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Talk:Monty Hall problem: Difference between revisions
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I've flagged the entry and hopefully someone will take a look and correct it. --[[User:Paddy3118|Paddy3118]] 16:19, 26 September 2009 (UTC)
==Someones theory==
I dont beleive everyone else, This is my theory:
door #'s: 1 2 3
doors: O O O
probability
So now say we
So now we open door 2 and it's a goat. Now some think that door one still has a 66% chance and door 3 has 33% chance,
but thats not true because now we only have '''2''' numbers; therefore it's 50/50 chance between the two
and I was looking at this and they were using larger numbers to try and explain it, this is what I think:
door #'s: 1 2 3 4 5 6 7 8 9 10▼
now say we
lets say we open all the doors except for door 7 and door 9.
door #'s: 7 9
Doors: O O
okay and now they both have a 50/50 chance. PROBLEM?▼
Well then,This is what I think, my friend tells me I'm crazy but I just don't know.
:The point about the problem is that the experimental result - as mirrored in the correct simulations, is often seen as un-intuitive. If you google you can find several explanations, maybe one will work for you? --[[User:Paddy3118|Paddy3118]] 06:50, 13 April 2011 (UTC)
:It's not clear to my why you think that opening that door changed the probability that you had originally guessed right. --[[User:Rdm|Rdm]] 11:47, 13 April 2011 (UTC)
:I can't tell you how it works, just that it works. All of these simulations and thousands of math professors can't be wrong. --[[User:Mwn3d|Mwn3d]] 13:24, 13 April 2011 (UTC)
▲ 1 2 3 4 5 6 7 8 9 10
▲ O O O O O O O O O O
▲each one has a 10% chance
▲now say we open door 9, at this point, you only have a 1 in 10 chance of getting it right. now next step:
: Conditional probabilities ''are'' difficult, but when you model the problem exactly with a contestant agent and a Monty Hall agent (and you can clearly see who has just how much knowledge) you get the counter-intuitive result. Because when you model it properly you get this, it means that if your theory gives any other result, it is ''your'' theory that must be wrong. IOW, Reality sees your theory, has a good chuckle, and spits it back out.
▲lets say we open all the doors except for door 7 and door 9 so looking at this problem with all the doors you would think it's in door 7, BUT thats not true now we just have two doors:
: Here's how I recommend thinking about it. You had a 1/3 chance of picking the right door initially (simple!) so there's a 2/3 chance you got it wrong to start out with. ''What Monty Hall does doesn't change that at all.'' Switching is like going from picking one door to picking two; yes, one has been subsequently eliminated, but you knew at least one of them didn't have the car anyway. BFD. You're switching from a 1/3 chance to the complement of a 1/3 chance (i.e., 2/3), which is a win. –[[User:Dkf|Donal Fellows]] 13:28, 13 April 2011 (UTC)
:: And if you still don't believe me, get a friend and play the game for real (well, maybe with a swig of beer instead of a car). 10 or 20 rounds with one strategy or the other shouldn't take too long. –[[User:Dkf|Donal Fellows]] 13:31, 13 April 2011 (UTC)
:"Now some think that door one still has a 66% chance and door 3 has 33% chance, but thats not true because now we only have 2 numbers; therefore it's 50/50 chance between the two". This is wrong. The fact that you know something after choosing does not change the fact that on average you got it wrong 66% times in the first place. [[User:Eoraptor|Eoraptor]] ([[User talk:Eoraptor|talk]]) 06:48, 18 February 2019 (UTC)
▲okay and now they both have a 50/50 chance.
==Not quite good enough to be a reference on the page==
* [https://www.youtube.com/watch?time_continue=153&v=AD6eJlbFa2I&feature=emb_logo The Monty Hall Problem | Brooklyn Nine-Nine]
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