[seqfan] Re: MH problem

Rick Shepherd rlshepherd2 at gmail.com
Thu Oct 29 14:42:52 CET 2015


Thanks, Bob.  I agree that the part you cited can be made clearer. I merely
meant that -- with p1 and p2 as you use below -- that the *difference* p2 -
p1 = (n-1)/(n(n-2)) - 1/n = ((n-1)-(n-2))/(n(n-2)) = 1/(n(n-2)) = 1/a(n).
I was just trying to note this second connection to a(n) itself.  My use of
"over that" also probably didn't help in trying to say this (perhaps
suggesting the division (p2-p1)/p1).

I'll update my comment in A067998 later today.

Ambiguity happens.  Seen on the internet this week: "Time flies like an
arrow; fruit flies like a banana."

Also, with all the fussiness that the Wikipedia article says is necessary
to state the Monty Hall problem clearly, I don't believe anyone actually
says there that the goat is never considered a prize (Hey, some people
(say, without fuel) may prefer goats to cars.).

Thanks again for your reply,
Rick

On Thu, Oct 29, 2015 at 12:57 AM, Bob Selcoe <rselcoe at entouchonline.net>
wrote:

> 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.
>
>
>


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