@mathforphysics Interesting. I am not arguing. Perhaps this is a forbidden thought experiment. You know. Maybe Choice allows us to choose a door between 2 and infinity, but never "randomly."
@mathforphysics I lost your last comment. Sorry about that. You wanted me to clarify that this is a random selection from an infinite number of doors. Yes, I am taking Monty Hall and extending it from three doors to infinite doors. I am then "privileging" door number one. You say that the prob. never reaches zero. It is paradoxical. Perhaps it is forbidden to randomly pick from an infinite number of choices, avoiding this paradox!
I would start by adding one door and seeing if the probabilties change. I think it ously would. Then, adding a large number of doors would further increase the probability that the tiger is behind door number 2. But, I believe, the whole exercise might be unnecesary because you could just ask "if the probability of an event is infinitsmally small can it still happen?" That is what your question reduces to and I think the answer is "yes."
I just reread my post and I see that it isn't clear at all. By "the probabilities are the same" I meant that you should switch doors. You should act the same as you would in the 3 door scenario.
Maybe I misunderstood your proposal, but I think your idea was to replace door number three with an infinite number of doors.
My original intuition was that the probabilities should reduce to the same numbers if you consider a larger number of doors and open all of them, but now I'm not sure.
I think you meant to say "all doors to the right of door 2." Door 3, Door 4 etc. Otherwise, there is only door 1 left to open and no choice between one and two. Assuming that is correct, then Iwould say the probablities are the same. The infinite number of doors (if they all have tigers) are equivalent to one door in that the lady can't be there.
@mathforphysics So you are suggesting that in the Monty Hall problem the actual answer is 50-50? Do a sample space and you will see otherwise. (Unless I have misunderstood you)
The probability might go to zero, but never reach zero. There should be a small chance that the Lady is behind door one.
mathforphysics 8 months ago
@mathforphysics Interesting. I am not arguing. Perhaps this is a forbidden thought experiment. You know. Maybe Choice allows us to choose a door between 2 and infinity, but never "randomly."
CHistrue 8 months ago
@CHistrue
I think i might have misunderstood. I thought all the infinite number of doors were opened?
mathforphysics 8 months ago
@mathforphysics I lost your last comment. Sorry about that. You wanted me to clarify that this is a random selection from an infinite number of doors. Yes, I am taking Monty Hall and extending it from three doors to infinite doors. I am then "privileging" door number one. You say that the prob. never reaches zero. It is paradoxical. Perhaps it is forbidden to randomly pick from an infinite number of choices, avoiding this paradox!
CHistrue 8 months ago
@CHistrue
I would start by adding one door and seeing if the probabilties change. I think it ously would. Then, adding a large number of doors would further increase the probability that the tiger is behind door number 2. But, I believe, the whole exercise might be unnecesary because you could just ask "if the probability of an event is infinitsmally small can it still happen?" That is what your question reduces to and I think the answer is "yes."
mathforphysics 7 months ago
I just reread my post and I see that it isn't clear at all. By "the probabilities are the same" I meant that you should switch doors. You should act the same as you would in the 3 door scenario.
Maybe I misunderstood your proposal, but I think your idea was to replace door number three with an infinite number of doors.
My original intuition was that the probabilities should reduce to the same numbers if you consider a larger number of doors and open all of them, but now I'm not sure.
mathforphysics 8 months ago
I think you meant to say "all doors to the right of door 2." Door 3, Door 4 etc. Otherwise, there is only door 1 left to open and no choice between one and two. Assuming that is correct, then Iwould say the probablities are the same. The infinite number of doors (if they all have tigers) are equivalent to one door in that the lady can't be there.
mathforphysics 8 months ago
@mathforphysics So you are suggesting that in the Monty Hall problem the actual answer is 50-50? Do a sample space and you will see otherwise. (Unless I have misunderstood you)
CHistrue 8 months ago
Excellent video.
mathforphysics 8 months ago
@mathforphysics A fine compliment from a fine gentleman. Thank you!
CHistrue 8 months ago