[seqfan] Re: MH problem

zbi74583.boat at orange.zero.jp zbi74583.boat at orange.zero.jp
Fri Oct 30 05:25:54 CET 2015


    Hi,Andrew

    I lacked the second term.
    It should be as the following

    p(n) : 1,1/2,2/3,5/8,11/15,....

    I think that the probabilities of your three options depend on whether Monty
is a friend of the player or not

    To Franklin
    I think that in the case you wrote Monty is not a friend of the player.

    To Seqfans
    First of all, I should have written a link of original MH problem.

    https://en.wikipedia.org/wiki/Monty_Hall_problem

    Read it and understand what it is.


    One more important thing
    I wrote :

    This is the probability that the player gets a car changing the doors every
time.

    Do you understand this definition?
    Is my English good?

    Read carefully the example which I wrote.
    It was the case of 5-MH and Monty is not a friend of the player.

    Probability is 4/15+1/5+4/15=11/15 ,so p(5)=11/15

    To Rick and Bob
    Your p2 is different from my p(n).
    p2  =Probability that the player gets a car changing a door at once
    p(n)=Probability that the player gets a car changing doors (n-2) times



    Yasutoshi


> On Wed, Oct 28, 2015 at 10:11 AM, Andrew Weimholt <andrew.weimholt at gmail.com
>> wrote:
>
>> On Tue, Oct 27, 2015 at 7:07 PM, <zbi74583.boat at orange.zero.jp> wrote:
>>>     Hi,Seqfans
>>>     Once I and my friend Kobayashi discussed about n-MH problem.
>>>     Where "n-MH" means n doors Monty Holl
>>>     Kobayashi is a scientist writer.
>>>     And I met a Sequence of probability related with n-MH problem
>> [...]
>>>     p(n) : 1,2/3,5/8,11/15,....
>> p(2) should be 1/2. With only two doors, you pick w/ 50/50 probability,
>> and there is no opportunity to switch doors. Either you've won, or you
>> didn't.
>> If you start the sequence at n=1, then p(1) would be 1
>> If you start the sequence at n=0, then p(0) would be 0
>> I've confirmed, by hand, that p(4) is 5/8.
> For n=5 and beyond, there are choices for the player to switch doors which
> are not symmetric in terms of odds of winning.
> There are multiple ways to handle this, which will [most likely] lead to
> multiple sequences.
>
> option 1) player's choice of which door to switch to is evenly distributed
> over the remaining closed doors (other than the one he's switching away
> from)
> option 2) player's choice of which door to switch to is based on the
> probability of the door containing the car. He choses the door with the
> highest probability.
> option 3) player's choice of which door to switch to is based on the
> probability of the door containing the car. He choses the door with the
> lowest probability (anticipating that he'll again switch).
>
> Andrew
>
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