RSL
<R> james
<_j> aye charles
<R> you know the monty hall problem
<R> old news, am I right?
<_j> which door
<R> three doors, you choose one, monty opens one, then you can switch
<_j> right
<R> if there are three doors, you always switch
<_j> switch the door
<_j> yup
<R> because 2 out of 3 times, monty is forced to not choose the car's door
<R> and he only has a choice 1 of 3 times
<R> so.
<R> what if there are four doors?
<_j> hmmm
<R> you choose one, then monty has to open one of the other three, and the door he opens must be a loser
<R> clearly, he always has a choice
<_j> i wonder if its able to e generalised to n-doors
<R> with three doors to choose from, and only one winner among the four
<R> he is never forced into a choice
<_j> switching takes on a new meaning
<R> so what is your switching strategy? does it still matter?
<_j> randomly choose one of the alternatives
<_j> it will always be better to switch for any number n
<_j> the betterness declines as n increases
<R> so you still always switch
<_j> yes
<_j> probbly we could generalise it mathematically if we tried
<R> with four doors, I came up with an increase from 1/4 chance to win, to 3/8 by switching
<_j> ok if there are n doors
<_j> initially your odds are 1 / n
<_j> odds odf being wrong, n-1 / n right
<R> yes
<R> initial odds to win, 1/n
<_j> the eye of god destroys one of the wrong choices
<_j> n-2 doors remain after the hand of god destroys a bad chice
<_j> the odds become,
<_j> heres where im bad at stats
<_j> in any case they become better than 1/n
<_j> its easy to see it generalises to always rechoosing
<_j> im not sure the equation
<R> (n-1)/(n)(n-2)?
<_j> could be id have to plug in numbers for the 3 door choice to be sure
<_j> to feel sure
<R> in the case of 5 doors, odds improve from 1/5 to 4/15
<_j> intuitive its wlays better to switch tho because you are choosing from a sample set with one fewer bad possibilities
<R> anyway. that much is pretty obvious
<R> what I was thinking about last night, is this:
<_j> its counterintuitive at the same time since you feel, oh but look he didnt destroy my choice maybe im right
<R> what if there are quite a few doors, and one bad choice (not your current pick) is removed at a time
<R> and you want to maximize your chance of picking the right door after every elimination
<R> first you switch, but after the second one is removed, switch again? including back to the first door?
<_j> yes
<_j> i dont see it makes any sense to keep a list of previous selections to avoid
<R> so, you need at least 5 doors to start with
<_j> in a sense each step is independent
<R> doors A-E, I choose A
<R> you say, ah but door E is a loser
<R> so I choose B
<R> my odds are currently 4/15
<R> you say, ah but door D is a loser
<R> so I should choose A or C? they're equal chances to win?
<R> a moment ago, A and C were NOT equal, but now they are?
<_j> sure they are
<R> a moment, james
<_j> its not any door that should take it personally
<R> suppose there are four doors. A B C D
<_j> its the destruction of a loser
<R> odds of the car are equal among them. with me?
<_j> k
<_j> hmmm on second thought
<_j> now im not so sure
<R> I choose B, you remove D
<_j> in a sense maybe a surviver door is a better choice
<R> now the odds of A and C are better than B, and A is equal to C
<R> you see where I'm going with this
<_j> yeah
<R> four doors, but they dont start out equal.
<R> suddenly, after you remove D, A and C are equal?
<R> they weren't equal a moment ago
<R> if they had been, they'd still be equal
<_j> yeah
<_j> huh
<_j> curious
<_j> you are onto something here charles
<R> so perhaps it makes sense to choose a door that hasn't been chosen before
<R> pending a simulation or counterexample, let's say that's the case
<_j> is there any literature on this
<R> I call it "reverse socratic" logic
<R> RSL
<R> but we're not quite done, james
<R> seven doors. A B C D E F G
<R> I choose A, you remove G.
<R> I choose B, you remove F.
<R> I choose C, you remove E.
<R> I choose D, you remove A.
<R> do I choose B or C?
<R> you follow?
<_j> so far
<R> I think the answer is C
<_j> why c
<R> it had better odds than B before the removal of A
<R> I'm going to post this. RSL.
<_j> seems to generalise
<R> you want your real name attached?
<_j> doesit only wor for odd numbers of doors
<_j> id be honored charles
<_j> i havent done anything worthy tho
<R> nah, it works for all cases
<R> even/odd makes no difference
<_j> just easier to illustrate
<_j> the odds have you computed the gain in using your method formulaicly
<R> no, 7 just happens to be the minimal number of doors for that last bit to matter
<_j> its an interesting result
<R> I have not yet computed the gain by using RSL in a general case. I'll get to it.
<_j> seems like someone would have studied this
<_j> yeah
cast of characters
_j James Price
R Charles Griffiths
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