahead of time (because as soon as you know you shouldn’t switch bc 3 is the wrong door, then you’re back in scenario 1)
Scenario 3: For completeness, let’s say the 3rd door can be correct sometimes. Then it doesn’t matter if you switch or not. It’s a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn’t matter at all and the second choice is the ‘real’ choice bc that’s the only time you’re able to choose from all real possibilities.
The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)
In scenario 1, legit or not, you said the chance is still 50-50. In other scenarios also you shouldn’t change or it wouldn’t matter. That’s what I say, just in the opposite direction. But the problem of probability depends on the wordings and phrases, which means I may not have understood the ques well.
Another angle: You explained the Monty Hall problem at the end that the probability changes because in second choice we have more information. So you are implying that the initial 1/3 probability of the now-open door adds to the door we did not choose - making the switch advisable.
Here I also say the probability does change from initial 1/3, but to 1/2-1/2 for each remaining doors; why should the probability be poured to the unselected single door?
In this example, there were 100 choices in the beginning, and later you reduced to 2 choices. Clearly an advantage. Does the same apply to the 3 door problem?
Let’s take this question in another angle. Instead of 3, there are only 2 doors. I am to choose one out of 2, which has a prize. After I choose one, you show me a third door which is empty. Now, should I change my option?
Yes, it’s the same concept. The same math/logic behind it doesn’t change. You’re choosing 1/3 or you are choosing 2/3 and I’ll tell you which of the two is incorrect. It’s just easier to visualize with 100 doors instead.
I’m not sure I’m following the other angle…there are 3 correct possibilities at the start but I can only choose 2? Or there are 2 possibilities and then you introduce a 3rd door that is never correct?
Yes that one. Similar to the one you did with 100 doors, just in opposite direction.
Do you know the third door is never correct? Because then the probability doesn’t change.
Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn’t a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)
Scenario 2: If you think it’s possible that 3 could be correct (but it actually never is) then, no, you wouldn’t want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there’s no way to know this ahead of time (because as soon as you know you shouldn’t switch bc 3 is the wrong door, then you’re back in scenario 1)
Scenario 3: For completeness, let’s say the 3rd door can be correct sometimes. Then it doesn’t matter if you switch or not. It’s a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn’t matter at all and the second choice is the ‘real’ choice bc that’s the only time you’re able to choose from all real possibilities.
The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)
In scenario 1, legit or not, you said the chance is still 50-50. In other scenarios also you shouldn’t change or it wouldn’t matter. That’s what I say, just in the opposite direction. But the problem of probability depends on the wordings and phrases, which means I may not have understood the ques well.
Another angle: You explained the Monty Hall problem at the end that the probability changes because in second choice we have more information. So you are implying that the initial 1/3 probability of the now-open door adds to the door we did not choose - making the switch advisable. Here I also say the probability does change from initial 1/3, but to 1/2-1/2 for each remaining doors; why should the probability be poured to the unselected single door?