Robert's "general issue" reminds me when many years ago I was asked
(obliged?) to compile that year's scholarship paper in mathematics at
Manchester University. The HoD told me that a good question should have a
two page solution, a half-page solution and a two line solution. Setting a
good question was not easy!
Quentin Burrell
*************************************
Dr Quentin L Burrell
Isle of Man International Business School
The Nunnery
Old Castletown Road
Douglas
Isle of Man IM9 4EX
via United Kingdom
[log in to unmask]
www.ibs.ac.im
*************************************
----- Original Message -----
From: "Robert Newcombe" <[log in to unmask]>
To: <[log in to unmask]>
Sent: Thursday, February 07, 2008 9:47 AM
Subject: Airplane riddle, and wider implications
>A week ago I sent the following to Allstat following the exchange re the
>airplane riddle. It never
> came back into my inbox, and it isn't in the Allstat archive, so I suspect
> it went into the wonted
> black hole. Possibly Allstat's spam filter rejects any message with RE: in
> the subject line - if so,
> we'd all better take note ...
>
> - - - -
>
> A McDonald's voucher seems scanty reward for Tim Earl's elegant proof!
> Others have rightly pointed
> out the trivial fact that this result depends critically on the
> presupposition that the numbers of
> seats and passengers are identical, call this number n. Equally, of
> course, it doesn't matter what n
> is. The result is trivially true when n = 2, and easily demonstrated for n
> = 3 and 4, which suggests
> both that the answer is 1/2 for all n, and also a way to develop a general
> proof - but it's the less
> neat one, and Tim has shown how all the detail can be bypassed.
>
> This seems to point to a general issue - there can often be two proofs of
> the same result, one
> neat, one more turgid and elaborate. In 1927 Edward Wilson formulated the
> score confidence interval
> for the single proportion - not that the concept of a score interval had
> been developed then. This
> interval resembles the default Wald standard error based interval, but
> with inversion so that the SE
> imputed to the proportion is based on the hypothesised value, not the
> empirical estimate. The Wald
> interval is of course symmetrical on an additive scale (unless truncated
> to ensure it lies within
> [0,1] - something that is often done with Wald intervals, in contrast to
> the Wilson interval which
> is boundary-respecting so that this is never needed). In 1994 I noticed,
> from empirical figures
> calculated to high precision, that the Wilson interval is symmetrical on a
> logit scale. I was
> stunned that apparently it was 67 years before anyone noticed this -
> seeing that the logit scale is
> such a natural one for proportions. But how to prove it? An hour later I
> had hacked out an inelegant
> proof involving surds. OK, but I was sure there must be a simpler proof.
> Two days later, I found it.
> The Wilson lower and upper limits, L and U, are roots of a certain
> quadratic. So we can express
> their product as a simple function of the coefficients in the quadratic.
> We get a similar expression
> for (1-L)(1-U), divide, and it all falls out.
>
> Or indeed, there may be more than two proofs. There are numerous proofs of
> Pythagoras' theorem, and
> also striking ways to demonstrate it that fall short of being actual
> general proofs. My favourite of
> these relates to shapes called P-pentominoes, which consist of 5 adjacent
> unit squares, 4 of them
> forming a 2 by 2 square, the other one stuck on to the side of one of the
> others. Difficult to draw
> in email, but it looks something like the following, hence shaped like the
> letter P.
>
> **
> **
> *
>
> An infinite 2-dimensional plane can be tesselated into an infinite array
> of these shapes, all in
> the same orientation. Each has area 5 units, and corresponding points on
> adjacent pentominoes are
> sqrt(2**2 + 1**2) = sqrt(5) units apart.
>
> Which all makes me wonder - did Fermat have a neat proof after all??
> Slightly less plausible, given
> that mathematicians had sought a simple proof of his theorem for
> centuries, whereas no-one had
> noticed the logit scale symmetry result. How reasonable is it to expect
> that a simple result should
> have a simple proof?
>
> Robert G. Newcombe PhD CStat FFPH
> Professor of Medical Statistics
> Department of Primary Care and Public Health
> Centre for Health Sciences Research
> Cardiff University
> 4th floor, Neuadd Meirionnydd
> Heath Park, Cardiff CF14 4YS
>
> Tel: 029 2068 7260
> Fax: 029 2068 7236
>
> Home page
> http://www.cardiff.ac.uk/medicine/epidemiology_statistics/research/statistics/newcombe
> For location see
> http://www.cardiff.ac.uk/locations/maps/heathpark/index.html
>
>
>>>> "Upton, Graham J" <[log in to unmask]> 31/01/08 10:30 >>>
>
> Slick solution from Tim Earl, a PhD student.
