[seqfan] Re: MH problem

[Bob Selcoe]

Hi Rick & Seqfans,

That's a good explanation, Rick.  One minor point though.  Your comment in 
A067998:

>> The increased probability of winning by switching over that of not 
>> switching, which is simply 1/n, is 1/a(n), approaching zero as n 
>> approaches infinity...

is a little hard for me to follow.  First, not sure what you mean by "1/n, 
is 1/a(n)...".  Second, I think the increased probability is 1/(n-2)  n>=3, 
not 1/n; or at least the meaning of "increased" can be interpreted in 
multiple ways.  For me,  "increased probability" means (p2-p1)/p1 where 
probability p2>p1 (which is 1/(n-2)), rather than p2-p1 (which is 1/n). 
So...

n=3 doors: p1=1/3, p2=2/3; (2/3-1/3)/(1/3)= 1
n=4 doors: p1=1/4, p2=3/8; (3/8-1/4)/(1/4) = 1/2
n=5 doors: p1=1/5, p2=4/15; (1/5-4/15)/(1/5) = 1/3
etc.

You may wish to clarify in the posting.

Regards,
Bob S.

--------------------------------------------------
From: "Rick Shepherd" <rlshepherd2 at gmail.com>
Sent: Wednesday, October 28, 2015 10:30 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: MH problem

> There are many variants of this problem. The English Wikipedia article I
> refer to in my comment in A067998 is useful and thought- provoking.  --  
> Rick
> On Oct 28, 2015 3:56 PM, "Frank Adams-Watters" <franktaw at netscape.net>
> wrote:
>
>> I think, when you switch away from a door, the next thing that happens is
>> that Monte shows you what you just passed up - you can never switch back 
>> to
>> that door. Under that interpretation, all probabilities for what you 
>> might
>> switch to are the same, and a "friendly" Monte can't actually help you.
>>
>> Franklin T. Adams-Watters
>>
>>
>> -----Original Message-----
>> From: Andrew Weimholt <andrew.weimholt at gmail.com>
>> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>> Sent: Wed, Oct 28, 2015 2:28 pm
>> Subject: [seqfan] Re: MH problem
>>
>>
>> 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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