November 15th, 2001
|05:27 pm - from john|
i don't think i quite buy that..
here is me and john discussing it via email than IM:
not exactly, because 2/3 of the time, the prize will be behind one of
the two you didn't pick, and he's telling you that it's definitely not
behind one of them. if you always choose a, it'll be a 1 of every 3
times, but it will be b or c the other two times, and he's telling you
which one it's definitely not.
i found a simulation you can try..
click a door to choose it, and then it will show another door with a
piggy to denote that door sucks. you may then click the same door again
if you wanna keep it, or choose the remaining door. if you do it many
times (and i've done it a few just now), the average should be about 1/3
for keeping your original door, and 2/3 for switching.
From: dan widrew [mailto:email@example.com]
Sent: November 15, 2001 4:32 PM
To: John Lavallée
Subject: RE: i want a math problem named after me..
that's true, but you could also say that once he takes out the 3rd door,
you now know that the correct door is the one you have or hte other one.
so now it's 1/2 likely to be either one. since you're making a new
choice, you get a new probability of it being correct
> the idea is when you choose originally, it's 1/3 likely to be whatever
> you chose, and 2/3 likely to be one of the other doors. now, if one
> of the other two is taken out, then your choice is between the 1/3
> chance that you originally took, and the 2/3 chance of the combined
> other doors. 1/3 of the time, it will be the door you originally
> chose, but 2/3 of the time, it's one of the other two, and by removing
> one of them from contention, he is in essence allowing you to choose
> the 2/3 probability by taking the other door.
> -----Original Message-----
> From: dan widrew [mailto:firstname.lastname@example.org]
> Sent: November 15, 2001 4:23 PM
> To: John Lavallée
> Subject: RE: i want a math problem named after me..
> hmmm that's weird. i dont know if i quite buy it. it seems to make
> sense that way, but the thing is.. after he takes 1 door out, then *no
> which* of the other two doors you'd originally chosen, you have a 2/3
> chance of the other door being correct. that's impossible, theyre
> if you choose A and he shows C is bad, then it's 2/3 likely to be B,
> if you choose B and C is bad, it's 2/3 likely to be A?
King Nixon (5:32:24 PM): that math thing is neat
King Nixon (5:32:27 PM): that you sent me
tiiznit (5:37:22 PM): i explained the math thingie, yo uexplain that
King Nixon (5:37:34 PM): hm?
King Nixon (5:37:41 PM): ooh new email
tiiznit (5:37:52 PM): new email?
King Nixon (5:38:24 PM): that you sent to me, explaining the math thing
tiiznit (5:38:43 PM): ah
tiiznit (5:38:46 PM): did it make sense that time?
tiiznit (5:39:02 PM): it dint make sense to me at first either, until it clicked
King Nixon (5:41:23 PM): see, it does make sense waht youre saying. but i can explain it to be 1/2 and that makes just as much sense to me
tiiznit (5:42:29 PM): the problem is that it isn't strictly 1/2.. it's 1/2 between your original choice and whatever is left of B and C.. and that one of them was removed isn't strictly arbitrary, it was because it is a zonk. so the one left over has a higher probability of not being a zonk
tiiznit (5:43:31 PM): (well, it is arbitary in the 1/3 of the time that A was right in the first place, but the other 2/3 of the time, it's whatever of B and C wasn't taken out)
King Nixon (5:43:31 PM): if you look at it as either keeping the choice you made or switching, it's 2/3 wrong. if you look at as an entirely new choice,it's 1/2. that he threw out a zonk is irrelevent, you know of the 2 doors you have left to choose from, 1 is right and 1 is wrong
tiiznit (5:44:37 PM): it's not an entirely new choice.. the whole basis of the argument is that there is a link
tiiznit (5:44:54 PM): did you try the java thingie btw?
King Nixon (5:45:22 PM): why shoudlnt it be a new choice? he changed the choices so you make a new choice
King Nixon (5:45:34 PM): no i havent yet. i'm being plato =) pure thought, experiment is less fun
King Nixon (5:45:42 PM): (that was plato, wasnt it?)
