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Talk:Monty Hall problem: Difference between revisions
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::Sorry for inserting misinformation. Could you please tell me why my program guesses correctly 50% of the time instead of 2/3 of the time? It's the Common Lisp one.
:::Your "show-goat" only depends on "round" (the state of which door has the car and which ones have the goat). In the actual Monty Hall problem, which goat the host shows also depends on the choice that the player makes (i.e. "initial" in your code) -- in particular, you need to make sure that the goat that the host shows is ''not'' the door that the player chose (i.e. "goat" should never be equal to "initial"). --[[User:Spoon!|Spoon!]] 07:36, 6 August 2009 (UTC)
:::Also, when the player chooses the winning door, preferably the goat door shown should be chosen randomly from the two available. If you always choose the one the has lower index then the player can use the ordering of the doors to get more information. --[[User:Spoon!|Spoon!]] 07:53, 6 August 2009 (UTC)
::::Thank you, now my program works correctly. [[User:Foobie-bletch|Foobie-bletch]] 08:10, 6 August 2009 (UTC)
:If you look at the Tcl solution, you'll see that the chance depends massively on the strategy being used. (I'm rather proud of the fact that I separated the player strategies so it was clearer what was going on.) Indeed, this whole game is great for sorting out the men from the boys when it comes to probabilistic analysis. If it makes you feel easier, think of it this way: would you rather choose one or two doors to start out with? Does the prize move around? —[[User:Dkf|Donal Fellows]] 08:07, 6 August 2009 (UTC)
::I, on the other hand, think that probability questions can easily have non-intuitive answers. I just have to take more care, and simulate where possible :-) --[[User:Paddy3118|Paddy3118]] 15:56, 6 August 2009 (UTC)
== Scheme Solution Problem? ==
I would have thought that with 1M trials, that the Scheme implementation would give a result closer to 1/3 and 2/3, instead of:
<pre>;; > (compare-strategies 1000000)
;; (stay-strategy won with probability 33.3638 %
;; and switch-strategy won with probability 51.8763 %)</pre>
Is their a problem? --[[User:Paddy3118|Paddy3118]] 01:44, 26 September 2009 (UTC)
: Sounds like they've got a bug; that's massively off the expected frequency. Probably due to incorrect problem modeling; it's very easy to get the problem model wrong or jump to a (wrong) solution. That's why the Tcl solution specifically models the whole thing, since only then can you know whether you've got it right or blundered. —[[User:Dkf|Donal Fellows]] 10:34, 26 September 2009 (UTC)
: sqrt(1M)=1000, so a deviation of this order of magnitude isn't exactly unexpected. Indeed, in this case the deviation is just below two standard deviations, i.e. a bit high, but still inside the 95% confidence interval. That is, for a truly random sequence with probabilities 1/3 and 2/3, there's a 5% chance of an even higher deviation. So, no clear indication of a problem. --[[User:Ce|Ce]] 10:39, 26 September 2009 (UTC)
:: Oops, I had looked only at the first number; now I notice that the second number is indeed far off. Given that winning with the switch strategy is the same as not winning with the stay strategy, I couldn't imagine that you could get one right and the other wrong ... --[[User:Ce|Ce]] 10:47, 26 September 2009 (UTC)
::: Looks to me like they're just randomly picking from two doors the second time. (Or their RNG is improbably distributed, which would be a much worse problem.) —[[User:Dkf|Donal Fellows]] 11:42, 26 September 2009 (UTC)
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:
alright, now lets say each O is a door-
door #'s: 1 2 3
doors: O O O
probability: 1/3 1/3 1/3
okay each door has a 1/3 chance; but the two on the left have 66% chance.correct,
So now say we pick one for example, door 3; awesome possum, so we say that it has a 33% chance of it being a car.
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
doors: O O O O O O O O O O
So each door has a 10% chance
now say we predict door 9; at this point, you only have a 1 in 10 chance of getting it right. now next step:
lets say we open all the doors except for door 7 and door 9. Now you would think it's in door 7, BUT thats not true now we just have two doors and TWO doors only not 10 but Two:
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)
: 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.
: 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)
==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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