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solution? wrote:Really, it shouldn't matter, right? You already increased your odds from 33% to 50%. So at this point your odds are equal on either door.
always switch it gives you better odds of winning since there are three different possiblilities. either you picked empty door 1 initially and switching would get you the kittens, or you picked empty door 2 initially and switching would get you the kittens, or you picked the kittens initially and switching would lose it for you. so by always switching you would improve your odds from 1/3 to 2/3
You pick the door you didn't pick.
The door you picked (say door A) had 1/3 chances of having girl (don't like kittens)
The doors together had 2/3. (say door B and C)
When the host eliminated on of those two doors (Say door C), the chances of the girl being behind the door you didn't pick (door B) is still 2/3 . Your door (door A) just has the 1/3 chance.
So depending on whether you want the girl or have a gf you pick appropriately.
Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.
Anyway, orthogonal to the thread but i thought it was appropriate.
Yakk wrote:You can fix this by specifying the exact algorithm the host follows. "The host always picks and opens a door that is empty, and you know this", or "The host always opens a door at random. This time you got an empty door."
Yakk wrote:"The host always opens a door at random. This time you got an empty door."
Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.
Anyway, orthogonal to the thread but i thought it was appropriate.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
skeptical scientist wrote:I once asked this as a riddle on a long car trip, and unsurprisingly nobody got the answer. However, after I revealed the correct answer and explained it, they kept arguing and wouldn't believe I had the correct answer (in fact, the debate lasted several days....
elminster wrote:There are 3 cups, 2 are upside down, 1 is the right way up. You are told one of the 3 cups contains a ball. If you know for a fact that the right way up cup does not, and will not contain the ball, even if it could, it wont for any particular asking of this simulation. The probability would be 50/50.
It represents the same probabilistic state as the game show though.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
EvanED wrote:Sygnon wrote:I recently had to explain this to a small group, I have found that when explaining the problem you can change the numbers and imagine that you start with 100 doors, select 1, open 98, This immediately keyed people in on what was happening.
Anyway, orthogonal to the thread but i thought it was appropriate.
I never found that argument compelling. I don't see how it works with 100 doors any better than 3 myself.
yellowtamarin wrote:I can't see any posts that have put it as simply as I see it:
(The scenario where we know a door without kittens will be opened after we choose.)
The probability is 50/50, as we are always only choosing between two doors, one with and one without the kittens. There is no third door to consider because it is always removed from the equation and we know this.
As for the other scenario where we don't know that a losing door will be opened, there is a neat pictorial represenation of this in the book The Curious Incident of the Dog in the Night-time. Although it is used to illustrate the first scenario, which makes it incorrect.
You choose one door. You have a 1/3 chance of that door being correct and a 2/3 chance one of the other doors is correct.
One of the other doors is revealed to not contain kittens.
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