You are correct just the probability of door 2 (the unopened one of the host's doors) having the car changes,
You are correct just the probability of door 2 (the unopened one of the host's doors) having the car changes,
ask yourself, if you were offered the opportunity to make your choice before the prizes are dealt, would the act of your choosing have any effect on the probabilities of the distribution of the prize?
of course not, the probabilities are one in three for each door. that never changes because of any action taken.
so, you have the opportunity to increase your access from one door to two, before the prizes are distributed, or after. makes no difference. logically. mathematically.
yes, that is the intuitive statement of the solution i and several others, likely including yourself, employed in the first pages of this thread.
but i think that way of stating it is problematic. because it imputes some sort of cause and effect on the action of revealing a door. which is exactly what the magician wants.
which is where nick gets confused, as he insists that because one door has been eliminated (with a flourish of the magician's hand) the probabilities now become 50/50 between the remaining doors.
the probabilities are set by the random distribution of prizes at the outset. it is more proper to say that both remaining doors still have a one in three chance. but, because by rule the dealer must open a goat door, his two choices are logically one probability. 1/3+1/3=2/3.
in choosing to switch to the dealer's remaining door, you have chosen his 2/3 probability of having the prize over your 1/3 probability.
your door does not magically develop a 1/2 chance because of the dealer opening a door. but, neither does the dealer's single remaining door magically develop a 2/3 chance. nothing changes by the door opening action, except confusion of the mind.
To beat this horse with a dead stick I'll ask:
What if you get to switch to the two doors the host holds before he opens one. After you make the switch, he opens one to reveal....
a goat.
Now what are your chances that the other door holds the car?
66.66%Originally Posted by Chris Woodward
"They have a lot of stupid people that vote in their primaries. They really do. I'm not really supposed to say that but it's an obvious fact. But when stupid people vote, you know who they nominate? Other stupid people." -- James Carville on the plethora of low-quality GQP candidates in the mid-term election.
66.66%Originally Posted by Chris Woodward
"They have a lot of stupid people that vote in their primaries. They really do. I'm not really supposed to say that but it's an obvious fact. But when stupid people vote, you know who they nominate? Other stupid people." -- James Carville on the plethora of low-quality GQP candidates in the mid-term election.
I think I'll keep reposting this again and again until the naysayers respond to it.
Like Peb said, if I always switch, I can only lose if my original choice held the car. This is just a statement of fact.
Everyone seems to agree that the first choice has a 1 in 3 chance of having the car, so by switching every time, I have only a 1 in 3 chance of losing.
Anyone still sticking to the 50/50 odds needs to reconcile this fact with their theory.
Ignoring it is just disingenuous.
I was hung up on this originally, as well. Opening one door reduces the odds to 50-50. However, as has been pointed out, the choice was already made, when the odds were 1/3. It's not a new game with the new odds. The other consideration not mentioned here, is that in a game show, a contestant only gets one chance to play the game, so his two thirds likelihood of winning will not be realized by playing over and over again. But, as pointed out above, while the odds are 50-50, the original choice made when the odds are 1 in 3 means that, correctly, by choosing to change means to forsake one's 1/3 chance of being wrong.
Opting instead for a 2/3 chance of being right. Lather, rinse, repeat.
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“That’s diplomats. We’re not gentlemen.”
“So you lie to save your hides.”
That’s politicians. Different game entirely.”
I don't understand your response at all. I don 't believe the odds are ever 50/50.
The odds are always 2/3 to 1/3 in favor of Monty's pair of doors. And we always know, with absolute certainty, that Monty has at least one goat.
So why would you think that consciously (selectively, NOT randomly) showing you that he has a goat change the odds at all?
I don't believe it does, and Peb's observation above supports my belief.
Let's say Monty has 100 doors; 99 goats, 1 car.
You choose 1, Monty reveals 98 goats. Are you really gonna choose to keep your first choice when faced with Monty's final offer???
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Trouble is you think the doors are equal when they're not.
The door you picked initially only had a 33% chance of being right. When one of the wrong doors is removed, the remaining door has a 66% chance of being right.
Yes - its a 50/50 choice, but the odds are not 50/50.
Its a two horse race and one of the horses is slow.
Some day Nick the penny will drop and you'll read back on this thread and probably cringe.
It's all fun and games until Darth Vader comes.
Monty's odds ARE 2/3 because he does not pick. The contestant picks one door and Monty gets whatever two remain.
We disagree often. And with matters of opinion I completely accept that, but the probabilities here are well established and documented. It is in essence: settled science. You need to let go of this one. If you can't grasp it then accept that it is beyond your capacity. Just sayin'