File: stay or switch.swf-(1.12 MB, Game)
[_] Stay or switch, /f/? Anonymous 01/09/13(Wed)21:06 No.1853435
Marked for deletion (old).
>> [_] Anonymous 01/09/13(Wed)21:24 No.1853447
fuck yeah i won a car
>> [_] Anonymous 01/09/13(Wed)21:25 No.1853448
wow this game is fucking retarded, why would it matter if you stay or switch
>> [_] Anonymous 01/09/13(Wed)21:34 No.1853452
>>1853448
Always switch. That way you have 1/2 instead of 1/3 chance. There were massive threads about this
on /b/ for fucking ever...
>> [_] Anonymous 01/09/13(Wed)21:35 No.1853454
actually it does matter, i cant explain it, a friend studying math explained it to me sometime
>> [_] Anonymous 01/09/13(Wed)21:38 No.1853456
>>1853452
The correct choice is to Stay. It is a well known probability example, and the removal of a door
does not change a 1/3 chance to a 1/2 chance.
>> [_] Anonymous 01/09/13(Wed)21:40 No.1853457
>>1853452
If you choose to stay you also have 1/2 of a chance to have picked the correct one
/b/ is retarded
>> [_] Anonymous 01/09/13(Wed)21:43 No.1853462
love that car winning melody
SOOOOOOOOOOO SATISFYING
>> [_] Anonymous 01/09/13(Wed)21:44 No.1853463
http://en.wikipedia.org/wiki/Monty_Hall_problem
let the doors be named x y and z.
let x be the door you choose first.
if x's chances of getting the car is equal to 1/3, then the chances of either of the other two
doors having the car is 2/3. (Y + Z = 2/3)
you are then shown which one of the other two doors actually has the goat - let's say it's door
y. thus, you know that door y now has a 0/3 chance of having the car because it doesn't have it
at all.
Y + Z still = 2/3, however, because there is a 1/3 chance that X has the car and there are 3
doors. if we know that Y = 0, then that means 0 + Z = 2/3.
thus, by simple algebra, Z has a 2/3 chance of having the car.
this is one of the few demonstrable paradoxes that people get consistently wrong because life is
not always intuitively correct
>> [_] Anonymous 01/09/13(Wed)21:50 No.1853469
>>1853463
All i know is I am now the proud owner of 3 cars and a goat.
>> [_] Anonymous 01/09/13(Wed)21:52 No.1853472
>>1853457
>>1853456
Switch is the proper choice.
It's really quite silly.
>> [_] Anonymous 01/09/13(Wed)21:54 No.1853475
>>1853472
>>1853463
Reading the Wikipedia article did make me reconsider my thoughts on it, but it blows my mind too
much
>> [_] Anonymous 01/09/13(Wed)21:56 No.1853480
>>1853475
Math has some things like that.
Like how .999 repeating is the same as 1. That one seems more reasonable, but still trips people
up sometimes.
>> [_] Anonymous 01/09/13(Wed)21:59 No.1853485
>>1853480
The Universe truly is a strange place to live in
>> [_] Anonymous 01/09/13(Wed)22:01 No.1853488
>>1853452
the answer's correct, but the reasoning's wrong. if you switch, you get a 2/3 chance of getting
the car.
>>1853456
>>1853457
common misconceptions. there's nothing compelling you to not choose the goat door that's revealed
- it's just that it now has a 0/3 chance of having the car. there are still three doors to choose
from.
>> [_] Anonymous 01/09/13(Wed)22:03 No.1853490
here's another explanation that doesn't involve algebra:
let's call the first goat "goat A" and the second goat "goat B."
at the start you can choose from any door. let's say you choose door number 1 for simplicity's
sake, but this applies to every door. we also have to assume for the sake of the problem that the
host will only reveal a goat - he'll never reveal the car.
if you choose door number 1 and it actually has a car, then the host will reveal either goat A or
B. if you choose to switch, then you get the goat.
if you choose door number 1 and it actually has goat A, then the host will reveal goat B. if you
choose to switch, then you'll get the car.
if you choose door number 1 and it actually has goat B, then the host will reveal goat A. if you
choose to switch, then you'll get the car.
thus, switching has a 2/3 chance of getting a car.