>
> Let A be the first person and Z be the last.
>
> Although there may be many misplaced people, the outcome is certain as
> soon as some passenger
> chooses to sit in the seat of either A or Z. At all times these seats are
> equally likely to be
> chosen by the currently displaced person.
>
> Thus the probability that Z sits in the correct seat is 1/2.
>
> Very surprising!
>
> Graham Upton
>
> Prof. G. J. G. Upton,
> Professor of Statistics/Environmetrics
> Dept of Mathematical Sciences/Centre for Environmental Science
> University of Essex, Colchester, Essex. CO4 3SQ
> http://www.essex.ac.uk/maths/staff/upton/
> Tel: 01206 873027; Sec:01206 872704; Fax: 01206 873043
>
>
>>>> "Leonid V. Bogachev" <[log in to unmask]> 31/01/08 11:51 >>>
>
> Dear Suhal & allstat,
>
> The answer is 0.5.
>
> The puzzle can be modelled using a Markov chain on the integers 1,...,100,
> with transition probabilities as follows:
> from state 1 it can jump to any state (including 1) with equal
> probability:
> p(1,i)=1/100 (i=1,...,100);
> from state 2 it can jump, with equal probability 1/99, either to state 1
> or to any of the subsequent states, 3,...,100:
> p(2,i)=1/99 (i=1 or i=3,...,100);
> .............
> in general, from state k it can jump, with equal probability 1/(100-k+1),
> either to state 1 or to any of the subsequent states, k+1,...,100;
> ............
> finally, from state 100, it jumps to state 1 with pobability 1.
>
> The idea behind this Markov chain is to follow the sitting allocations of
> "displaced" passengers: the 1st passenger can choose any seat; if he takes
> say seat 10 then passengers 2,...,9 will take their own seats, but
> passenger 10 will have to go to seat 1 or to any of the remaining seats
> 11,...,100, etc.
>
> Now, in order that the 100th passenger gets his own seat, all we need is
> that our Markov chain hits state 1 prior to state 100. In other words, we
> need to find the return probability f_1 (i.e., starting from state 1 to
> get back to 1 before getting to state 100). Introducing the similar
> probabilities f_k (k=1,...,100), we have a set of difference equations:
>
> f_1=1/100 +(1/100)*(f_2+...+f_100),
> f_2=1/99 + (1/99)*(f_3+...+f_100),
> f_3=1/98 + (1/98)*(f_4+...+f_100),
> ........
> f_k=1/(100-k+1) + (1/(100-k+1))*(f_{k+1}+...+f_100),
> ........
> f_98=1/3 + (1/3)*(f_99+f_100),
> f_99=1/2 + (1/2)*f_100,
> f_100=0 (boundary condition).
>
> Solving this system "in the reverse order" we obtain (by induction)
>
> f_99=1/2,
> f_98=1/3+(1/3)*(1/2)=1/2,
> ...
> f_k=1/(100-k+1) + (1/(100-k+1))*(100-k-1)*(1/2)=1/2,
>
> ...
> hence
>
> f_1=1/2, QED.
>
> Given such a simple answer, I don't quite like the solution above, as it
> is too technical. There must be a "smart" solution, but knowing the answer
> may help find it!
>
> Many thanks for a beautiful puzzle.
>
> Dr Leonid V. Bogachev
> Reader in Probability
> Department of Statistics Tel. +44 (0)113 3434972
> University of Leeds Fax +44 (0)113 3435090
> Leeds LS2 9JT bogachev'at'maths.leeds.ac.uk
> United Kingdom www.maths.leeds.ac.uk/~bogachev
>
>
> -----Original Message-----
> From: A UK-based worldwide e-mail broadcast system mailing list on behalf
> of Suhal Bux
> Sent: Wed 1/30/2008 16:23
> To: [log in to unmask]
>
> I know this isn't strictly the right sort of post but it's interesting
> and is doing my head in...
>
> 100 passengers are queuing to get on a plane.
> Each has a ticket with a seat number and they are standing in order.
> The first man in the queue is crazy and will sit anywhere at random.
> The rest of the passengers will sit in their own seat, unless it is not
> available, in which case they will sit in another seat at random.
> What is the probability that the 100th passenger sits in his own seat?
>
> I'd appreciate proofs or good reasoning.
|