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.