>> [_] Anonymous 01/10/13(Thu)00:15 No.1853623
>>1853488
FUCK
I finally get it. You aren't choosing between two doors, you are choosing between two unknown
doors and a known goat. That's what makes the switch door more likely to be a car. Because the
known door will never be a car.
>> [_] Anonymous 01/10/13(Thu)00:17 No.1853624
the math on this one always makes my head hurt.
>> [_] Anonymous 01/10/13(Thu)00:30 No.1853632
>>1853624
Right? It seems like all you've done is eliminate a door and so are left with a 50/50 choice,
except that the door eliminated is based upon what door you chose. So if you pick goat A, they
reveal goat B 100% of the time, and visa versa, and if you pick the car then they reveal either
goat with a 50% chance. But that means that you still are picking between unknown 1, unknown 2
and a goat.
>> [_] Anonymous 01/10/13(Thu)00:32 No.1853633
Why won't this work right?
I havn't stayed once, and I've played 25 games, won 15, so I have 60%
Shouldn't it be 66%?
If I play to 100 should it balance out?
>> [_] Anonymous 01/10/13(Thu)00:35 No.1853635
Another way of looking at this problem is this: you have a 2/3 chance of initially picking a
goat, in which case the host effectively shows you precisely where the car is by eliminating the
other goat. The only time the host does not show you the car is if you have picked a car to start
with, which only happens 1/3 of the time.
>> [_] Anonymous 01/10/13(Thu)00:41 No.1853643
>>1853633
Probability is never truly clean or exact, if the sample is truly random (or close enough to it).
There will be runs, even though switching gives you a car 2/3 of the time, where you will get 10
goats in a row.
>> [_] Anonymous 01/10/13(Thu)00:44 No.1853645
>>1853635
Exactly. Staying means you only have the same 1/3 chance of getting the car you had when you made
the first choice, while switching converts that to only a 1/3 chance of losing the car you had.
>> [_] Anonymous 01/10/13(Thu)00:48 No.1853649
played 20 games won 10 times ...
>> [_] Anonymous 01/10/13(Thu)00:52 No.1853650
>>1853633
Random number generator
>> [_] Anonymous 01/10/13(Thu)01:05 No.1853660
fuck yeah i won a goat
>> [_] Anonymous 01/10/13(Thu)01:08 No.1853662
As long as a goat is revealed (which it always is), you have a 50/50 shot whether you switch or
stay.
FUCK EVERYTHING!
>> [_] Anonymous 01/10/13(Thu)01:29 No.1853678
I think that the 1,2/3rd people forget that since one goat is ALWAYS shown it means that one of
the doors is now eliminated for a chance at a car.
Door X, Y, Z.
X has a goat
Y and Z are the only 2 unknown doors left
Ergo 1/2
>> [_] Anonymous 01/10/13(Thu)01:33 No.1853683
This is the best analogy I've seen for those who still don't understand why it's advantageous to
switch:
Imagine now that the game has 100 doors. You pick a door, and the host reveals that behind 98
other doors, there are all goats. Now, do you assume that your first guess, one door randomly
picked out of 100, is correct, or do you switch to the one other door that hasn't been opened yet?
>> [_] Anonymous 01/10/13(Thu)01:34 No.1853685
dicks
>> [_] Anonymous 01/10/13(Thu)01:38 No.1853688
>>1853683
I am enjoying this. Imagine it with 30,000 doors, one for each person in a major stadium, and you
pick one, and all but one other door is revealed to be empty.
Do you stick with your original door?
BALLS OF STEEL.
>> [_] Anonymous 01/10/13(Thu)01:39 No.1853689
>>1853678
No, not at all. It's all about the initial choice before a goat is revealed.
If you select a goat initially, the only door that can be revealed will be the other goat, so
when you switch you will get the car. The probability of selecting a goat initially is 2/3.
>> [_] Anonymous 01/10/13(Thu)01:41 No.1853690
>>1853463
i see your math there, but if you think about wouldn't z or y have a 1/2 probablity since one of
them is always goat then that door '1/3' is not part of the equation either i.e goat=x car=y
door=z (2x+y/3z) - (x/z)= x+y/2z or 1goat+1car/2doors
>> [_] Anonymous 01/10/13(Thu)01:48 No.1853694
>>1853490
I get it now.
