The probability that a specific person celebrates their birthday the same week as you is, obviously, 1/52 (if we understand by “the same week” the same interval from Monday to Sunday; if we talk about birthdays not separated by more than seven days the probability is practically double). But for any two people from a group of seven to have their birthday in the same week, as was proposed in the previous installment, is much higher, contrary to what intuition suggests. And even greater is the probability that two of them share a zodiac sign.
As in other problems of this type, it is easier to calculate the probability that something does not occur to determine its complementary probability. Let’s focus on the zodiacal case, with the same approach as the weekly one, but with more manageable numbers. Let’s start by arbitrarily numbering people from 1 to 7. The probability that 2 is not the same sign as 1 is 11/12; the probability that 3 is not of the same sign as 1 or 2 is 10/12…, and the probability that 7 is not of the same sign as any of the other 6 is 6/12. Therefore, the probability that none of those matches occurs will be:
11/12 x 10/12 x 9/12 x 8/12 x 7/12 x 6/12 = 0.11 approx.
Thus, the probability that there are at least two people of the same sign will be complementary, that is, almost 90%. The next time, in a small gathering, two magufas see a sign of destiny in the extraordinary coincidence of sharing a zodiac sign, you can annoy them with the unappealable verdict of mathematics (or not).
Using the same procedure, although with somewhat more cumbersome operations, you can verify that the probability that in a group of 23 people there are at least two who celebrate their birthday on the same day is slightly higher than 50% (this is known as the “paradox of the birthday” due to its counterintuitive result).
Also in the case of the cards that we name when we put them on the table, it is easier to calculate the probability that none of them match their invocation, since for each one that probability is 39/40. Therefore, the probability that no card matches your name will be 39/40 to the 40th power, approximately 0.363. Consequently, the probability that at least one card appears when named is approximately 1 – 0.363 = 0.637. Two out of three times you try, at least one of the cards will “magically” appear when you say its name.
An isosceles triangle and a beer can
Intuition can not only deceive us by making the probable seem improbable, but also by estimating the level of difficulty of a problem. Let’s look at a couple of examples (which apparently – just apparently – have nothing to do with each other):
- Given an isosceles triangle whose equal sides measure 10 cm, how long must the third side be for its area to be maximum? It looks like a typical maximum and minimum problem that is solved by expressing the area of the triangle as a function of the third side and differentiating the function. And it seems so because it is; but, with a little ingenuity, it can be solved without resorting to differential calculus.
- In a full beer can, if we consider it perfectly cylindrical and homogeneous, the center of gravity is the midpoint of the axis of the cylinder. As it is emptied, the center of gravity drops and the balance of the can on its base becomes, therefore, more stable (as if to compensate for the presumed loss of stability of the consumer of its contents). But, once it is completely empty, the center of gravity of the can returns to be at the midpoint of its axis, so there must be a moment of emptying in which said center reaches its lowest point, moment by moment. from which it rises again. If the can is 20 cm high, weighs 45 g when empty, and when full contains 360 g of beer, how much beer is in the can at the moment when the center of gravity is at the lowest height?
Once again there seems to be a problem to be solved using differential calculus, and this is how Walter B. Roberts, from Princeton University, solved it when he posed it during a picnic; but then (and, presumably, once the fumes of the contents of the can had dissipated) he realized that it could be easily solved without functions or derivatives.
The two previous problems do not seem to be related at all; However (and this is a good clue for the second, more difficult than the first) both require the same change of perspective for their resolution. Literally, it is about looking at them from a different angle.
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#point #emptying #beer #stable