## take a random integer

**A** weird puzzle from FiveThirtyEight: what is the probability that the product of three random integers is a multiple of 100? Ehrrrr…, what is a random integer?! The solution provided by the Riddler is quite stunning

Reading the question charitably (since “random integer” has no specific meaning), there will be an answer if there is a limit for a uniform distribution of positive integers up to some number . But we can ignore that technicality, and make do with the idealization that since every second, fourth, fifth, and twenty-fifth integer are divisible by and , the chances of getting a random integer divisible by those numbers are , , , and .

as it acknowledges that the question is meaningless, then dismisses this as a “technicality” and still handles a Uniform random integer on {1,2,…,N} as N grows to infinity! Since all that matters is the remainder of the “random variable” modulo 100, this remainder will see its distribution vary as N moves to infinity, even though it indeed stabilises for $N$ large enough…

February 16, 2019 at 4:45 pm

Tres amusant. On remplace $100$ par $n$ produit de $k$ nombres premiers $p$ a la puissance $a_p$. On resout le probleme d’abord pour $n=p^a$ et dans l’anneau $A=Z/p^a/Z$. Une va uniforme $X$ dans $A$ a une valuation $v(X)$ en $p$ aleatoire qui suit une loi de Pascal de parametre $1/p$ tronquee en $a$ et la probabilite P(p_a)pour que $v(X_1)+v(X_2)+v(X_3)\geq a$ se calcule sans trop de peine. Et pour finir dans le cas general, on a

$$P(n)=\prod_pP(p^a).$$

February 16, 2019 at 5:41 pm

Ce qui définit correctement le contexte probabiliste?