King Nixon (5:45:52 PM): or socrates? i dont remember
King Nixon (5:45:56 PM): or some other jerk
tiiznit (5:46:03 PM): heh, yeah, i ignored that until i had figured it out myself, and then amused myself by proving it
tiiznit (5:46:07 PM): mm jerk
tiiznit (5:46:38 PM): anyway, you can't look at the whole thing as two choices. if you look at it as being three choices, even when it's narrowed to two, then you can see where the 2/3 / 1/3 comes in
tiiznit (5:47:26 PM): stuff like this was pretty much the entire first three months of my phil class last year. i love that.
King Nixon (5:50:36 PM): but why would you look at it as 3 choices when its narrowed? thats like choosing between 2 closed box and one box you can see is empty. you KNOW the empty one is wrong, so that cant beweighed the same as the other 2
King Nixon (5:52:01 PM): youre choosing between A or (B + C) when you know either B or C is wrong
tiiznit (5:52:06 PM): its weight is that it was C instead of B or B instead of C. in 2/3 of the possible outcomes, the one that was removed is relevant because hte answer is the other. in the third case, it's a random choice, but that's only 1/3
King Nixon (5:52:17 PM): which since youre looking for hte right one, makes them essentially equal one
tiiznit (5:52:46 PM): why do you like arguing the incorrect argument? =)
King Nixon (5:53:08 PM): i dont think the two choices should be related. if he removes 1 of the doors the probability has to change, theres only 2 options left
King Nixon (5:53:48 PM): youve got A B or C. you know C is wrong. its still an option, but you have 50/50 of it being either A or B.
King Nixon (5:54:06 PM): if you chose A or B originally, its still 50/50 between them. otehrwise you'd be affecting it by your choice which is impossible
np: disturbed - stupify
|Date:||November 15th, 2001 03:04 pm (UTC)|| |
i'm insulted that you have demanded more proof from me.. :p
|Date:||November 15th, 2001 03:45 pm (UTC)|| |
Let's assume that you chose A. There are three scenerios that could occur, depending on which is the actual winning door:
It is actually A
(B or C) is removed
Your choice is A or (B or C)
You choose A => you win :)
You choose (B or C) => you lose :(
It is actually B
C is removed
Your choice is A or B
You choose A => you lose :(
You choose B => you win :)
It is actually C
B is removed
Your choice is A or C
You choose A => you lose :(
You choose C => you win :)
Out of the three scenarios, there is only one case where choosing A will result in you winning. However, there are two scenarios where not choosing A will result in you winning. In the first scenario, which will occur 1/3 of the time if everything is nice and perfectly random (i'm not getting into that), which of B and C are removed from contention does not matter, because neither alternate door is the correct door. However, in 2/3 of the possible cases (scenarios 2 and 3), where A is not the correct choice, the door that is removed will point to the other available door as being the correct choice, and therefore, that door will have a 2/3 probability of being correct.
just for the helluvit, i went to that sim page and tried it out. after 60 tries of staying with my 1st choice, i got it right 24 times, which is 40%. after 60 tries of switching, i got 35, 58%. so the results lean towards your answer, but they're close to being halfway between us. of course, 60 tries isn't exactly enough to base anything on, but i got bored.
hmm, on further examination, i dont' know how random that page actually is. i tried it some more, only counting the results if C was the removed door, which should have effectively eliminated it from the odds. but i came out with close to the same results, which makes even less sense. it may be biased against A, which i was using every time cuz i'm lazy. which is also why i'm not going to investigate that any further. staring at all those little doors is putting me to sleep.
that's true, but only because you're taking the 1st choice into account, which has been my point all along. you're grouping it by [chosen door] vs. [removed door and unknown door]. you could just as easily do it by [unknown door] vs. [removed door and chosen door] and get the exact opposite results. no matter which one you focus on, it's more likely to be the other two, it's nothing to do with which one you chose in the previous round.
The simple answer is thus:
Choosing the first is a preliminary in order to get Monty to take out one of the bad ones and narrow your chances. Now you have a one in two chance of getting the correct one. It doesn't matter if you switch. Half the time you will get it and half you do not. Switching does not improve your chances.