savoy082

Pt En In this twitter proof we will see how the exponential function can mess up with objects from our daily lives!.. Claim: with less than $50$ folds, a piece of paper will be so thick that it will cover the distance from the Earth to the Moon. Twitter proof: a common sheet of paper is $0.1$mm thick. If we fold it once, it becomes $0.2$mm thick. Folding twice, $0.4$mm. Folding $49$ times, the paper becomes $2^{49}\times 0.1$mm thick, which is around $5.63\times 10^{13} $mm or $5.63\times10^7$km, $141$ times the distance from the Earth to the Moon ($398818$km). Neste post vamos ver como a função exponencial pode interagir com objetos do nosso quotidiano e criar resultados inesperados. Proposição: com menos de $50$ dobras, uma folha de papel fica com uma grossura superior à distância da Terra à Lua. Prova num tweet: uma folha de papel normal tem $0.1$mm de grossura. Se a dobrarmos uma vez, fica com $0.2$mm de grossura. Dobrando de novo, fica com $0.4$mm. Dobrando $49$ vezes,...

Pt En It has become quite common for online services to provide some form of 2-factor authentication when logging in from unknown devices. For example, whenever I try to access my Gmail account from a computer I never used, I get a text message with a one-time use 6-digit code. One day I was using that same service to log in into my email, when I noticed that one of the digits in the security code appeared twice, like the $1$ in $315641$. But when I read the other text messages from Google, I noticed that there were plenty more security codes with repeated digits than security codes that had six different digits. I found that weird and then decided to compute the probabilities of these events, just to check whether my intuition was tricking me or not... We are about to compute some probabilities regarding these $6$-digit codes - which I will start calling PINs for the sake of brevity - with the rather intuitive formula $$P(\text{some property}\ A) = \frac{\text{# PINs that satis...

Pt En This post has the purpose of presenting a result that may seem counterintuitive and that can provide a really nice excuse for a wager between you and one or more of your friends. For this post, when I talk about a birthdate I am only referring to the day and month of birth, and not the year. What is the probability that you and your best friend have the same birth day and month? Even without an exact number one knows that you are much more likely to have different birthdates than having equal birthdates. Assuming all $366$ days are equally likely, the probability that two people have the same birthdate is $\frac{1}{366} \approx 0.27\%$ and the probability that the birthdate is different is $\frac{365}{366} \approx 99.73\%$. How many people do you need so that the probability of existing at least two sharing the birthdate is higher than the probability of everyone having different birthdates? What would your guess be? It only takes $23$ people. If you have a group of $23$...