Archive for métro

simulation in Gare du Nord [jatp]

Posted in Statistics with tags , , , , , , , , on January 30, 2018 by xi'an

don’t run with 21run [for locals only]

Posted in pictures, Running with tags , , , , on January 22, 2017 by xi'an

This (whining) post is of little interest to anyone but French runners: last week I ordered a new pair of running shoes as mine had suffered too many kilometres to keep going and came upon a sale offer on that suited my needs. Checking similar offers on other running sites made this sale the best choice and hence I ordered the shoes. As is often the case with running shoes, the size varies with the brand and the pair of Asics I received was too small. Nothing unusual with that, but I then found out that the company is actually located in Germany, despite the website being integrally in French, plus advertising in the Paris métro, which again is not an issue per se, except that it charges returns outside Germany. Meaning that I ended up paying 17% of the overall price just to return shoes that were not my size.

The Magicians [#2]

Posted in Books, Kids, pictures, Travel with tags , , , , , , , on October 1, 2016 by xi'an

magicienIn a minor key coincidence with my book review last week, I spotted this advertising board on my métro platform this morning. Announcing the diffusion of the first season of The Magicians on the French SyFy [sic] channel. (Next to this board, a more relevant one from Droit au Logement.)

random walk on a torus [riddle]

Posted in Books, Kids, pictures with tags , , , , , , , , , on September 16, 2016 by xi'an

Galgate, Lancastershire, July 19, 2011The Riddler of this week(-end) has a simple riddle to propose, namely given a random walk on the {1,2,…,N} torus with a ⅓ probability of death, what is the probability of death occurring at the starting point?

The question is close to William Feller’s famous Chapter III on random walks. With his equally famous reflection principle. Conditioning on the time n of death, which as we all know is definitely absorbing (!), the event of interest is a passage at zero, or any multiple of N (omitting the torus cancellation), at time n-1 (since death occurs the next time). For a passage in zero, this does not happen if n is even (since n-1 is odd) and else it is a Binomial event with probability

{n \choose \frac{n-1}{2}} 2^{-n}

For a passage in kN, with k different from zero, kN+n must be odd and the probability is then

{n \choose \frac{n-1+kN}{2}} 2^{-n}

which leads to a global probability of

\sum_{n=0}^\infty \dfrac{2^n}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}} 2^{-n}


\sum_{n=0}^\infty \dfrac{1}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}}

Since this formula is rather unwieldy I looked for another approach in a métro ride [to downtown Paris to enjoy a drink with Stephen Stiegler]. An easier one is to allocate to each point on the torus a probability p[i] to die at position 1 and to solve the system of equations that is associated with it. For instance, when N=3, the system of equations is reduced to

p_0=1/3+2/3 p_1, \quad p_1=1/3 p_0 + 1/3 p_1

which leads to a probability of ½ to die at position 0 when leaving from 0. When letting N grows to infinity, the torus structure no longer matters and the probability of dying at position 0 implies returning in position 0, which is a special case of the above combinatoric formula, namely

\sum_{m=0}^\infty \dfrac{1}{3^{2m+1}}  {2m \choose m}

which happens to be equal to

\dfrac{1}{3}\,\dfrac{1}{\sqrt{1-4/9}}=\dfrac{1}{\sqrt{5}}\approx 0.4472

as can be [unnecessarily] checked by a direct R simulation. This √5 is actually the most surprising part of the exercise!

Le Monde puzzle [#907]

Posted in Books, Kids, Statistics, University life with tags , , , on September 18, 2015 by xi'an

A combinatorics (?) Le Monde mathematical puzzle:

Each day of 2014, more than half of the 365 Paris métro drivers are at work. What is the minimal number of drivers one should consider to be sure to include at least a driver for each day of the year?

I may be missing an item of information from the puzzle: since at least 183 drivers are at work every day, if I select 183 drivers at random, there remain 182 further drivers. Even in the most extreme case where the 182 further drivers are at work every day of the year, there will be at least one of the 183 selected drivers at work every day. Conversely, if I select 182 or less drivers, one configuration is that the 183 or more remaining drivers are the ones always at work…

art brut

Posted in pictures, Travel with tags , , , , , , on August 5, 2015 by xi'an


snapshot from Boston [guest shot]

Posted in pictures, Travel with tags , , , , on July 4, 2015 by xi'an