## The reflection principle

**I**n the weekend magazine of ** Le Monde**, there always is one mathematical puzzle that often is easy to solve but sometimes is harder (like the puzzles involving arcane triangular geometry!) and occasionally just impossible (because the puzzle is missing one crucial assumption). The puzzle of this week is a direct application of the reflection principle exposed in Chapter III of Feller’s

*, which is my favourite probability book. (More precisely, the answer is (almost) given in problem 1, page 95 of this book.)*

**An Introduction to Probability Theory and Its Applications, Vol. 1****G**iven a sequence of heads and tails ending with exactly heads, what is the probability that the sequence always remained in favour of heads, i.e. that the number of heads was larger than or equal to the number of tails from trial one to trial ? The number of non-negative paths of length ending up in zero is the same as the number of (strictly) positive paths of length ending up in one (see Figure 1, page 69, in Feller). Now, by the reflection principle (Feller, Lemma, page 72), this number is

with

being the number of paths of length ending up in (and being the number of paths hitting the zero value at least once after the starting time). The probability of a non-negative sequence is therefore

This has a potential connection with Laplace’s succession rule, but I cannot make it straight away because the underlying events of not visiting -1 are not independent…

**I**f you want to check this by simulation, a simple R program is

Nsim=10^6; n=4

cont=0;gain=0

while (gain<Nsim){

vale=cumsum(sample(c(-1,1),2*n,rep=TRUE))

gain=gain+(vale[2*n]==0)

cont=cont+(vale[2*n]==0)*(min(vale)>-1)

}

print(cont/Nsim)

**I**f you have never read * An Introduction to Probability Theory and Its Applications, Vol. 1*, you should consider doing it: it makes for a very pleasant read, while making combinatorics enjoyable and it also constitutes a wealth of problems that I use extensively when teaching at Polytechnique.

February 21, 2010 at 7:35 am

[…] Monde puzzle The puzzle in Le Monde is quite straightforward (!) this weekend: it can be rewritten as to figure out the […]

February 7, 2010 at 12:17 am

There is a nice paper by MARC RENAULT,

“Four Proofs of the Ballot Theorem”,

about this problem and some generalizations.

I gave it to my students in the Stochastic Processes course and three out of ten solved it

February 7, 2010 at 7:21 am

Thank you, Brunero! The link reflection principle actually points to Renault’s webpage about the ballot theorem. Very nicely done…