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I really hate the Monty Hall problem. I just can't accept that changing doors makes a difference.
If a cat can be half dead, and half alive,Does it still need to be fed?
About the Monty Hall problem.1. we have differently sized infinities defined mathematically. Would now this reasoning give me a better probability of finding something in a lesser infinity, as you can see that as open and then taking aways 'some room'
2. Imagine i have two doors to choose between. Somebody tells me that there was three doors before I got there, but one got taken away containing a goat, to the right side. Will this guarantee better odds for me if I too pick the right side? If it does, isn't this a hidden parameter? Information?
Alternatively getting informed of which door he opened to then remove, leaving two choices for me in which I first choose one door of two, to then switch it. Would my chances improve?
Why wouldn't it give the same odds?the situation is the same as if I stood before those three doors, seeing him open the one to the right, finding a goat? Instead of being there I get informed of what door that was. Ahh, I think I see, I didn't make that choice before getting informed
You can't have it both ways
Either the way I describe is equivalent to the original experiment, only one door missing as I arrive to switch my original choice, or it isn't equivalent. To me it actually is equivalent.
If one want to define the mathematics on what doors that really is existent at the time I arrive you're putting a lot of weight on what exist, less on the mathematics being equivalent.
As the situation is the exact same, except that instead of me standing there, watching him choose a door, I'm on my way to the game. You could imagine me seeing him on a television, or someone informing me per telephone. Otherwise it should be the exact same as it seems to me, although he remove the door he opened before I arrive.
If you now assume that the odds change because of the removal of a door that we both know to be wrong, then it seems to me that you also have to assume that 'kismet' steps in, to rearrange what's behind the two doors that's left, somehow?
Well, the mathematics won't care where you, or the door, are. As long as you're informed about the game as I see it. That simple..
"You can't have it both ways" referred to both your posts before, misread you there, and commented on that in the post, take a second look under the "=".
His choice of doors you mean?
Assume him to be informed of my choice then. That leaves him the same choices as in the original experiment.
But the parameters didn't change, you know them just as good as if you had been standing in front of three doors the whole time, and that's my point.
What I'm wanting to discuss is whether you can assume the odds to still be there after that 'one door' is gone too.
And I presume that you should be able to, assuming that you have the same information as if standing in front of those doors the whole time. Otherwise it becomes a example of a mathematics based not on 'information', instead based on? Tactile reality? As you then should need all of those doors existing, to be able to 'switch' door for getting those better odds, in the end.
Umm...When I first joined this debate about the Monty Hall problem, I said that the main problem that I had with it was why so many authorities were saying that the odds increased from 1 in 3 to 1 in 2 if you swapped. In fact it increases from 1 in 3 to 2 in 3.Wikipedia is one source that has the right answer:http://en.wikipedia.org/wiki/Monty_Hall_problemMy reply #23 on this thread clearly points out (to my way of thinking) why this is the case. I simply cannot see why others cannot see it this way.
That's a weird one alright. Saw someone explain it with a hundred doors instead, suddenly making it make sense. You have a hundred doors to pick from, you pick one of them. Then the game leader opens 98 of the other doors showing you nothing in them but goats. Now the question becomes one of keeping your original one that you picked randomly out of a hundred, not opened, or use the last door out of 99 that the game leader opened? It's a question of odds, and you picked one randomly from a hundred closed doors, but the 'other side' of it is the one where 98 doors was opened to find nothing, one left. It's like two games, the one you had from the beginning being the hardest to guess, wheres the one the game leader had being the 'foolproof' one. In reality it can't be foolproof as it could be your door too, but imagining it as two separate games makes it easier to see the reasoning.and it is weird as you could imagine yourself not choosing any of those doors, waiting until the game leader opened 98 of them, then having two doors left to choose between. In that case you would have a 50/50 % probability of getting the right one, as I see it. Statistics as magic?
Thankyou Yor_on! your example just highlights my reasoning! Because you only had a 1 in 100 chance of getting the first door right, then you are 99% sure if you change doors that you will be right after the game leader has opened 98 of them
Quote from: CliffordK on 06/05/2013 06:35:21If a cat can be half dead, and half alive,Does it still need to be fed?If it's Schrodinger's cat, it's both dead and alive, so you only need to feed the living version. You bury the dead version.I once learned how to mentally calculate the day of the week of any given Georgian calendar date. It was too complicated and wasn't useful enough, even as a party trick, to remember once the novelty wore off (on a Thursday).