This evening, I found a nice probabilistic puzzle on http://www.futilitycloset.com/“A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?”
To be honest, I did not understood the answer on the blog, but if we write it down, we want to solve
Let us count: if 
is the total number of balls, and if 
is the number of white 
balls then
I.e. we want to solve a polynomial equation (of order 2) in 
, or to be more precise, in 
If is equal to 16, then is either 6 or 10. It can be visualized below
> balls=function(n=16){ + NB=rep(NA,n) + for(k in 2:(n-2)){ + NB[k]=(k*(k-1)+(n-k)*(n-k-1)) + } + k=which(NB==n*(n-1)/2) + if(length(k)>0){ + plot(1:n,NB,type="b") + abline(h=n*(n-1)/2,col="red") + points((1:n)[k],NB[k],pch=19,col="red")} + return((1:n)[k])} > balls() [1] 6 10
But more generally, we can seek other ‘s and other pairs of solutions of such a problem. I am not good in arithmetic, so let us run some codes. And what we get is quite nice: if 
admits a pair of solutions, then is the squared of another integer, say 
. Further, the difference between 
and 
is precisely . And will be one of the answers when the total number of balls will be 
. Thus, recursively, it is extremely simple to get all possible answers. Below, we have , , 
and the difference between and ,
> for(s in 4:1000){ + b=balls(s) + if(length(b)>0) print(c(s,b,diff(b))) + } [1] 9 3 6 3 [1] 16 6 10 4 [1] 25 10 15 5 [1] 36 15 21 6 [1] 49 21 28 7 [1] 64 28 36 8 [1] 81 36 45 9 [1] 100 45 55 10 [1] 121 55 66 11 [1] 144 66 78 12 [1] 169 78 91 13 [1] 196 91 105 14 [1] 225 105 120 15 [1] 256 120 136 16 [1] 289 136 153 17 [1] 324 153 171 18 [1] 361 171 190 19 [1] 400 190 210 20 [1] 441 210 231 21 [1] 484 231 253 22 [1] 529 253 276 23 [1] 576 276 300 24 [1] 625 300 325 25 [1] 676 325 351 26 [1] 729 351 378 27 [1] 784 378 406 28 [1] 841 406 435 29 [1] 900 435 465 30 [1] 961 465 496 31
Thus, given 
, consider an urn with 
balls. We draw two balls at the same time. It is equally likely that the two will be the same color as different colors. Then the number of colors within the bag are respectively
Finally, observe that the ‘s are well known, from Pascal’s triangle,
also known as triangular numbers,
Maths can be magic, sometimes…
, for some 
, if
. An this concept can be related to sums and maxima (see 
: let
and 
. Then it can be shown easily that
if and only if
for some 
if and only if the exists a non-degenerate variable 
such that
if and only if
, define 
and 
. Then analogous results can be derived,
if and only if
if and only if the exists a non-degenerate variable 
such that
. Alors que si je gagne, je gagne une cravate qui vaut plus cher que la mienne, disons 
, avec 
. Aussi, mon espérance de gain est
Via 
était un estimateur Gaussien. En particulier, chacun des estimateurs est alors Gaussien, au sens où 
, ce qui peut se visualiser sur le graphique suivant (la constante est en abscisse, et la pente en ordonnée), avec un intervalle de confiance à 95%,
such that
where 
‘s are i.i.d. with cumulative distribution function 
. They had supporting arguments, but no (rigorous) proof. Nevertheless, the obtained that the only possible types for G were
as possible limit. Richard von Mises gave in 1936 sufficient, but not necessary conditions for their (max) domain of attraction, i.e. characterization of function 
such that the maxima converges to some specific function 
is regularly varying at infinity, with index 
(even if the term “regular variation” was not mentioned in the paper). Similar results were derived to characterize functions in the (max) domain of attraction of Weibull. For the (max) domain of attraction of 
, Boris Gnedenko obtained that a necessary and sufficient condition was that there exists a function 
such 
(with a much shorter proof).
. For bounded distribution, consider e.g. the uniform distribution on the unit interval, i.e. 
on the unit interval. Let 
and 
. Then, for all 
and 
,
. Let 
and 
, then
. Let 
and 
, then
sample. We can use the following approximation of the cumulative distribution function (based on l’Hopital’s rule)
. Let 
and 
. Then we can get
. I.e. the limiting distribution of the maximum of a Gaussian sample is Gumbel’s. But what we do not see here is that for a Gaussian sample, the convergence is extremely slow, i.e., with 100 observations, we are still far away from Gumbel distribution,
, if
et (comme je l’expliquais dans un autre 
. La règle d’affectation est alors simple: on se fixe un seuil 
, et
, alors 
, alors 
Un professeur rédigeant son plan de cours veut mentionner un examen surprise, qui compterait pour 10% de la note finale. Il annonce donc à ses étudiants, au premier cours, qu’un examen surprise (et un seul) se tiendrait à une date (surprise) parmi les 12 séances de cours. Mais comme le fait noter un élève, l’examen ne pourrait avoir lieu lors de la dernière séance (la 12ème), car cela signifierait qu’il n’y a pas eu d’examen les 11 premières séances, et donc l’examen surprise devrait forcément se tenir au 12ème cours (mais alors il ne serait plus surprise du tout). Bref, impossible d’avoir un examen surprise ce jour là. Mais dans ce cas, surenchérit une étudiante, il ne peut non plus avoir lieu à l’avant dernière séance (la 11ème), car là encore cela signifierait qu’il n’y a pas eu d’examen les 10 premières séances, et donc l’examen surprise devrait forcément au 11ème cours (car il ne peut pas avoir lieu au 12ème comme on vient de le voir, et là encore, il ne serait plus surprise du tout). Bref, de manière récurrente, on peut montrer que l’examen surprise a alors lieu forcément le premier jour. Mais là encore, il n’y a pas de surprise. Et donc impossible qu’il y ait un examen surprise si ce dernier a été annoncé. Moralité, pour faire un examen surprise, il ne faut surtout pas évoquer qu’il y en aura un. Mais comme on ne peut pas faire un examen non annoncé dans les modalités de contrôle des connaissances, il est impossible de faire des surprises aux étudiants à l’université… C’est triste, non ?